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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , +2, PHYSICS, BOOK BACK, PUBLIC (MAR 2006 – MAR 2017), PTA & OTHER IMPORTANT QUESTIONS, , www.Padasalai.Net, COTENTS, , UNIT, 1., , ELECTROSTATICS, , 2., , CURRENT ELECTRICITY, , 44, , 3., , EFFECTS OF ELECTRIC CURRENT, , 79, , 4., , ELETRO MAGNETIC INDUCTION AND, ALTERNATING CURRENT, ELECTROMAGNETIC WAVES AND WAVE, OPTICS, ATOMIC PHYSICS, , 109, , DUAL NATURE OF RADIATION AND MATTER RELATIVITY, NUCLEAR PHYSICS, , 224, , SEMICONDUCTOR DEVICES AND THEIR, APPLICATIONS, COMMUNICATION SYSTEMS, , 292, , 5., 6., 7., 8., 9., 10., , 1, , PAGE, NO., 03, , J.SHANMUGAVELU, , [P.G. T. in Physics], , 141, 186, , 248, , 326, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 3 : www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, BLUE PRINT, UNIT, , OBJECTIVES, , 1, MARK, 4, , 3, MARK, 2, , 5, MARK, 1, , 10, MARK, 1, , TOTAL, MARKS, 25, , 1., , Electrostatics, , 2., , Current Electricity, , 1, , 3, , 2, , 3., , Effects of Electric current, , 2, , 1, , 1, , 1, , 20, , 4., , Electromagnetic Induction &, Alternating Current, Electro Magnetic Waves &, Wave optics, Atomic Physics, , 4, , 2, , 1, , 1, , 25, , 4, , 2, , 1, , 1, , 25, , 4, , 2, , 1, , 1, , 25, , Dual Nature of Radiation &, Matter – Relativity, Nuclear Physics, , 2, , 1, , 2, , 4, , 2, , 1, , 1, , 25, , Semi conductor devices &, their Applications, Communication Systems, , 3, , 4, , 1, , 1, , 30, , 2, , 1, , 1, , 1, , 20, , 5., 6., 7., 8., 9., 10., , 20, , 15, , www.Padasalai.Net, TOTAL MARKS, , 30, , 60, , 60, , 80, , E ß Ú õÀ • i ² ®, E ß Ú õÀ © m k ÷ © • i ² ®, E ß Ú õÀ • i ¯ õu x J ß Ö ª À ø », , Prepared by, J. SHAAM BREEZE SHANMUGAVELU, Contact for subject doubt, Email :
[email protected], Phone No: 9952223467, , JEYAM TUITION CENTRE, [PARAMAKUDI], Contact for tuition, Email :
[email protected], Phone No: 9944888512, 2, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html, , 230
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 1.ELECTROSTATICS, PREPARED BY, J.SHANMUGAVELU M.Sc, B.Ed. [P.G. Assist in Physics], , ONE MARK QUESTIONS - (4 Questions), BOOK BACK ONE MARKS:, 1. A glass rod rubbed with silk acquires a charge of +8 × 10−12C. The number of, electrons it has gained or lost, a) 5 × 10−7(gained), b) 5 × 107(lost), c) 2 × 10−8(lost), d) –8 × 10−12(lost), Sol: n =, , =, , 107 (lost), , =5, , ., , 2. The electrostatic force between two point charges kept at a distance d apart, in a, medium εr= 6, is 0.3 N. The force between them at the same separation in vacuum is, a) 20 N, b) 0.5 N, c) 1.8 N, d) 2 N, Sol: Fm =, ⇒ F = Fm, , = 0.3, , 6 = 1.8 N, , www.Padasalai.Net, 3. Electric field intensity is 400 V m−1 at a distance of 2 m from a point charge. It, will be 100 V m−1 at a distance?, a) 50 cm, b) 4 cm, c) 4 m, d) 1.5 m, ⇒ E∝, , Sol: E =, , 400 ∝, , ……….1 ;, , 100 ∝, , ……….2, , ⇒4=, , equation, , ;, , = 16 ; r = 4 m, 4. Two point charges +4q and +q are placed 30 cm apart. At what point on the line joining, them the electric field is zero?, a) 15 cm from the charge q, b) 7.5 cm from the charge q, c) 20 cm from the charge 4q, d) 5 cm from the charge q, , Sol :E1 =E2, , ⇒, , =, , ", ", , ⇒, , ", , #", , =, =, , #", #", , 2r - 2x = x, , 60 = 3x ;, , x = 20, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 5. A dipole is placed in a uniform electric field with its axis parallel to the field. It, experiences, a) only a net force, b) only a torque, c) both a net force and torque, d) neither a net force nor a torque, Sol :, F = qE + (- qE) =0 and, % = pE sin&, , = pE sin0° = 0 (since & ( 0°), 6. If a point lies at a distance x from the midpoint of the dipole, the electric potential at this, point is proportional to, a) 1/x2, b)1/x3, c)1/x4, d)1/x3/2, Sol:, , + ,-. /, , V=, , ;, , 0, , i.e. V∝, , ", , 7. Four charges +q, +q, −q and –q respectively are placed at the corners A, B, C and D of a, square of side a. The electric potential at the centre O of the square is, , a), , b), , 23 4, , A+q, , +q B, , 23, , c), , 4, , 23, , 4, , d)zero, , www.Padasalai.Net, -q, D, , a, , -q, C, , Sol:, , V=, , =, , 56 7 6 8 6 8 69, 5, , #, 6, , 9=, , 56 9 = 0, , 8. Electric potential energy (U) of two point charges is, ;< ;=, a), b), c)pEcos θ, >? @, , d) pE sin θ, , 9. The work done in moving 500 µC charge between two points on equipotential surface is, a) zero, b) finite positive, c) finite negative, d) infinite, BCD, sol:, VA-VB =, ; since VA = VB,, , WAB = 0, 10. Which of the following quantities is scalar?, a) dipole moment, b) electric force, c) electric field, , d) electric potential, , 11. The unit of permittivity is, a) C2 N−1 m−2, b) N m2 C−2, , d) N C−2 m−2, , c) H m−1, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 12. The number of electric lines of force originating from a charge of 1 C is, a) 1.129 × 1011, b) 1.6 × 10−19, c) 6.25 × 1018, d) 8.85 × 1012, Sol : N =, , =, , =1.129, , . E, , 1011, , 13. The electric field outside the plates of two oppositely charged plane sheets of charge, density σ is, a), , FG, , #I, , b, , 3, , c), , 3, , G, , 3, , d)zero, , Sol: E1 and E2 are equal magnitude and acts on opposite direction ., , E1 - E2 =, , G, , 8, , G, , =0, , 14. The capacitance of a parallel plate capacitor increases from 5 µf to 60 µf when a dielectric, is filled between the plates. The dielectric constant of the dielectric is, a) 65, b) 55, c) 12, d) 10, , JKL: N, , PQ, OP, , =, , E, , =12, , 15. A hollow metal ball carrying an electric charge produces no electric field at points, a) outside the sphere, b) on its surface, c) inside the sphere, d) at a distance more than twice, , www.Padasalai.Net, PUBLIC ONE MARKS:, , 16. The unit of electric flux is, a) Nm2C-1, b)Nm-2C-1, , c)Nm2C, , d)Nm-2C, , 17. The work done in moving 4μC charges from one point to another in an electric field is, 0.012J.The potential difference between them is, a) 3000 V, b) 6000 V, c) 30 V, d) 48 103V, B, Sol: W=Vq; V =, =, , ., , R, , =, , S, , =3000 V, 18. Torque on a dipole in a uniform electric field is maximum when the angle between P and, E is, a) 00, b) 900, c) 450, d) 1800, Sol: τ = pE sinθ, τ = pE, 19. The potential energy of two equal point charges of magnitude 2 μC placed 1 m apart in air, is, a) 2 J, b) 0.36 J, c) 4 J, d) 0.036 J, Sol:, , =, , T, , R, , R, , =36 10-3= 0.036 J, , 5, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 20. The unit of electric field intensity is, a) NC-2, b) NC, , c) Vm-1, , 21. The value of permittivity of free space is, a)8.854 1012C2N-1m-2, c) 1/9 109C2N-1m-2, , b) 9 109C2N-1m-2, d) 1/4π 9 109 C2N-1m-2, , Sol:, , 1, 4UN0, , =9, , d) Vm, , 109, , N =1/4π 9, , 109, , 22. A lightning arrestors works on the principle of, a) corona discharge, b) diffusion of charge, c) discharge of electricity, d) separation of charges, 23. The unit of electric dipole moment is, a) volt/metre (v/m), b) coulomb/metre (c/m) c) volt . metre, , d) Coulom. metre (Cm), , 24. Electric potential energy of an electric dipole in an electric field is given as, a) pEsinθ, b) – pEsinθ, c) pEcosθ, d) –PEcosθ, 25. Which of the following is not a dielectric?, a) Ebonite, b) Mica, , c) Oil, , d) Gold, , 26. In the given circuit, the effective capacitance between A and B will be, , www.Padasalai.Net, a) 3 μF, Sol:, , b) 36/13 μF, , Cs, , C C, = C 17C 2, 1, 2, 3, , 6, , Cs1= 376 =, Cs2 =, , F, , =, , T, , c) 13 μF, , d) 7 μF, , =2 μF ;, =1μF ;, , Cp = Cs1+ Cs2 = 2+1 =3μF, 27. The direction of electric field at a point on the equatorial line due to an electric dipole is, a) along the equatorial line towards the dipole, b) along the equatorial line away from the dipole, c) parallel to the axis of the dipole and acts opposite to the direction of the dipole moment, d) parallel to the axis of the dipole and in the direction of dipole moment, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 28. The effective capacitance of two capacitors connected in series is 1.5 μF. If the, capacitance of one capacitor is 4 μF , then the capacitance of the other is, a) 2.4 μF, b) 0.24 μF, c) 0.417 μF, d) 4.17 μF, Sol:, Cs = 1.5 ; C1 = 4 μF, PW, , =P +P, , 1, , 1, , ⇒X (X P, Y, 2, P, P, , ⇒, , =, , .E, , -, , # .E, , =, , C2 =, , .E, , .E, , =, , = 2.4 μF, , 29. The law which govern the forces between the charges is, a) Ampere’s law, b) Faraday’s law, c) Coulomb’s law, , d) Ohm’s law, , 30. An electric dipole is placed at an angle Z in a non-uniform electric field experiences, a) only a net force, b) only torque, c) both force and torque, d) Neither a net force and a torque, 31. A capacitor of capacitance 6 μF is connected to a 100 V battery. The energy stored in the, capacitor is, a) 30 J, b) 3 J, c) 0.03 J, d) 0.06 J, , www.Padasalai.Net, Sol :, , U = CV2, =, , 6, , 10-6, , 1002, , =, , 6, , 10-2 =3, , 10-2 = 0.03 J, , 32. The potential energy of an electric dipole of dipole moment P aligned in the direction of, electric field E is, a)PE, b) zero, c) –PE, d) PE /√2, Sol :, U = -pE cos&; then & ( 0, ∴U = - pE, 33. The quantization of electric charge is given, a) q = ne, b) q = CV, 34. An example of conductor is, a) glass, b) human body, , c) q = e/n, c) dry wood, , d) q = C/V, d) ebonite, , 35. An electric dipole is placed in a non-uniform electric field with its axis at an angle θ with, the field experiences, a) only a net force, b) only torque, c) both a net force and torque, d) Neither a net force and a torque, , 7, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 36. The magnitude of force acting on a charge of 2, of 10 Vm-1 is, a) 2 10-9N, b) 4 10-9N, c) 2, Sol :, F= qE, 2 10-10 = 2, , = 10, , 2017-2018, , 10-10C placed in a uniform electric field, , 10-10N, , d) 4 10-10N, , 10-9 N, , 37. The capacitance of a parallel plate capacitor increases from 5µF to 50 µF when a, dielectric is filled between the plates. The permittivity of dielectric is, a) 8.854 10-24C2N-1m-2 b) 8.854 10-11C2N-1 m-2, c) 12, d) 10, JKL:, , N (, , PQ, P, , E, , =, , = 10 ;, , E, , N(N N, 10-12, , = 8.854, , 10 = 8.854, , 38. The negative gradient of potential is, a) electric force, b) torque, , 10-11 C2N-1 m-2, c) electric current, , d) electric field intensity, , 39. The torque (τ) experienced by an electric dipole placed in a uniform electric field (E) at, an angle θ with the field is, a) PEcosθ, b) –PEcosθ, c) PEsinθ, d) 2 PEsinθ, 40. When a point charge of 6 µC is moved between two points in an electric field, work done, is 1.8 10-5 J. The potential difference between the two points is, a) 1.08 V, b) 1.08 µV, c) 3 V, d) 30 V, Sol : W = Vq, , www.Padasalai.Net, \, , ., , ⇒V=, , R, , =, , =3V, , ^^_ and, 41. Torque on a dipole in a uniform electric field is maximum when the angle between ], ^`, ^_ is, a) 00, b) 900, c) 450, d) 1800, 42. Three capacitances 1 μF, 2 μF and 3 μF are connected in series. The effective capacitance, of the capacitors is, a) 6 μF, b) 11/6 μF, c) 6/11 μF, d) 1/6 μF, Sol:, , 1, XY, , = P + P + Pa, = 7 7a, FaF, , =, ⇒ Cs =, , =, , ;, , μF, , ^^_ is placed in a uniform electric field of intensity `, ^_ at an, 43. An electric dipole of moment ], angle θ with respect to the field. The direction of torque is, ^^_, ^^_, a) along the direction of ], b) opposite to the direction ], ^_, ^^_and `, ^_, c) along the direction of `, d) perpendicular to the plane containing ], , 8, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 44. The electric field intensity at a distance r from an infinitely long uniformly charged, straight wire is directly proportional to, a) r, b) 1/r, c) r2, d) 1/r2, Sol :, , E(2U b (, ⇒ E=, , cd, c, , ⇒ E∝, , 45. The ratio of electric potentials at points 10 cm and 20 cm from the centre of an electric, dipole along its axial line is, a) 1:2, b) 2:1, c) 1:4, d) 4:1, Sol :, , + ,-. /, , V=, , 6, , V1 ∝, , ⇒V1∝, , …………(1) ;, , V2 ∝, , ⇒V2∝, , …………(2) ;, , e, , ⇒ e1 =, , Equ:, , ;, , ∴ 4:1, , =, , 2, , 46. The intensity of electric field at point is equal to, a) the force experienced by a charge q, b) the work done in bringing unit positive charge from infinity to that point, c) the positive potential gradient, d) the negative gradient of potential V, , www.Padasalai.Net, 47. The capacitance of a capacitor is, a) directly proportional to charge q given to it, b) inversely proportional to its potential, c) directly proportional to charge q and inversely proportional to its potential V, d) independent of both charge q and potential V, Hints: q ∝ V ; if charge q is increases and v is also increases ., C=, , f, , so capacitance does not change, , 48. Intensity electric field produces a force of 10-5N on a charge of 5 µC is, a) 5 10-11NC-1, b) 50 NC-1, c) 2 NC-1, d) 0.5 NC-1, Sol :, , E=, , g, h, , =, =, , \, , R, , E, E, , = 2 NC-1, , 49. The unit of number of electric lines of force passing through a given area is, a) no unit, b) NC-1, c) Nm2C-1, d) Nm, Sol : N =, =i, , 9, , j, , P, , k, , = Nm2C-1, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 50. A dielectric medium is placed in an electric field E0. The field induced inside the medium, is, a) acts in the direction of electric field E0, b) acts opposite to E0, c) acts perpendicular to E0, d) is zero, 51. A non-polar dielectric is placed in an electric field (E). Its induced dipole moment, a) zero, b) acts in the direction of E, c) acts opposite to the direction of E, d) acts perpendicular to E, 52. n capacitors of capacitance C connected in series. The effective capacitance is, a)n/C, b) C/n, c) nC, d) C, Sol:, , (P 7, , PW, , (P7, , PW, PW, , (, , P, P, , ………7 P, , l, , since (X ( X … … ( Xm ), , ………7 P, , F F ……F, P, , m, , X, , (P;, , Cs = n, , 53. The unit of relative permittivity is, a) C2N-1m-2, b) Nm2C-2, , JKL:, , c) No unit, , d) NC-2m-2, , N (, =, , P o, P o, , p, p, , = No unit, , www.Padasalai.Net, 54. The value of relative permittivity of air is, a) 8.854 10-12C2N-1m2 b) 9 109 N-1m-2, , c) 1, , d) 8.854, , 1012, , 55. An electric dipole of dipole moment ‘p’ is kept parallel to an electric field of intensity ‘E’., The work done in rotating the dipole through 900 is :, a) Zero, b) –pE, c) pE, d) 2pE, Sol :, dw = %. q& ( take integral on both sides), T, , w = r st sin & q&, , xYynz{ % ( st Yyn&|, , = pEx8z}Y&|T, = - pE cos 900 + pE cos 00 = 0 + pE, = pE, 56. The total flux over a closed surface enclosing a charge ‘q’ ( in Nm2C-1), a)8πq, b)9 109 q, c)36π 109 q, d)8.854 10-12 q, , =9, , Sol:, , ⇒, By Gauss law, , 10, , 109, , = 4π 9, ɸ=, , 109 = 36π, , 109, , = 36π 109 q, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 57. Two point charges +q and –q are placed at points A and B respectively separated by a, Small distance. The electric field intensity at the midpoint of AB, a) is zero, b) acts along AB, c) acts along BA, d) acts perpendicular to AB, Hints: +ve→ outward direction; 8ve → inward direction. so it acts along AB, 58. The electric field inside (between) the plates of two oppositely charged plane sheets each of, charge density σ is, Ž, G, G, a)+, b) c), d) zero, Sol:, , @, , E1 and E2 are equal magnitude and acts on same direction., •, •, E1+ E2 = 2N 7 2N, 0, , (, , 0, , G, , 59. The electric field intensity at a short distance r from a uniformly charged infinite plane, sheet of charge is, a) proportional to r, b) proportional to 1/r, c) proportional to 1/r2, d) independent of r, Sol :, , E=, , G, , . So independent of r, , 60. The number of lines that radiate outwards from one coulomb charge is, a) 1.13 1011, b) 8.85 10-11, c) 9 109, d) infinite, , www.Padasalai.Net, Sol :, , =, , n=, , . E, , =1.129, , 1011, , 1011, , =1.13, , 61. When the charge given to the capacitor is doubled, its capacitance, a)increases twice, b) decreases twice, c) increases four times d) does not change, Hints: q ∝ V ; if charge q is increases and v is also increases ., C = so does not change, f, , 62. On moving a charge of 20 C by 2 cm, 2J of work is done, then the potential difference, between the points is, a) 0.5 V, b)0.1 V, c) 8 V, d) 2 V, Sol :, W= V q, •, , ⇒ V=h =, , = 0.1V, , 63. The repulsive force between two like charges of 1 coulomb each separated by a distance of, 1 m in vacuum is equal to, a)9 109 N, b) 109N, c)9 x 10-9N, d) 9 N, Sol:, , F=, =, , π, , [, , π, , (9, 11, , 6, π, , (9, , 109 ], , 109 N, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 64. What must be the distance between two equal and opposite point charges (say +q and –q), For the electrostatic force between them to have a magnitude of 16 N?, ;, •, a) 4‘’h metre, b) √” metre, c)4 kq metre, d) metre, Sol:, , >, , F=k, ⇒ 16 = k, ⇒ r2 =, , k ⇒ r = √’ metre, , ^_. The induced dipole moment, 65. A non-polar molecule is placed in an external electric field –, acts, ^_, a) in the direction of –, b) opposite to the direction of ^E_, c) perpendicular to the direction of ^E_, d) at random, 66. Van de Graaff generator works on the principle of, a) electromagnetic induction and action of points, b) electrostatic induction and action of points, c) electrostatic induction only, d) action of points only, 67. For which of the following medium, the value of relative permittivity = 1:, a)mica, b)air, c)glass, d)water, 68. Point charges +q, +q, -q and –q are placed at the corners A, B, C and D respectively of a, square. O is the point of intersection of the diagonals AC and BD. The resultant electric, field intensity at the point O, a) acts in a direction parallel to AB, b) acts in a direction parallel to BC, c) acts in a direction parallel to CD, d) is zero, C -q, Sol: B +q, , www.Padasalai.Net, +q, -q, A, D, Hints:, If the charge is positive the direction of electric field is outward direction., If the charge is negative the direction of electric field is inward direction., Reason:, The direction of electric field intensity of charge of corners A and C is A to C, and, Corners B and D is B to D., So the resultant direction of electric field intensity of A to C and B to D is acts in the direction, of parallel to BC and AD, 69. The unit of molecular polarisability is, b)Nm2C-1, a) C2N-1m, sol : p ∝ E ⇒ p ( ™t, ™=, =, , c)N-1m-2C2, , d)C-1m2V, , +, š, , Pp, , oP, , = C2N-1m, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 70. Two point charges +q1 and +q2 are Placed in air at a distance of 2 m apart. One of the, charges is moved towards the other through a distance of 1m. the work done is :, a), , c), , b), , ;< ;=, , ›? @, , d), , Sol: (work done is stored as electrostatic potential energy) W = U=, , Initial ( r = 2m ) ⇒ Wi = Ui =, , =, , ⇒ Wf = Uf =, , =, , Final (r = 1m ), , 8, , W = Uf - Ui =, =, , 51 8 9, , =, , 5 9=, , 71. The capacitances 0.5 œF and 0.75 œF are connected in parallel. Calculate the effective, capacitance of the capacitor, a) 0.80 •F, b) 0.70 •F, c) 0.25 •F, d) 1.25 œF, sol:, Cp = C1 + C2, , www.Padasalai.Net, = 0.5 •F + 0.75 •F, , = 1.25 •F, 72. When a dielectric slab is introduced between the plates of a charged parallel plate, capacitor, its :, a) potential increases, b) electric field decreases, c) charge increases, d) capacitance decreases, Hints:, G, Electric field in between parallel plate capacitor E =, ., When a dielectric slab is introduced between the plates of a charged parallel plate capacitor,, The Electric field E′ =, , G, , =, , G, Ÿ, , =, , š, Ÿ, , So electric field decreases N times., , 13, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 73. A and B are two hollow metal spheres of radii 50 cm and 1 m carrying charges 0.6 µC and, 1 µC respectively. They are connected externally by a conducting wire. Now the charges, flows from :, a) A to B till the charges become equal, b) A to B till the potential become equal, c) B to A till the charges become equal, d)B to A till the potential become equal, Sol:, , Sphere A, = 4π, = 50 10# m, h = 0.6 10# C, • =, , e =, , ¡, , =, , Sphere B, = 4π, =1m, h = 1 10# C, • =, , 2, , e =, , 2, , •, , ¡, , =, , 2, , 2, , ( •, , www.Padasalai.Net, =, , 2, , =, , 2, , 2, , ⇒ ¢< ( ¢=, , 2, , =, ., , R, , ., , R, , E, E, , 0.6, , =, =, , 10# = 50, , R, , R, , 10#, , 10#, , 0.6 ( 0.5, Charge can flows high potential to low potential, A→B, 74. The equipotential surface of an electric dipole is :, a) a sphere whose centre coincides with the centre of the electric dipole, b) a plane surface inclined at an angle of 45 ° with the axis of the electric dipole, c) a plane surface passing through the centre of the electric dipole and perpendicular to, the axis of the electric dipole, d) any plane surface parallel to the axis of the electric dipole, Hints:, The equipotential surface is always perpendicular to the lines of force, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 75. Force between two charges situated in a medium of permittivity ′ £ is :, ε, T, a), b)9 10T ε6, c) 9 10T, d), 6, 6, ε¤, π 6, Sol :, , Fm =, =, Fm =, , ∴ xN ( N, , 6, , π3, π, , [, , 6, , Ÿ, , T, , N |, , (9, , π, , 6, , 109 ], , 6, , ε¤, , 76. The work done in moving 6 œ¥ charge between two points is 1.2 10-5 J. find the potential, difference between the two points is, a) 6 V, b) 2V, c) 12 V, d) 72 V, Sol :, W= V q, •, , ⇒ V=h, =, , \, , ., , R, , =2V, , =, , 77. Point charges q and q are placed in air at a distance 'r'. The ratio of the force on charge, q by charge q and force on charge q by charge q is:, , www.Padasalai.Net, d) §, , c) 1, , b), , a), , Sol : Force exerted on charge q due to charge q, , F21 =, , Force exerted on charge q due to charge q, , F12 =, , π3, , 6, , π3, , 6, , ¨, , ratio of the force on charge q by charge q and force on charge q by charge q, ©, π3 6, =, ;, [r, ( r ], π3 6, ©, ©, ©, , =1, , 78. The electric field at a point 2 cm from an infinite line charge of linear charge density 10−7, Cm−1 is :, a) 4.5 × 104 NC−1, b) 9 × 104 NC−1, c) 9 ×102 NC−1, d) 18 × 104 NC−1, ª, Sol :, E=, 23 6, , =, , «, , §∵, , 2πε0, , ( 18, , 10T ¨, , λ = 10-7 Cm-1, , 15, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PTA One mark questions:, Choose the best answer:, , 79. SI Unit of electric charge is, a) coulomb, b) ampere second, , c), , ¯-°± . ,-²³, -´k, , 80. Electric dipole moment always acts in the direction from, a) +q to –q, b) –q to +q, c) ∞ to +q, , d) all the above, , d) ∞ to –q, , 81. An electric dipole in a uniform electric field experiences a, a) force, b) torque, c) momentum, d) neither force nor torque, 82. When a dipole is aligned with field, then potential energy is given as, a) PE sin&, b) 0, c) –PE, , d) –PE cos&, , 83. Work done in moving an electric charge on an equipotential surface is, a) 0, b) minimum, c) maximum, , d) infinity, , 84. Which one of the following is a non – polar molecule?, a) H2O, b) CO2, c) HCL, , d) O2, , 85. Potential at a point due to point charge is given by, , www.Padasalai.Net, ;, , a)>?, , b), , c), , d), , 86. Dielectric is also called as, a) conductor, b) inductor, , c) resistor, , d) insulator, , 87. The permittivity of any medium is given by, a) N 0/ Nr, b) N 0+ N r, , c), , d) N 0 - N r, , 88. Electric field intensity at any point is given by, a) E= Fq, b) E=F/q, , c) E= q/F, , @, , 0, , r, , d) E= F-q, , 89. A device not working with the principle of electrostatic induction is, a) Vande Graff generator b) microwave oven, c) lightning arrestor d) (a) and (c), 90. When two capacitors are connected in series to a source of emf, then each one of them will, have same, a) voltage, b) electric field, c) both (a) & (b), d) charge, 91. Four charges +q, +q, -q and –q, respectively are placed at the corners A,B,C and D of a, square of side ‘a’. The electric potential at the centre ‘O’ of the square is, a), , µ, , b), , µ, , c), , µ, , d)Zero, , 92. If the medium between two charges is replaced by air, then the force between them, a) increases, b) decreases, c) becomes zero, d) remains constant, , 16, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 93. An electric dipole consists of two, a) like and equal charges., c)unlike and equal charges, , 2017-2018, , b) like and unequal charges, d) unlike and unequal charges, , 94. If a Gaussian surface encloses a dipole of moment 2qd, then the total flux through the, surface is, a) q/N 0, b) 2q/N 0, c) q/2N 0, d) 0, 95. When air medium in capacitor is replaced by a medium of dielectric constant, capacitance, a) decreases N r times, b) remains the same, c) increase r times, d) increased N r2 times, 96. An example for polar molecule, a) N2, b) H2, , c) H2O, , r,, , the, , d) O2, , 97. Two charges 10-6 C and 10-7 repel each other with a force of 400N. The distance between, the charge is, a) 0.15 mm, b) 1.5 mm, c) 15 mm, d) 1.5 m, 98. The potential difference between two parallel plates is 100 V and the electric field between, them is 104 V/m. Then the distance between the plates., a) 1 mm, b) 1 m, c) 10 cm, d) 1 cm, 99. The plates of a parallel plate capacitor are separated by a distance of 1 mm. If the, capacitance is 8.854œ , then the area of the plates is, a) 10-3 m2, b) 10 m2, c) 103m2, d) 102m2, , www.Padasalai.Net, 100. If a capacitor of capacitance 55 PF is charged to 1.6 V, then the number of electrons, on its negative plate is, a) 55 <@7, b) 5.5 107, c) 550 107, d) 0.55 107, 101. The workdone in moving a charge of 2œ¥ ¶·¸¹··º two points having different, potential of 110 V and 220 V is, a) 22 10-4 J, b) 2.2 104 J, c) 22 10+4 J, d) 2.2 <@-4 J, 102. Two charges +4 C and +1 C are separated by a distance of 3 m. To keep these charges, in equilibrium, a third charge is to be placed at, a) 2m from the charge 4 C, b) 2m from the charge 1 C, c) 1.5 m from the charge 4 C, d) 2.5 m from the charge 1 C, 103. Equivalent capacitance of two capacitors when connected in parallel is 8œ and when, connected in series is 15/8 œ .Then the valves of the two capacitors are, a) 7 μF and 1 μF, b) 6 μF and 2 μF, c) 4 μF and 4 μF, d) 5 ½¾ ¿ÀÁ  ½¾, 104. Two capacitors of capacitance 200 PF and 600 PF are connected in parallel and then, charged to a potential of 120 V. Then the value of the total charge on the capacitors is, a) 24 10-9 C, b) 96 <@-9 C, c) 48 10-9 C, d) 72 10-9 C, 105. The intensity of the electric field that produces a force of 10 N on a charge of 5C is, a) 2 NC-1, b) 50 NC-1, c) 5 NC-1, d) 0.5 NC-1, , 17, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 19 : www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 106. If two identical point charges separated by 3 m experience a force of 10 N, then the, value of each charge is, a) 10-1 C, b) 10 C, c) 1C, d) 10-4 C, 107. Two identical metal spheres have charges + 15 œ¥ and +25 œ¥ and are separated by a, distance. If the spheres are first brought into contact and then separated by a distance,, then the ratio of the new force between them to the previous force is, a) 15:16, b) 3:5, c) 16:15, d) 5:3, 108. If the moment of an electric dipole is 1.2X10-9 Cm and the distance between the, charges is 3 mm then the charge of the dipole is, a) 3.6 •X, b) 40 •X, c) 3.6 10-12 C, d) 0.4 œ¥, 109. A Parallel plate capacitor consists of two circular plates of radius 3 cm separated by a, dielectric material of thickness 0.5 mm and dielectric constant 4. Then the capacitance of, the capacitor is, a) 50 PF, b) 200 PF, c) 2 PF, d) 0.5 PF, 110. A parallel plate capacitor connected to a 12 V source is charged to 21 œ¥ ÃÄ the, capacitor is filled with an oil of dielectric constant 3, then the charge stored is, a) 7 •X, b) 63 œ¥, c) 14 •C, d) 57 •X, 111. The equivalent capacitance of two capacitors in series is 1.5 œ . The capacitance of one, of them is 4 œ . The value of capacitance of other is, a) 2.4 œ, b) 0.24 •g, c) 0.417 •g, d) 4.17 •g, , www.Padasalai.Net, 112. Three capacitors 2 œ , Æœ and 3 œ are in parallel across 5 V suppy. The charges on, each of the m respectively are, a) 12μC, 30 μC, 18μC, b) 10½Ç, =Æ ½Ç, <Æ ½Ç, c) 8 μC, 20 μC, 12μC, d) 6μC, 15μC, 9 μC, 113. If the distance between two protons in uranium atom is 9 <@-15 m, then the mutual, electric potential energy between them is, a) 9 10-14 J, b)1.44 10-15 J, c) 2.56 <@-14 J, d) 1.6 10-5 J, , Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], Contact for subject doubt, Email :
[email protected], Phone No: 9952223467, , JEYAM TUITION CENTRE, [PARAMAKUDI], Contact for tuition, Email :
[email protected], Phone No: 9944888512, 18, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 3 MARK QUESTIONS (2 – questions: Q.No: 31, 32), PUBLIC ‘3’ MARKS:, 1. What do you mean by additive nature of charges ? Give an example. (O– 07), , The total electric charge of a system is equal to the algebraic sum of, electric charges located in the system., For example, if two charged bodies of charges +2q, −5q are brought in contact,, the total charge of the system is –3q., 2. State Coulomb’s law in electrostatics. ( J – 07, J – 10, O – 11, J – 12,M-15 ), , Coulomb’s law states that the force of attraction or repulsion between two, point charges is directly proportional to the product of the charges and inversely, proportional to the square of the distance between them., The direction of forces is along the line joining the two point charges., F∝, ^_, In vector form F, , F=, , 6, , = 4πε, , 0, , 6, , 23 π 6, , rÈ, , 3. Define: Coulomb on the basis of Coulomb’s law. ( M – 06, M – 10, O – 10,M-, , www.Padasalai.Net, 13 ), , One Coulomb is defined as the quantity of charge, which when placed at a, distance of 1 metre in air or vacuum from an equal and similar charge,, experiences a repulsive force of 9 ×109N., , 4. Give any three properties of electric lines of force. ( J – 10, M-16 ), , i) Lines of force start from positive charge and terminate at negative charge., ii) Lines of force never intersect., iii) The tangent to a line of force at any point gives the direction of the electric, field (E) at that point., 5. What is an electric dipole? Define: dipole moment. ( O – 09, J – 11,M-14 ,M-17), , Dipole: Two equal and opposite charges separated by a very small distance, constitute an electric dipole., Examples : Water, ammonia, carbon−dioxide and chloroform molecules, Dipole moment: The dipole moment is the product of the magnitude of the one, of the charges and the distance between them., ∴ Electric dipole moment, p = q2d or 2qd., It is a vector quantity and acts from – q to +q., The unit of dipole moment is C m., , 19, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 6. Define electric dipole moment. Give its unit. (M-14), , Dipole moment: The dipole moment is the product of the magnitude of the one, of the charges and the distance between them., ∴ Electric dipole moment, p = q2d or 2qd., It is a vector quantity and acts from – q to +q., The unit of dipole moment is C m., 7. Give the principle of working of a microwave oven. ( J – 08,O-14 ), , It is used to cook the food in a short time. When the oven is operated, the, microwaves are generated, which in turn produce a non-uniform oscillating, electric field. The water molecules in the food which are the electric dipoles are, excited by an oscillating torque. Hence few bonds in the water molecules are, broken, and heat energy is produced. This is used to cook food., 8. Define: Electric potential at a point in an electric field. ( M – 07, J – 09,O-13 ), , The electric potential in an electric field at a point is defined as the amount, of work done in moving a unit positive charge from infinity to that point against, the electric forces. It is a scalar quantity. Its unit is volt, 9. Define: Electric flux. Give its unit. ( J – 08, J – 12 ), , www.Padasalai.Net, The electric flux is defined as the total number of electric lines of force,, crossing through the given area. i.e. ɸ = ∮ qɸ = ∮ ^E_ . ^^^^_, ds, 2 -1, Its unit is N m C ., , 10. State Gauss’s law in electrostatics. ( M – 09, J – 06 , O – 06, M – 11,J-15,M-16 ), , The total flux of the electric field E over any closed surface is equal to 1/ε, times the net charge enclosed by the surface., , ( i.e.) ϕ =, , 3, , 11. What is electrostatic shielding? ( M – 08 ), , It is the process of isolating a certain region of space from external field., It is based on the fact that electric field inside a conductor is zero., 12. Why, is it safer to be inside a car than standing under a tree during, , lightning ? ( M -06, J- 06, J – 09, M-10,J-14,J-15,M-17 ), , (i) The metal body of the car provides electrostatic shielding,, (ii) Inside the car electric field is zero., (iii)During lightning the electric discharge passes through the body of the car., , 20, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 13. What is a capacitor? Define its capacitance. ( M – 09 ), , i) A capacitor is a device for storing electric charges., ii) The capacitance of a Capacitor (conductor) is defined as the ratio of the, charge given to the Capacitor (conductor) to the potential developed in the, conductor., C=, , Ë, iii) unit is farad, 14. What are polar molecules? Give an example. ( M – 07,M-13 , J-16), , i) A polar molecule is one in which the centre of gravity of the positive, charges is separated from the centre of gravity of the negative charges by a, finite distance., ii) Examples : N2O, H2O, HCb , NH3., iii) They have a permanent dipole moment, 15. What are non-polar molecules? Give an example. ( O – 10, J – 11,O-13 ), , i) A non-polar molecule is one in which the centre of gravity of the positive, charges coincide with the centre of gravity of the negative charges., ii) Example: O2, N2, H2., iii) The non-polar molecules do not have a permanent dipole moment., , www.Padasalai.Net, 16. Distinguish between polar and non polar molecules, , Polar molecules, 1. A polar molecule is one in which, the centre of gravity of the, positive charges is separated from, the centre of gravity of the, negative charges by a finite, distance., 2. Examples : N2O, H2O, HCb , NH3., 3. They have a permanent dipole, moment, , Non - polar molecules, A non-polar molecule is one in, which the centre of gravity of the, positive charges coincide with the, centre of gravity of the negative, charges., Example: O2, N2, H2., The non-polar molecules do not, have a permanent dipole moment., , 17. What is dielectric polarization? ( O - 06, O – 09, O – 11,J-14 ), , The alignment of the dipole moments of the permanent or induced dipoles in, the direction of applied electric field is called polarisation or electric polarisation., The magnitude of the induced dipole moment p is directly proportional to the, external electric field E., ∴p ∝ E or p = ™E, where α is the constant of proportionality and is called, molecular polarisability., The unit of molecular polarisability is C2N-1m, , 21, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 18. Write the applications of a capacitor. ( O – 07, M – 11, M – 12, O – 16 ), , i) They are used in the ignition system of automobile engines to eliminate, sparking., ii) They are used to reduce voltage fluctuations in power supplies and to, increase the efficiency of power transmission., iii) Capacitors are used to generate electromagnetic oscillations and in tuning the, radio circuits., 19. What is action of points (corona discharge)? Give its application ? ( J – 07, O –, , 08,O-14, J-15,O-15), The leakage of electric charges from the sharp points on the charged, conductor is known as action of points or corona discharge., This principle is used in the electrostatic machines for collecting charges and in, lightning arresters., , OTHER IMPORTANT ‘3’ MARKS:, 20. What are insulators and conductors?, , Bodies which allow the charges to pass through are called conductors., E.g. metals, human body, Earth etc., Bodies which do not allow the charges to pass through are called insulators., E.g. glass, mica, ebonite, plastic, etc., , www.Padasalai.Net, 21. Define permittivity and relative permittivity ?, , Permittivity: The ability of a medium to permit the electric lines of force to, pass through it. The unit of permittivity C2N-1m-2, Relative permittivity: The ratio of permittivity of medium to that of permittivity, of air or vacuum. N =, , ; The value of N for air or vacuum is 1., , It has no unit., 22. What is known as quantisation of electric charge ?, , The fundamental unit of electric charge (e) is the charge carried by the, electron and its unit is coulomb. e has the magnitude 1.6 × 10−-19 C. The electric, charge of any system is always an integral multiple of the least amount of, charge. It means that the quantity can take only one of the discrete set of, values., The charge, q = ne. where n is an integer., , 22, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 23. State the law of conservation of electric charge ?, , The total charge in an isolated system always remains constant. But the, charges can be transferred from one part of the system to another, such that the, total charge always remains conserved., For example, 92U238 → 90Th234 + 2He4, Total charge before decay = +92e, total charge after decay = 90e + 2e., Hence, the total charge is conserved. i.e. it remains constant., 24. Give the basic properties of electric charge ?, , (i) Quantisation of electric charge, (ii) Conservation of electric charge, (iii) Additive nature of charge, 25. Define electric field intensity., , Electric field intensity at a point, in an electric field is defined as the force, experienced by a unit positive charge kept at that point. It is a vector quantity., Ì©_ Ì, Ìt^_ Ì = . The unit of electric field intensity is N C−1, 26. What is known as electric lines of force?, , www.Padasalai.Net, Electric line of force is an imaginary straight or curved path along which a, unit positive charge tends to move in an electric field., , isolated charge, , unlike charges, , like charges, , 27. Define potential difference, , The potential difference between two points in an electric field is defined as, the amount of workdone in moving a unit positive charge from one point to the, other against the electric force. The unit of potential difference is volt., 28. Distinguish between electric potential and potential difference ., , Electric potential, 1. The amount of work done in, moving a unit positive charge, from infinity to that point against, the electric forces., 2. Unit is volt, 3. Scalar quantity, , Potential difference, The amount of work done in, moving a unit positive charge, from one point to the other, against the electric force., Unit is volt, Scalar quantity, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 29. Define electric potential energy, , The electric potential energy of two point charges is equal to the work done, to assemble the charges., Potential energy (U) =, 30. What does an electric dipole experience when kept in a uniform electric field, , and non uniform electric field ?, , When a dipole is kept in a uniform electric field at an angle θ, the net force, F is zero., , ^_, It experiences a torque %_ = s_ t, τ = pE sinθ, In a non uniform electric field in addition to a torque τ, it also experiences a, force F, 31. What is equipotential surface?, , If all the points of a surface are at the same electric potential, then the, surface is called an equipotential surface., i) In case of an isolated point charge, equipotential surfaces will be a series of, concentric spheres with the point charge as their centre., ii) In case of uniform field, equipotential surfaces are the parallel planes with, their surfaces perpendicular to the lines of force, , www.Padasalai.Net, 32. What is called electrostatics induction?, , It is possible to obtain charges without any contact with another charge., They are known as induced charges and the phenomenon of producing induced, charges is known as electrostatic induction. It is used in electrostatic machines, like Van de Graaff generator and capacitors., , 33. Define farad., , A conductor has a capacitance of one farad, if a charge of 1 coulomb given, to it, rises its potential by 1 volt., Farad is the unit of capacitance., 34. Define one volt., , The potential difference between two points is 1 volt if 1 joule of work is, done in moving 1 Coulomb of charge from one point to another against the, electric force., 35. What are dielectrics or insulators and give an example., , A dielectric is an insulating material in which all the electrons are tightly, bound to the nucleus of the atom. There are no free electrons to carry current., Ebonite, mica and oil are few examples of dielectrics., , 24, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 3. Calculate the potential at a point due to a charge of 4×10-7 C located at 0.09 m away, from it. (M–12,M-14), Given data: q = 4, Sol:, , 108Ï C;, , r = 0.09 m = 9, , The electric potential V =, , =, , T, , 108 m;, , V=?, , Ð, 8«, , T, , V = 4 × 104volt, , 4. Three capacitors each of capacitance 9 pF are connected in parallel. Find effective, capacitance., Given data: X ( X ( Xa ( 9, Sol:, , 108 F, , The effective capacitance CP = C1 + C2 + C3, =9, , 108 + 9, , 108, , = ( 9 + 9 + 9), , 10#, , 9, , 108, , www.Padasalai.Net, CP = 27 pF, , 5. A sample of HCl gas is placed in an electric field of 2.5 × 104 N C−1. The dipole, moment of each HCl molecule is 3.4 × 10−30 C m. Find the maximum torque that, can act on a molecule. [ M – 15], , Data : E = 2.5 × 104 N C−1, p = 3.4 × 10−30 C m., Solution : Torque acting on the molecule, , τ = pE sin θ for maximum torque, θ = 90o, = 3.4 × 10−30 × 2.5 × 104, , Maximum Torque acting on the molecule is = 8.5 × 10−26 N m., , 26, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 6. A point charge causes an electric flux of –6 × 103 Nm2 C−1 to pass through a, spherical Gaussian surface of 10 cm radius centred on the charge., If the radius of the Gaussian surface is doubled, how much flux will pass, (i), through the surface?, (ii) What is the value of charge? ( O-15), , Data : ɸ = −6 × 103 N m2 C−1;, , r = 10 cm = 10 × 10−2 m, , Solution :, (i), , If the radius of the Gaussian surface is doubled, the electric flux, through the new surface will be the same, as it depends only on the, net charge enclosed within and it is independent of the radius., ɸ = −6 × 103 N m2 C−1, , (ii), , ɸ=, q = ɸ N = - (6 × 103, , 8.854 × 10-12), , q = − 5.31 × 10−8 C, , www.Padasalai.Net, 7. Three capacitors each of capacitance 3 pF are connected in series. What is total, capacitance of the combination . (J – 16), , 10# F, , Given data: X ( X ( Xa ( 3, , Sol: The total capacitance of the combination, PÎ, PÎ, , =, =, , Cs =, , P, , 7, , =1, , 7, 7, , a, a, , P, , PS, a, , 7, , a, , =, , a, , a, , 10# F, , Cs= 1 pF, , 27, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html, , a
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 10. Three charges -2 × 10-9 C, +3 × 10-9 C and -4 × 10-9 C are placed at the vertices, of an of an equilateral triangle ABC of side 20 cm. calculate the workdone in, shifting the charges from A, B and C to A1, B1 and C1 respectively. Which are, the mid-points of the sides of triangles? [J-11], , Given data: q1=-2 × 10-9C; q2=+3 × 10-9C; q3=-4 × 10-9C., Sol: The triangle is equilateral triangle, AB = BC = CA = r = 20 cm = 0.20 m, The potential energy of the system of charges,, U=, , 23, , 5, , 6, , 7, , 6, , S, , 7, , S, , 6, , 9, , www.Padasalai.Net, Work done in displacing the charges from A, B and C to A1, B1and C1 respectively, W = Uf – Ui, , Ui and Uf are the initial and final potential energy of the system., Ui =, , TÔ, , ., , Õ86, , 10#, , – 12, , 10#, , 7 8, , 10# ×, , – 12, , 10#, , 7 8, , 10# ×, , = − 4.5 × 10−7 J, Uf =, , TÔ, ., , Õ86, , 10#, , = −9 × 10−7J, , ∴work done =Uf – Ui = −9 × 10−7 – (−4.5 × 10−7), W = − 4.5 × 10–7J, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 14. A parallel plate capacitor has an area 200 cm2 and the separation between the, plates is 1mm. Calculate i) the potential difference between the plates if 1 nC, charge is given to the capacitor. ii) With the same charge (1nC) if the separation, is increased to 2 mm, what is the new potential difference and iii) the electric, field between the plates. [M-06], , Given data: A = 200 cm2;, , 10-3;, , d = 1mm = 1, , q = 1 nC = 1, , 10-9C;, , Sol: Capacitance of a parallel plate capacitor, C=, , 3 Ü, ³, , Ó, , . E, , =, , S, , C = 0.177 × 10−9 F = 0.177 nf, (i), , The potential difference between the plates, V=, , i, , =, , . ÏÏ, , www.Padasalai.Net, V = 5.65 V, , (ii), , If the plate separation is increased from 1 mm to 2 mm, the capacitance is, decreased by 2, the potential difference increases by the factor 2, , ∴ New potential difference is 5.65 × 2 = 11.3 V, (iii), , Electric field is,, , E=, =, , I, , 3, , =, , 5since σ ( 9, , Ü3, , Ü, , Ó, , . E, , E = 5650 N C−1, , 32, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 15. Three capacitors each of capacitance 9 pF are connected in series i) What is the, total capacitance of the combination? ii) What is the potential difference across, each capacitor if the combination is connected to 120 V supply? [J-06,O-06,J-11], 10-12 F;, , Given data: C1 = C2 = C3 = 9 pF = 9, Sol:, , V = 120 V, , i) If Cs is the capacity of the series combination, iØ, , =, , i, , =, , 7, , T, , Cs = Â, , 7, , iS, , 7, , T, , =, , i, , a, , =, , 7, , T, , T, , a, , <@#<= F, , ii) V1, V2 and V3 be the potential difference across the three capacitors then,, V1 + V2 + V3 = 120 V. If q is the charge on each capacitor., , www.Padasalai.Net, q5, , i, , 7, , i, , 7, , i, , 7, , i, , 7, , q5, , iS, , iS, , = 120, , 9= 120, , 9= 120, , a, , q = 360, , <@#<= Ç, , The potential across each capacitor, i.e. V =, V=, , i, , a, , T, , = 40 V, , C = 3 pF, each of 40 V., , 33, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 36 : www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 17. Three charges +1μC, +3μC and –5μ C are kept at the vertices of a equilateral, triangle of sides 60 cm. Find the electrostatic potential energy of the system of, charges. [M-16] [compulsory], Given data: q1=1 × 10-6C; q2=+3 × 10-6C; q3=-5 × 10-6C., Sol: The triangle is equilateral triangle, AB = BC = CA = r = 60 cm = 0.60 m, , The potential energy of the system of charges,, U=, , 5, , 23, , = 9 X 10, =, =, , 6, 9, , 7, 1, , 10, , 5, , 6, , 86, , S, , 3, , 7, , S, , 86, , 10, , 0.60, , TÔ, , 10#, , Õ3, , ., , TÔ, , 6, , 9, , 7, , 3, , – 15, , 10, , 86, , 86, , 85, , 10, , 0.60, , 10#, , 85, , 7, , 85, , 86, , 10, , 1, , 0.60, , 86, , 10, , 9, , 10# ×, , (-17 10#, , ., , www.Padasalai.Net, =, , # Ea, , S, , ., , = - 255 10#a, , U = - 0.255 J, , Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], Contact for subject doubt, Email :
[email protected], Phone No: 9952223467, , JEYAM TUITION CENTRE, [PARAMAKUDI], Contact for tuition, Email :
[email protected], Phone No: 9944888512, 35, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, FIVE MARKS:(1-question: Q.No: 51), , 1. Write the properties of electric lines of forces. (M - 07, O - 07, M - 08, M - 10,, , M -11, O - 11, J - 12, M -13,M-15 ), 2. Define electric potential at a point. Obtain an expression for electric potential due, , to a point charges. (M – 09, O-16 ), 3. Derive an expression for the torque acting on the electric dipole placed in a, , uniform electric field. (O - 10, O -12,O-14,J- 16 ), 4. What is electrostatic potential energy of a system of two point charges? Deduce, , an expression for it. (O - 09), 5. What is capacitor ? Explain the principle of a capacitor.(J-14), 6. Deduce an expression for the capacitance of the parallel plate capacitor.(J- 10,, , J-15,M-17), 7. Explain the effect of introducing a dielectric slab between the plates of parallel, , plate capacitor. (J - 13), 8. Deduce an expression for the effective capacitance of capacitors of capacitances, , C1, C2 and C3 connected in series.(M – 14), 9. Prove that the energy stored in a parallel plate capacitors E=, , i, , (M-12, J-15,O-15), , www.Padasalai.Net, TEN MARKS: (1-question: Q.No: 63), , 1. What is an electric dipole? Derive an expression for the electric field due to an, , electric dipole at a point on its axial line. (M - 06, J - 06, M - 09, J - 10, O -10,, M – 11,O-13,J-14,M-16), , 2. Derive an expression for electric field due to an electric dipole at a point along, , the equatorial line. (M - 07, J – 09,J-15), 3. Derive an expression for electric potential at a point due to an electric dipole., , Discus the special cases.(O-06,M-08,J-08,M-10,O-11, M – 13, M-15), 4. State Gauss’s law. Using this, derive an expression for electric field due to, , infinitely long straight charged wire with uniform charge density . (J - 11, M - 12,, J – 13, O – 16 ), 5. Explain the principle of a capacitor. Deduce an expression for the capacitance of, , a parallel plate capacitor. (J -12,O-14), 6. Derive an expression for the capacitance of a parallel plate capacitor with a dielectric, , medium between the plates. Write the application of capacitor. ( J – 16 ), 7. Deduce an expression for equivalent capacitance of capacitors connected, , (i) parallel (ii) series. (J - 07, O - 07), 8. State the Principle and explain construction and working of Van de Graaff, , generator. What is its use? (O - 08, O - 09, O – 12, M -14,O-15,M-17), , 36, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , Only for slow bloomers, ( 5 & 10 mark answers ), 1. Properties of electric lines of forces., i) Start from positive charge and terminate at negative charge., ii) Never intersect., iii) The tangent to a line of force at any point gives the direction of the electric field, iv) The number of lines per unit area is proportional to the magnitude of E., v) Each unit positive charge gives rise to, , 3, , lines of force in free space., , 2. Torque acting on the electric dipole placed in a uniform electric field:, A dipole AB of dipole moment p placed at an angle θ, in an uniform electric field E., The charge +q experiences a force qE and the charge, –q experiences an equal force in the opposite, direction. Thus the net force is zero., % = One of the forces x perpendicular distance between the forces, = F x 2d sin θ, = qE x 2d sin θ = pE sin θ, (∵ q × 2d = P), ^_ t^_, In vector notation, %_ = ß, , www.Padasalai.Net, 3. Electric potential due to a point charges.:, , Let +q charge situated at O. P is a point at a distance r from +q. Consider two points A, and B at distances x and x + dx from the point O, dV = −E dx, E=, dv = V = 8 râ, , à, , à, , dx, , à, , dx =, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 7. Applications of superconductors:, i), ii), iii), iv), v), , 1., , Superconductors form the basis of energy saving power systems., Used to levitate trains above its rails., Used to launch satellites into orbits directly from the earth without the use of, rockets, Used for transmission lines., Used as memory or storage elements in computers., , Electric dipole. Electric field at a point on its axial line, Dipole : Two equal and opposite charges separated by a very small distance constitute, an electric dipole., , The charges –q and +q at A and B constitute an electric dipole. P is a point at a, distance, r from the centre of the dipole along the axial line, Electric field at P due to + q, , www.Padasalai.Net, Electric field at P due to – q, , The resultant electric field at P is, E = E1 + (-E2), , d << r, , p = q x 2d, E acts in the direction of dipole moment., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 4., , 2017-2018, , Gauss’s law. Electric field due to infinitely long straight charged wire, Law : The total flux of the electric field E over any closed surface is equal to, , times, , the net charge enclosed by the surface., Consider an uniformly charged wire of infinite length, having a constant linear charge density λ., Let P be a point at a distance r from the wire and E be, the electric field at the point P., A cylinder of length l, radius r, closed at each end by, plane caps normal to the axis., The magnitude of the electric field will be the same at, all points and directed radially outward., The electric flux (ɸ) through curved surface = ∮ E ds cos θ, , ɸ = ∮ E ds, , [θ = 0; cos θ = 1], , ɸ = E (2U b), Electric flux through the plane caps = 0, , www.Padasalai.Net, q = λb, , ɸ (, By Gauss’s law,, , E (2U b ) =, E =, , ªd, , ª, , The direction of E is outward, if line charge is positive and inward, if the line charge is, negative., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 44 : www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 5., , 2017-2018, , Van de Graaff generator., It produces large potential difference of the order of 107 V., Principle : Electrostatic induction and action of points., Diagram :, Construction :, A hollow metallic sphere A is mounted on insulating, pillars. A belt made of silk moves over the pulleys B ,C., The pulley C is driven continuously by an electric, motor., Two comb shaped conductors D and E are mounted, near the pulleys., The comb D is maintained at a positive potential of the, order of 104 volt by a power supply., Comb E is connected to the inner side of the hollow, metal sphere., , Working : Because of the high electric field near the comb D, the air gets ionised due to, action of points, The positive charges stick to the belt, moves up and reaches near the comb, E., As a result of electrostatic induction, the comb E acquires negative charge and the sphere, acquires positive charge and distributed on the outer surface of the sphere. The descending, belt will be left uncharged. The machine, continuously transfers the positive charge to the, sphere, , www.Padasalai.Net, Precaution : The leakage of charge from the sphere can be reduced by, enclosing it in a gas filled steel chamber at a very high pressure., , Uses : The high voltage can be used to accelerate positive ions (protons, deuterons)., , Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], Contact for subject doubt, Email :
[email protected], Phone No: 9952223467, , JEYAM TUITION CENTRE, [PARAMAKUDI], Contact for tuition, Email :
[email protected], Phone No: 9944888512, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 2 - CURRENT ELECTRICITY, PREPARED BY, J.SHANMUGAVELU M.Sc, B.Ed. [P.G. Assist in Physics], , ONE MARK QUESTIONS (1-question), BOOK BACK ONE MARKS:, 1. A charge of 60 C passes through an electric lamp in 2 minutes. Then the current in the, , lamp is, a) 30 A, , I=, , Sol :, , c) 0.5 A, , b) 1 A, , =, , ë, , =, , d) 5 A, , = 0.5 A, , 2. The material through which electric charge can flow easily is, , a) quartz, , b) mica, , c) germanium, , d) copper, , 3. The current flowing in a conductor is proportional to, , a) drift velocity, c) 1/no of electrons, Sol :I = neAVd ; I ∝ Vd, , b) 1/ area of cross section, d) square of area of cross section., , www.Padasalai.Net, 4. A toaster operating at 240V has a resistance of 120 Ω. The power is, , a) 400 W, , P=, , Sol :, , =, , c) 480 W, , b) 2 W, , f, , d) 240 W, , ì, , = 480 W, , 5. If the length of a copper wire has a certain resistance R, then on doubling the length its, , specific resistance, a) will be doubled, b) will become 1/4th, c) will become 4 times, d) will remain the same., Hinds: The length of the wire is increases the resistance is also increases. But specific resistance, of the wire will remain the same. Because specific resistance of the particular metal is, constant., 6. When two 2Ω resistances are in parallel, the effective resistance is, , a) 2 Ω, Sol:, , c) 1 Ω, , b) 4 Ω, , ìí, , d) 0.5 Ω, , (ì 7ì, î î, , ⇒ îs ( î 17î2, 1, 2, , ( 22722 ( 44 =1 Ω, 7. In the case of insulators, as the temperature decreases, resistivity, , a) decreases, , 44, , b) increases, , J.SHANMUGAVELU, , c) remains constant, , [P.G. T. in Physics], , d) becomes zero, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 12. The color code of carbon resistor is red-red-black. The resistance of the resistor, , b) 22 Ω, , a) 2.2 Ω, Sol : 22, , c) 220 Ω, , d) 2.2 k Ω, , 0, , 10 = 22Ω, , 13. The brown ring at one end of a carbon resistor indicates a tolerance of, , a) 1%, b) 2%, c) 5%, d) 10%, Hints: The tolerance of silver, gold, red and brown rings is 10%, 5%, 2% and 1% respectively., If there is no coloured ring at this end, the tolerance is 20%., , 14. The unit of conductivity is, , a) mho, , b) ohm, , 15. The transition temperature of mercury is, , a) 4.2 0C, , b) 4.2 K, , c) ohm-m, , d) mho-m-1, , c) 2.40C, , d) 2.4 K, , 16. The relation between current and drift velocity is, , b) I = nA¢ò e, , a) I = nAVd/e, , c) I = neVd/A, , d) I = nAVd, , 17. When the diameter of a conductor is doubled, its resistance, , a) decreases twice, b) decreases four times, c)decreases sixteen times, d) increases four times, Hints: when the diameter is doubled the radius is also doubled, Sol:, , R=, , ó°, , ;, , radius = 2r;, , ó°, , ó°, , www.Padasalai.Net, R1 =, , =, , ⇒ R1 =, , ô, , 18. The electrical resistivity of a thin copper wire and a thick copper rod are respectively, , ρ1Ω m and ρ2Ω m. Then, a) ρ1< ρ2, b) ρ1> ρ2, , ó, , c) ρ1= ρ2, , Hints: The resistivity is constant for a particular (copper) material, , d) ó, , =∝, , 19. The resistance of the filament of a 110 w, 220 V electric bulb is, , a) 440 Ω, Sol: I=, , +, , f, , R=, , f, õ, , b) 220 Ω, , =, =, , c) 484 Ω, , d) 848 Ω, , = 0.5A;, , .E, , = 440 Ω, , 20. A cell of emf 2.2 V sends a current of 0.2 A through a resistance of 10 Ω. The internal, , resistance of the cell is, a) 0.1 Ω, Sol : Ir = E – V, , ⇒r=, , ö# ¯, õ, , c) 2 Ω, , d) 1.33 Ω, , [P.G. T. in Physics], , Ph. No:9952223467, , ……..… 1 ;, , V = IR ( 0.2, Sub V in equ (1), , 46, , b) 1 Ω, , 10 = 2 V, , r=, , . #, ., , =, , J.SHANMUGAVELU, , ., ., , = 1Ω, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 25. The effective resistance between points A and B in the given network is :, , a) 2.5 Ω, b) 10 Ω, c) 0.4 Ω, Sol : R1 and R2 is connected in series combination, , d) 11 Ω, , Rs = R1 + R2 = 2 + 3 = 5Ω, R3 is parallel to the Rs, ìí, , (, , ìS, , 7, , î î, , ìÎ, 5, , 5, , 25, , ⇒ îs ( î 37î ( 575 ( 10 = 2.5 Ω, 3, 26. The colour code of a carbon resistor is, Brown, Black, Brown and Red. The value of the, resistor is:, a) 10 Ω ± 5%, b) 1 kΩ ± 2%, c) 100 Ω ± 2%, d) 10 Ω ± 2%, , Sol : The first yellow ring corresponds to 1., The next violet ring corresponds to 0., The third orange ring corresponds to 101., The silver ring represents 2% tolerance., The total resistance is 10 × 101 ± 2% = 100 Ω ± 2%, , www.Padasalai.Net, Choose the correct answer from the options given, , 27. The external energy necessary to drive the free electrons in a definite direction is called, , a) current, , b) resistance, , c) emf, , d) power, , 28. If a charge q coulomb passes through any cross section of a conductor in time ‘t’ second,, , then the current is given by, a) I=qt, b) I=t/q, , c) I=q/t, , d) I=1/qt, , 29. Acceleration experienced by an electron of mass ‘m’ and charge ‘e’ in an electric field ‘E’, , is a =, , a) p, , b), , ·`, , 30. Expression for mobility is, œ (, , a), , š, , p, , b), , š, , p, , 31. Drift velocity of electrons is proportional to, , a) electric field intensity b) charge of protons, , 32. Drift velocity of electrons is of the order of, , a) 0.2 cm s-1, , b) 0.1 cm s-1, , 33. The unit of current density is, , a) A m-1, , b) A m2, , c), , c), , š, , p, ·, , d), , d), , šp, , p, , c) area of the conductor d) none of these, c) 0.1 m s-1, , d) 1 cm s-1, , c) A m-2, , d) A m, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 34. Expression for electric resistance R is,, , a), , (, , º ·=, , m, , b) î ( p¡, , 35. Reciprocal of resistance is, , a) resistivity, , b) conductivity, , 36. The unit of conductance is, , b) mho, , a) ohm, , c) î (, , m¡, , p f, , d) î ( m¡, , p, , c) inductance, , d) conductance, , c) mho-1, , d) mho m-1, , c) mho m, , d) ohm m, , 37. The unit of conductivity is, , b) mho m-1, , a) ohm m-1, , 38. The conductivity of a material is obtained by the formula, , a) • (, , ì¡, , b) • (, , d, , d¡, , c) Ž (, , ì, , d) • (, , 39. Discontinuous change in specific heat of a material occours at, , a) transition temperature, , b) high temperature, , 40. Resistivity of mercury is zero at, , b) 4.20C, , a) 2.4 K, , c) 0 K, , ð¡, ì, , d) room temperature, d) 2.40C, , c) 4.2 K, , 41. The tolerance of silver, gold, red and brown rings in carbon resistors are respectively., , a)1%, 2%, 5% and 10%, c)10%, 5%, 1% and 2%, , b)10%, 2%, 5%, and 1%, d)10%, 5%, 2% and 1%, , www.Padasalai.Net, 42. The colour code for 1 in carbon resistors is, , b) Brown, , a) Black, , c) Silver, , d) Red, , 43. In a carbon resistor the third coloured ring indicates, , a) first significant figure, c) powers of 10 to be multiplied, , b) tolerance, d) second significant figure, , 44. The effective resistance of two resistances (R1, R2) connected in parallel is, , a), , (, , < =, <, , =, , b) îè (, , ì Fì, ì, , c) îè ( ì 8 ì, , d) Rp=R1+R2, , 45. If R0 and Rt are the resistances of a conductor at 00C and t0C respectively, then the, , temperature coefficient of resistance is, ì ë, ì ì, a) ™ ( ì, b) ™ ( ì ë, , c), , (, , ¸, , @¸, , d) ™ (, , @, , ì, , ì, , ì ë, , 46. If the resistance of a material increases with increase in temperature then its temperature, , coefficient of resistance is, a) zero, b) negative, , c) positive, , d) none of these, , 47. The temperature coefficient of resistance of insulators and semiconductors is, , a) positive, , b) negative, , 48. The temperature coefficient of manganin is, , a) infinity, , 49, , b) high, , J.SHANMUGAVELU, , c) low, , d) zero, , c) zero, , [P.G. T. in Physics], , d) low, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 49. The condition for bridge balance in wheatstone’s bridge is, +, , a) (, , ì, , b) PR=QS, , c) PS=QR, , d) PQ=RS, , 50. The positive ions which are mostly formed from metals or hydrogen are called, , b) cations, , a) anions, , c) positive particles, , d) atoms, , c) Cu, C, , d) Fe, Zn, , 51. The electrodes used in voltaic cell are, , a) Cu,Zn, , b) Cu, Fe, , 52. The potential difference between the two electrodes of voltaic cell is, , a) 1.5 V, , b) 1.8 V, , 53. The emf of Leclanche cell is about, , b) 1.5 V, , a) 1.08 V, , 54. Leclanche cell can supply a current of, , a) 0.25 A, , b) 0.5 A, , 55. Leclanche cell is used to supply, , a) Current of the order 2.5 A, c) Very high current, , d) 1.58 V, , c) 1.05 V, , d) 1.1 V, , c)2.5 A, , d) 5.2 A, , b) high current, d) intermittent current, , 56. Daniel cell produces an emf of, , b) 1.08 V, , a) 1.5 V, , c) 1.08 V, , c) 1.8 V, , d) 2.0 V, , www.Padasalai.Net, 57. Electrolyte used in lead-acid accumulator is, , a) lead acid, , b) Hcl, , c) dil. H2SO4, , d) HNO3, , 58. The emf of the lead-acid accumulator under freshly charged and discharged conditions, , are, a) 2.2V, 2V, , b) 2V, 2.2V, , c) 2.2V, 1.35V, , d) 2V, 1.35V, , 59. The values of emf developed and specific gravity of a freshly charged lead-Acid, , accumulator respectively are, a) 1.9V and 1.35, b) 2.2V and 1.28, , c) 1.35V and 1.9, , d) 2.2V and 1.9, , 60. A charge of 60C passes through an electric lamp in 2 minutes. Then the current in the, , lamp is, a) 30 A, , c) 0.5 A, , b) 1 A, , d) 5 A, , 61. If a charge of 1 C passes through an electrical equipment in 10 s, then the current flowing, , through it is, a) 0.5 A, , c) 0.1 A, , b) 1 A, , d) 10 A, , 62. The markings at the bottom of a tape recorder are as follows. 9V, 450 mA. The net, , resistance of the tape recorder is, b) 200 Ω, a) 20 Ω, , c) 1/20 Ω, , d) 2 Ω, , 63. An electrical instrument of resistance 30 Ω is operated at 240 V. The power is, , a) 240 W, , 50, , b) 1290 W, , J.SHANMUGAVELU, , c) 920 W, , [P.G. T. in Physics], , d) 1920 W, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 64. If two resistors of resistances 200 Ω and 0.1 kΩ are connected in series then the effective, , resistance of the system is, a) 200.1 Ω, b) 300 Ω, , c) 201 kΩ, , d) 2.1 kΩ, , 65. The resistance of a wire of 1 m length and 0.034 mm2 area cross section having a specific, , resistance of 1.7X10-8 Ω is, b) 5.0 Ω, a) 0.5 Ω, , c) 2 Ω, , d) 0.05 Ω, , 66. A wire of resistance 0.1 Ω having a length of 30 m has a specific resistance of, , 2.7X10-8Ω . The area of cross-section of the wire is, a) 0.81X10-6m2, b) 8.1X10-5m2, c) 8.1X10-4m2, , d) 8.1X10-6m2, , 67. The resistance of a conductor of 10 m long and 0.1 mm2 area is 1.7 Ω. The specific, , resistance of the material of the conductor is, a) 2.7X108Ω, b) 1.7X10-8Ω, c) 17X10-8Ωm, , d) 1.7X10-6Ωm, , 68. The number of electrons flowing per second through a conductor, when a current of 3.2 A, , flows through it is, a) 2X1019, , b) 3X1018, , c) 6.25X1018, , d) 6.25X1019, , 69. A 1.15 kW, 230 V water heater can draw a current of, , a) 0.2 A, , c) 5 A, , b) 2 A, , d) 0.5 A, , 70. The ratio of the diameters of two copper wires of lengths 2m and 8 m having equal, , resistance is, a) 2:1, , www.Padasalai.Net, b) 2:8, , c) 1:4, , d) 1:2, , 71. A current of 0.3 A from a cell of emf 1.5 V is passed through a resistance of 4Ω. The, , internal resistance of the cell is, a) 0.1Ω, b) 1Ω, , c) 10Ω, , d) 0.01 Ω, , 72. If charge per unit volume of a conductor is 600 C and the current density is 1.2 Am-2,then, , the drift velocity of the electron is, b) 7.2X10-3 m/s, a) 0.2X10-2 m/s, , c) 200 m/s, , d) 5X10-3 m/s, , 73. Three resistances of values 10 Ω, = Ω and 3 Ω are connected to form the sides of a triangle, , AB, BC and CA respectively. The effective resistance between A and B is, a) 3.33 Ω, b) 2.33 Ω, c) 3.5 Ω, d) 3.9 Ω, , 74. A cell of emf 9 V and internal resistance 1 Ω is connected to an external resistance of 8 Ω,, , the potential difference across the cell is, a) 9 V, b) 1 V, , c) 6 V, , d) 8 V, , 75. In Wheatstone’s bridge, under bridge balance condition, the four resistances of the four, , arms in cyclic order are, a) 5, 10,4,8, b) 5,10,8,4, , c) 5,8,10,4, , d) 5,4,8,10, , 76. Two resistances 6 Ω ¿ÀÁ >Ω ¿re connected in parallel and the combination is connected in, , series with a resistance of 2.6 Ω ¿ÀÁ ¿À ¿, #À "þ #rcuit is, a) 5/2 A, b) 5/4 A, , 51, , J.SHANMUGAVELU, , ü L¿ K K þü = Ω. !"þÀ "þ, c) 2/5 A, , [P.G. T. in Physics], , þÀ, , d) 5A, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 77. In a metre bridge, with a standard resistance of 5 ohm in the right gap, the ratio of, , balancing lengths is 3:2. The value of the other resistance is, a) 10/3 Ω, b) 10/9 Ω, c) 15/2 Ω, , d) 3/5 Ω, , 78. The balancing lengths of two cells are 250 cm and 750 cm respectively, in a potentiometer, , experiment. If the emf of the first cell is 2 V, the emf of the second cell is, a) 6 V, b) 4 V, c) 2/3 V, d) 3/2 V, -6, , 2, , 79. A copper wire of 10 m area of cross section carries a current of 1 A. The current density, , is, a) 2X106 A/m2, , b) 0.1X106 A/m2, , c) 1X10-6 A/m2, , d) 1X106 A/m2, , 80. A 750 W power iron box is used for 4 hours. If the cost per unit is 75 paise, the total, , expense is, a) Rs. 22.50, , b) Rs. 5.25, , c) Rs.2.25, , d) Rs.3.00, , 81. The value of a carbon resistor with the colour code of yellow, violet and orange is, , a) 37 k Ω, , b) 4.7 k Ω, , c) 47 k Ω, , d) 3.7 k Ω, , 82. The value of a carbon resistor is 33 k Ω. Then the colour code is, , a)Yellow, Orange, Red, c)Red, Blue, Orange, , 83. If 6.25X10, , 18, , b)Brown, Yellow, Orange, d)Orange, Orange, Orange, , electrons flow through a given cross section in unit time, then the current is, b) 2 A, c) 0.1 A, d) 0.2 A, , a) 1 A, , www.Padasalai.Net, 84. An incandescent lamp is operated at 240 V and the current is 0.5 A, then the resistance of, , the lamp is, a) 840 Ω, , b) 480 Ω, , c) 240 Ω, , d) 380 Ω, , 85. The resistance of nichrome wire at 00C is 10 Ω. If its temperature coefficient of resistance, , is 0.004/0C, then its resistance at 1000C is, a) 4 Ω, b) 12 Ω, , c) 14 Ω, , d) 18 Ω, , 86. A cell has a potential difference of 6 V in an open circuit, but it falls to 4V when a current, , of 2 A is drawn from it. Then the internal resistance of the cell is, a) 1 Ω, b) 10 Ω, c) 0.1 Ω, , d) 2 Ω, , 87. In a Wheatston’s bridge, P =1000 Ω, Q=10,000 Ω and R =20 Ω. If the galvano-meter, , shows zero deflection, the value of S is, a) 20 Ω, b) 200 Ω, , c) 2 Ω, , d) 2000 Ω, , 88. An electric iron of resistance 80 Ω is operated at 200 ¢ for two hours. The electric energy, , consumed is, a) 1 W h, , 52, , b) 10 kWh, , J.SHANMUGAVELU, , c) 1 kWh, , [P.G. T. in Physics], , d) 0.1 kWh, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , ‘3’ MARK QUESTIONS, , 2017-2018, , ( 3- Questions: Q.No: 33, 34, 35), , PUBLIC ‘3’ MARKS:, 1. Define: Drift velocity. Give its unit (M – 07,O – 08, J - 09, O– 09, M–10, M -11,O –, , 11,J-15 ), Drift velocity is defined as the velocity with which free electrons get drifted, towards the positive terminal, when an electric field is applied., Its unit is ms-1, , vd =, , š, , p, , τ = µE, , 2. Define: mobility. Give its unit. ( O – 06 , M – 08, M – 09,O-15,M-16 ), , The mobility is defined as the drift velocity acquired per unit electric field., The unit of mobility m2v-1s-1., , µ=, , p, , 3. Distinguish between drift velocity and mobility?, , www.Padasalai.Net, Drift velocity, , 1. Drift velocity is defined as the, , velocity with which free electrons, get drifted towards the positive, terminal, when an electric field is, applied., , 2., 3., , vd =, , š, , p, , τ = µE, , Unit is ms, , -1, , Mobility, Mobility is defined as the drift, velocity acquired per unit electric, field., , µ=, , p, , Unit is m2v-1s-1., , 4. Define: current density ? Give its unit. ( O-16), , The quantity of charge passing per unit time through unit area, taken, perpendicular to the direction of flow of charge at that point is called current, density, J=, , Ü, , Its unit is A m-2, 5. State Ohm’s law. ( M – 06 , O – 07, O – 09, M -10,M-13,O-14,J-16,M-17 ), , At a constant temperature, the steady current flowing through a conductor is, directly proportional to the potential difference between the two ends of the, conductor., ( i.e.) I ∝ V or V = IR, , 53, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 6. Define electrical resistivity of a material?(O- 13 , M-15), , The electrical resistivity of a material is defined as the resistance offered to, current flow by a conductor of unit length having unit area of cross section., The unit of ρ is ohm – m (Ω m). It is a constant for a particular material., , ρ=, , ì¡, d, , 7. What is superconductivity?(O-13), , The ability of certain metals, their compounds and alloys to conduct, electricity with zero resistance at very low temperatures is called, superconductivity. The materials which exhibit this property are called, superconductors., 8. Define critical temperature or Define transition temperature in, , superconductivity . ( M – 12,J-16 ), , The temperature at which electrical resistivity of the material suddenly drops, to zero and the material changes from normal conductor to a superconductor is, called the transition temperature or critical temperature TC, 9. What are the changes observed at the superconducting transition, , temperature?(J–10), , www.Padasalai.Net, i) The electrical resistivity drops to zero., ii) The conductivity becomes infinity, iii) The magnetic flux lines are excluded from the material., , 10. Give any three applications of the superconductors. ( J – 07,O – 07,J – 06 ,O –, , 06, O – 07, M-15), , i) Superconducting magnetic propulsion systems may be used to launch, satellites into orbits directly from the earth without the use of rockets., ii) Since the current in a superconducting wire can flow without any change in, magnitude, it can be used for transmission lines., iii) Superconductors can be used as memory or storage elements in computers., 11. Define: Temperature coefficient of resistance. ( J – 08, M – 11,J-14,M-16 ), , The ratio of increase in resistance per degree rise in temperature to its, resistance at 0o C is called as temperature coefficient of resistance ., Its unit is per oC, , α=, , ì #ì, ì ë, , 12. State Kirchoff’s first (current) law in electricity. ( J – 06 , M – 08 ), , The algebraic sum of the currents meeting at any junction in a circuit is, zero. This law is a consequence of conservation of charges., , 54, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 13. State Kirchoff’s second (voltage) law in electricity. ( M – 07,J – 06,M – 08,M –, , 09,J– 11 ), , The algebraic sum of the products of resistance and current in each part of, any closed circuit is equal to the algebraic sum of the emf’s in that closed, circuit. This law is a consequence of conservation of energy., 14. State Kirchoff’s, , (i) current law, , (ii) Voltage law (O-15), , current law:, The algebraic sum of the currents meeting at any junction in a circuit is, zero. This law is a consequence of conservation of charges., Voltage law:, The algebraic sum of the products of resistance and current in each part of, any closed circuit is equal to the algebraic sum of the emf’s in that closed, circuit. This law is a consequence of conservation of energy., , 15. Compare the emf and the potential difference. (J– 07,O– 08,J– 11,M-13, M-15), , www.Padasalai.Net, Emf, 1. The difference of potentials between, the two terminals of a cell in an, open circuit is called the, electromotive force (emf) of a cell., 2. The emf is independent of external, resistance of the circuit,, , Potential difference, The difference in potentials between, any two points in a closed circuit, is called potential difference., , Potential difference is proportional to, the resistance between any two, points., , 16. Distinguish between electric power and electric energy. ( J – 08 , J – 09,O-13,J-, , 14,O-14 ), , Electric power, 1. Electric power is defined as the, , rate of doing electric work., 2. Electric power is the product of, , Electric energy, Electric energy is defined as the, capacity to do work, Electrical Energy = power x time, E =P t, , potential difference and current, strength., $-6÷ ³-², ¯±, =, ( VI, Power =, ±%k, ±, 3. Unit: watt, Its unit is joule., , 55, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 57 : www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 17. State Faraday’s laws of electrolysis. ( M – 06, J – 06, J – 10,O -10, J-15,M-17 ), , First Law : The mass of a substance liberated at an electrode is directly, proportional to the charge passing through the electrolyte., or, m = zIt, m∝q, Second Law :The mass of a substance liberated at an electrode by a given, amount of charge is proportional to the chemical equivalent of the substance., m∝E, 18. Give any three uses of secondary cells. ( O – 08, O – 11,M-12 ), , i) The secondary cells are rechargeable., ii) The chemical reactions are reversible., iii) They can deliver a high current if required., iv) They are used in all automobiles like cars, two wheelers, trucks etc., 19. Distinguish between primary cell and secondary cell ( O - 16 ), , 1., 2., 3., , Primary cell, These cells cannot be recharged, electrically., Chemical actions are irreversible, The active materials can not be, reproduced, Daniel Cell and Leclanche cell, , Secondary cell, they are rechargeable., chemical reactions are reversible., The active materials can be reproduced, , www.Padasalai.Net, 4., , lead acid accumulator and alkali, accumulator., , Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], Contact for subject doubt, Email :
[email protected], Phone No: 9952223467, , JEYAM TUITION CENTRE, [PARAMAKUDI], Contact for tuition, Email :
[email protected], Phone No: 9944888512, , 56, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , OTHER IMPORTANT ‘3’ MARKS:, 1. What are the salient features of secondary cells?, i) They are rechargeable., ii) The chemical reactions that take place in secondary cells are reversible., iii) They have very low internal resistance., iv) The active materials that are used up when the cell delivers current can be, reproduced by passing current through the cell in opposite direction., Examples: lead acid accumulator and alkali accumulator., 2. Why copper wire is not suitable for a potentiometer?, Copper has low resistivity and high temperature coefficient of resistance., So copper is not suitable for potentiometer., 3. Define electric current, The current is defined as the rate of flow of charges across any cross, sectional area of a conductor., The current I =, , ë, , The unit of electric current is ampere, , www.Padasalai.Net, 4. What is resistance of a conductor?, , Resistance of a conductor is defined as the ratio of potential difference across, the conductor to the current flowing through it., The unit of resistance is ohm (Ω)., R=, , f, õ, , 5. What is carbon resistor ?, Carbon resistor consists of a ceramic core, on which a thin layer of, crystalline carbon is deposited. These resistors are cheaper, stable and small in, size., 6. A carbon resistor has yellow, violet and orange coloured rings and a silver, at the end of It. Find the value of resistor?, The first yellow ring corresponds to 4., The next violet ring corresponds to 7., The third orange ring corresponds to 103., The silver ring represents 10% tolerance., The total resistance is 47 × 103 ± 10% i.e. 47 k Ω, 10%., , 57, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 7. Define the effective resistance (Rs, Rp) of a number of resistances, connected (a) in series and (b) in parallel., (a) in series: The equivalent resistance of a number of resistors in series, connection is equal to the sum of the resistance of individual resistors., Rs = R1 + R2 +……….+Rn, , (b) in parallel: The number of resistors are connected in parallel, the sum of the, reciprocal of the resistance of the individual resistors is equal to the reciprocal of, the effective resistance of the combination., ìí, , (, , ì, , 7, , ì, , …..+, , ìl, , 8. What is meant by internal resistance of a cell?, During the process of flow of current inside the cell, a resistance is offered, to current flow by the electrolyte of the cell. This is termed as the internal, resistance of the cell., 9. What is the sign convention followed in the application of kirchoff’s second, law., In the application of Kirchoff’s second law, we follow that the current in, clockwise direction is taken as positive and the current in anticlockwise direction, is taken as negative, , www.Padasalai.Net, 10. Calculate the current flowing in the circuit given below., , Taking the current in the clockwise direction along ABCDA as positive, 10 I + 0.5 I + 5 I + 0.5 I + 8 Ι + 0.5 I + 5 I + 0.5 Ι + 10 I = 50 – 70 – 30 + 40, I ( 10 + 0.5 + 5 + 0.5 + 8 + 0.5 + 5 + 0.5 + 10) = −10, 40 I = −10, , ∴I=, , 8, , = -0.25 A, , The negative sign indicates that the current flows in the anticlockwise, direction., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 11. What is the principle of potentiometer?, The emf of the cell is directly proportional to its balancing length., , Eαl, 12. Define electrochemical equivalent of a substance ?, The electrochemical equivalent of a substance is defined as the mass of, substance liberated in electrolysis when one coulomb charge is passed through, the electrolyte. Its unit is kg C–1 (or), z=, , p, õë, , (, , p, , ÷ø, , 4kù 6 . ,-²³, , ., , 13. Write short notes on Wattmeter, A wattmeter is an instrument used to measure electrical power consumed. i.e, energy absorbed in unit time by a circuit. The wattmeter consists of a movable, coil arranged between a pair of fixed coils in the form of a solenoid. A pointer, is attached to the movable coil. The free end of the pointer moves over a, circular scale. When current flows through the coils, the deflection of the pointer, is directly proportional to the power., , www.Padasalai.Net, 14. Why automobile batteries have low internal resistance?, , In automobile batteries like, lead acid accumulator, the electrodes are, separated by suitable insulating materials and assembled in a way to give low, internal resistance. Since the cell has low internal resistance and hence it can, deliver high current, , 59, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PUBLIC ‘3’ MARK PROBLEMS:, , 1. If 6.25 1018 electrons flow through a given cross - section of a, conductor in unit time, find the current. ( given charge of electron, is 1.6 10-19) [M-10,J-11], , Given data:, , Sol: Current I =, =, =, , 10# T X, , e = 1.6, , 1018 ;, , n = 6.25, ë, m, , [ since q = ne ), , ë, , ', , . E, , ., , I =1A, 2. A manganin wire of length 2 m has a diameter of 0.4 mm with a, resistance of 70Ω. Find the resistivity of the material. [J-06,M-13], , Given data: b = 2 m;, , ä, , r= =, ì¡, , ì, , S, , ., , = 0.2 10#a ;, , R= 70Ω; ρ = ?, , www.Padasalai.Net, sol:, , ρ=, , =, , d, , Ï, , =, , a., , = 4.396, , ρ = 4.396, , d, , ., , S, , ., , S, , 10 -6, , 10 -6Ωm, , 3. The resistance of nichrome wire at 00C is 10Ω. If the temperature, coefficient of resistance is 0.004/0C, find its resistance at boiling point, of water. Comment on the result. [J-07,O-07,M-08,O-10,11,O-12,O-13,J15], , Given data: Ro = 10Ω;, , ™ = 0.004/ 0C;, , t = 1000C; Rt = ?, , Sol: Resistance at boiling point of water Rt = Ro (1+ ™t), = 10 [1 + (0.004 × 100)], , Rt= 14 Ω ., Result : As temperature increases the resistance of wire also increases., , 60, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 8. Find the magnitude and direction of the current in the following circuit, (M–11), , Sol:, , The current 5I + 10 I + 5 I = 10 + 20, 20 I = 30, , I = 1.5 A, Current flows along the path ABCD, 9. In the following circuit, calculate the current through the circuit., Mention its direction. [O-06,J-10,], , www.Padasalai.Net, Sol:, , The current 7 I + 3 I + 5I + 5 I = 10 + 8 – 2, 20 I = 16, , I = 0.8 A, Current flows along the path ABCD., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 15. In the circuit find the magnitude and direction of the current . (O-14), , Sol:, , when the current flows in the clock wise direction, 5I + 10 I + 5 I = 10 - 20, 20 I = -10, I = -10/20, , I = - 0.5 A, The negative sign indicates the current flows in the anti clock wise, direction, 16. A cell has a potential difference of 6 V in an open circuit, but it falls to 4 V, when a current of 2 A is drawn from it. Find the internal resistance of the cell., (O-15), , Given data, Sol:, , E = 6V;, R=, , V= 4V;, , I = 2 A;, , r=?, , f, , www.Padasalai.Net, õ, , =, , =2 Ω, , r=§, , š#f, , ¨R, , #, , ¨2, , f, , =§, , r = 1Ω, 17. How much time 1020 electrons will take to flow through a point in a conductor, so that the current is 200 mA [e = 1.6 × 10−19 C]? ( M-17), Data: n= 1020 electrons ; I = 200 mA = 200 10#a A; t = ?, , Sol :, , I=, t=, , ë, m, , =, , m, , ë, , õ, , =, , ., , S, , t = 80 S, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PUBLIC ‘5’ MARK PROBLEMS:, , 18. What is the drift velocity of an electron in a copper conductor having area, 10 10-6 m2 and carrying a current of 2A. Assume that there are 10 1028, electrons/m3. [J-10] [compulsory], 10-6 m2;, , Given data: A = 10, Sol:, , I = 2A;, , n = 10, , 1028;, , Vd =?, , I = neAVd, Vd =, , ² Ü, , Vd =, =, , ', , ., ., , R, , \, , <@#Æ ms-1, , Vd= 1.25, , www.Padasalai.Net, 19. A copper wire 10-6 m2 area of cross section, carries a current of 2 A. If the, number of electrons per cubic metre is 8 1028, calculate the current density, and average drift velocity. (Given e = 1.6 10-19 C)[M-09], , Given data: A = 10-6 m2;, , I = 2A;, , n=8, , 1028;, , e = 1.6, , 10-19C, , Sol:, Current density J =, , Ü, , =, Average drift velocity, Vd =, S, , ., , R, , *, , ², , = 2 106 Am-2, , ', , =, , = 0.1562, , R, , ., , 10#a, , Vd = 15.62 ms-1, , 67, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 20. The resistance of a filed coil measures 50 Ω at 20 0C and 65 Ω at 70 0C. Find, the temperature coefficient of resistance. [J-13], Given data : At 20℃, R20 = 50 Ω;, At 70℃, R70 = 65 Ω; α = ?, Rt = R0 ( 1 + ™t ), , Sol :, , R20 = R0 (1 + ™ 20 ), 50 = R0 (1 + ™ 20 ), , ……….(1), , R70 = R0 (1 + ™ 70 ), 65 = R0 (1 + ™ 70 ), , ⇒, , E, , E, , =, , ……….(2), , FÏ ,, , ,, , F, , 65 + 1300 ™ = 50 + 3500 ™, 2200 ™ = 15, , = 0.0068/℃, , www.Padasalai.Net, 21. An iron box of 400 W power is used daily for 30 minutes. If the cost per unit is, 75 paise, find the weekly expense on using the iron box. [J-12], , Data : Power of an iron box P = 400 W, rate / unit = 75 p, consumption time t = 30 minutes / day, cost / week = ?, , Solution :, Energy consumed in 30 minutes = Power × time in hours, = 400 × ½, = 200 W h, Energy consumed in one week = 200 × 7, = 1400 Wh = 1.4 unit, Cost / week = Total units consumed × rate/ unit, = 1.4 × 0.75, = Rs.1.05, , 68, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 23. In a metre bridge, the balancing length for a 10 Ω resistance in left gap is 51.8, cm. Find the unknown resistance and specific resistance of a wire of length, 108 cm and radius 0.2 mm. [O-10, J-12] [compulsory], , 10# m; b ( 51.8 z ; b ( 48.2 z ;, , Given data: b ( 108, r = 0.2 10#a m;, Sol:, , ì, , à, , à, , d, , (, , d, 51.8 1082, , =, , 48.2 1082, ., , x = 10, , E ., , x = 9.305Ω, ρ=, , d, , à, , a., , S, , ., , ., , S, , T.a E, , www.Padasalai.Net, ρ=, =, , ., , R, , ., , ρ = 1.082, , 70, , J.SHANMUGAVELU, , 10-6 Ωm, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , FIVE MARKS: (2 –question: Q.No: 52, 53), 1. Define mobility. Derive the relation between drift velocity and the current., , (M-06,O-14), 2. What is drift velocity ? Derive the relation between drift velocity and the current., , ( O-16), 3. Write any five applications of superconductors., , (O-08,M-09,J-11,O-12,M-13,J-14,O-14,M-16), 4. If two or more resistors are connected in parallel, derive the expression for the, , effective resistance.(O-06,O-15), 5. Explain the variation of resistance with temperature using a graph. (J-12), 6. Explain the method to find the internal resistance of a cell using the voltmeter., , (J-08,O-09,J-11,M-13,J-13,J-15), 7. State and explain Kirchoff’s second law for electrical network. (J-07,O-16), 8. Obtain the condition for bridge balance in Wheatstone’s bridge., , (M-06,J-06,O-06,M-08,J-09,M-10, M-15,M-17), , www.Padasalai.Net, 9. Explain the principle of the potentiometer with a neat diagram., , (O-07,J-14,J-15,J - 16), , 10. How will you compare the emfs of the two given cells using Potentiometer., , (M-07,O-10,O-11,M-12,M-14,M-17), 11. State Faraday’s first law of electrolysis. How is it verified experimentally?, , (J-08,O-09), 12. State and verify Faraday’s second law of electrolysis through an experiment., , (J-06,M-08,M-11,O-13,M-16,O-16), 13. Explain the construction and the working of a Daniel cell., , (O-08,J-09,J-10,M-11, M - 14), 14. Explain the working of Leclanche cell. ( J-07,O-12,O-13), 15. Explain the working of lead acid accumulator. (O-07,O-15), , 73, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 4. Determination of internal resistance of a cell using voltmeter:, With key K open, the emf of cell E is found by connecting, a high resistance voltmeter across it. A small value of, resistance R is included in the external circuit and key K is, closed. The potential difference across R is equal to the, potential difference across cell (V)., The potential across R,, , V = IR …….1, , Due to internal resistance r of the cell, V is less than the emf of cell., Then V = E – Ir or Ir = E−V ……..2, Dividing (2) by (1), , r=§, , š#f, f, , ¨R, , 5. Wheatstone’s bridge:, , Applying current law to junction B,, I1 – Ig – I3 = 0, , junction D, , I2 + Ig – I4 = 0, Applying voltage law to ABDA, I1 P + IgG – I2 R = 0, , www.Padasalai.Net, Applying voltage law to ABCDA, I1P + I3Q – I4S – I2R = 0, , For balancing condition Ig = 0., , Û, , 6. Daniel cell:, , 3, , (, , ô, Ú, , It cannot supply steady current for a long time., Construction :, Anode → Copper, Cathode → Zinc, Electrolyte → copper sulphate solution ,, Diluted sulphuric acid, Vessel → Copper, , Working : The zinc rod reacting with dilute sulphuric acid produces Zn++ ions, and 2electrons and thus becomes negative., Zn++ ions pass through the porous pot and reacts with copper sulphate solution,, producing Cu++ ions ., The Cu++ ions deposit on the copper vessel ., When Daniel cell is connected in a circuit, the two electrons on the zinc rod pass, through the external circuit and neutralizing the ions., Current passes from Cu to Zn, in the external circuit . emf → 1.08V, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 7. Leclanche cell, , Construction :, Anode → Carbon rod, Cathode → Zinc rod, Electrolyte → Ammonium Chloride, solution, Vessel → Glass, Working : At the zinc rod, due to oxidation reaction Zn atom is converted in to, Zn++ ions and 2electrons. Zn++ions reacting with ammonium chloride produces, zinc chloride and ammonia gas., Zn++ + 2 NH4Cl → 2NH3 + ZnCl2 + 2 H+ + 2e–, The positive charge of hydrogen ion is transferred to carbon rod. The two electrons, from the zinc rod move towards carbon and neutralizes the positive charge., Thus current flows from carbon to Zn., , emf → 1.5V, Current → 0.25 A, , 8. Lead – Acid accumulator:, , Construction :, Anode → Lead oxide, Cathode →Lead, Electrolyte →Diluted Sulphuric acid, Vessel → Rubber or Glass, , www.Padasalai.Net, Working : Spongy lead reacting with dilute sulphuric acid produces lead sulphate and, two electrons., At the positive electrode, lead oxide on reaction with sulphuric acid produces lead, sulphate and the two electrons are neutralized., The conventional current to flow from positive electrode to negative electrode in the, external circuit., The emf → 2.2 Volt and the specific gravity → 1.28., It has low internal resistance and hence can deliver high current., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 9. Principle of potentiometer:, A battery Bt is connected between the ends A and B, of a potentiometer wire through a key K. This forms, the primary circuit., A primary cell is connected in series with the positive, terminal A of the potentiometer, a galvanometer,, high resistance and jockey., If the potential difference between A and J is equal to the emf of the cell,, no current flows through the galvanometer. If the balancing length is b,, r is the resistance per unit length of the potentiometer wire, E = Ir b,, E∝ b, emf of the cell is directly proportional to its balancing length., 10. Comparison of emfs of two given cells using potentiometer:, , Construction : Wire AB is connected in series with, a battery , Key, rheostat. The cell of emf E1 is, connected between terminals C1 and D1 and the cell of, emf E2 is connected between C2 and D2 of the DPDT, switch., The potential difference across the balancing length b 1, E1 = Ir b l, , www.Padasalai.Net, The DPDT switch is pressed towards E2. The balancing length b 2 The potential, difference across the balancing length is l2, E2 = Ir b 2, ö, ö, , b, , (b, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 79 : www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 11. Faraday’s first laws of electrolysis:, Law: The mass of a substance liberated at an electrode is directly proportional to the, charge passing through the electrolyte., , Construction : A battery, a rheostat, a key and an, ammeter are connected in series to an electrolytic, cell., A current I is passed for a time t, the mass m of, the substance is obtained., Current I is passed for the same time t. The mass, m is obtained., I, m, (, m, I, , Experiment is repeated for current I but for different times t and t . If the masses of, the deposits are ma and m, kS, kÓ, , (, , ±, ±, , m ∝ It, m∝ q, 12. Faraday’s second laws of electrolysis:, Law: The mass of a substance liberated at an electrode by a given amount of charge is, proportional to the chemical equivalent of the substance., , www.Padasalai.Net, Construction : Two cells containing different, , electrolytes, CuSO4 and AgNO3 solution are, connected in series with a battery, a rheostat and an, ammeter. Copper electrodes in CuSO4 and silver, electrodes in AgNO3., , The current is passed for some time. Then masses of copper and silver are found as m, and m ., Rh, , k, k, , (, , ö, ö, , m∝ t, , Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], For contact, Email :
[email protected], Phone No: 9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 3– EFFECTS OF ELECTRIC CURRENT, PREPARED BY, J.SHANMUGAVELU M.Sc, B.Ed. [P.G. Assist in Physics], , ONE MARK QUESTIONS: (2-questions), BOOK BACK ONE MARKS:, 1. Joule’s law of heating is, , b) H = V2 Rt, , a) H = t, ô, , d) H = IR2t, , c) H = VIt, , 2. Nichrome wire is used as the heating element because it has, a) low specific resistance, b) low melting point, c) high specific resistance, d) high conductivity, 3. Peltier coefficient at a junction of a thermocouple depends on, a) the current in the thermocouple, b) the time for which current flows, c ) the temperature of the junction, d) the charge that passes through the thermocouple, 4. In a thermocouple, the temperature of the cold junction is 20℃, the neutral temperature is, 270℃. The temperature of inversion is, a) 520℃, b) 540℃, c) 500℃, d) 510℃, , www.Padasalai.Net, Θn =, , Sol :, , /4 F/5, , ⇒ θi = 2θn – θc, =2, , 270 -20, , = 540 – 20 = 520℃, 5. Which of the following equations represents Biot-savart law?, œ@ õäd, ^^^^^_ = œ@ õäd Í7m/, a) dB =, b)q6, , ^^^^^_, c)q6, , =, , ^^^^^^_, œ@ õäd, , d) ^^^^^_, q6 (, , _, , ^^^^^^_, œ@ õäd, , >?, , Â, , _, , 6. Magnetic induction due to an infinitely long straight conductor placed in a medium of, permeability µ is, , a), , œ@ õ, , µ, , b), , œ@ õ, , µ, , c), , õ, , µ, , d), , œÃ, , =?8, , o, , 7. In a tangent galvanometer, for a constant current, the deflection is30 . The plane of the, coil is rotated through 900. Now, for the same current, the deflection will be, a) 300, b) 600, c) 900, d) 00, 0, Hints: The plane of the coil is rotated through 90 . The earth and horizontal magnetic fields are, parallel to each other. So the needle suspended between the two fields, , 79, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 8. The period of revolution of a charged particle inside a cyclotron does not depend on, a) the magnetic induction, b) the charge of the particle, c) the velocity of the particle, d) the mass of the particle, Hints:, , p, , the time period is does not depend onthe velocity of the particle and radius of the, , circular path., , 9. The torque on a rectangular coil placed in a uniform magnetic field is large, when, a) the number of turns is large, b) the number of turns is less, c) the plane of the coil is perpendicular to the field d) the area of the coil is small, Hints: τ = nBIA sin θ ; when the no. of turns ,magnetic field, current passing through the coil, area of, the coil increases torque is also increases and the coil is parallel to the magnetic field the torque is, maximum, 10. Phosphor – bronze wire is used for suspension in a moving coil galvanometer, because it has, a) high conductivity, b) high resistivity, d) small couple per unit twist, c) large couple per unit twist, 11. Of the following devices, which has small resistance?, a) moving coil galvanometer, b) ammeter of range 0 – 1A, c) ammeter of range 0–10 A, d) voltmeter, Hints : ammeter has small resistance. But option (c) has small resistance. Because the range of, the ammeter is high. ( Current is inversely proportional to the resistance ), 12. A galvanometer of resistance G Ω is shunted with S Ω. The effective resistance of the, combination is Ra. Then, which of the following statements is true?, a)G is less than S, b) S is less than Ra but greater than G., d) S is less than both G and Ra, c) Ra is less than both G and S, Hints:, The increasing order of resistance Ra < S < G, , www.Padasalai.Net, 13. An ideal voltmeter has, a) zero resistance, b) finite resistance less than G but greater than Zero, c) resistance greater than G but less than infinity, d) infinite resistance, , PUBLIC ONE MARKS:, 14. The unit of reduction factor of tangent galvanometer is, a) no unit, b) tesla, c) ampere, , d) ampere/degree, , 15. The galvanometer can be converted into voltmeter by connecting, a) low resistance in series, b) high resistance in parallel, c) high resistance in series, d) low resistance in parallel, 16. In a thermocouple, the temperature of the cold junction is 20℃, the temperature of, inversion is 600℃, then the neutral temperature is, a) 310℃, b) 320℃, c) 300℃, d) 315℃, Sol :, , Θn =, , /4 F/5, , =, , 80, , F, , =, , = 310 ℃, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 17. In a TG a current 1 A, produces a deflection of 30°. The current required to produce a, deflection of 60° is, a) 3 A, b) 2 A, c) √3 A, d) 1/√3 A, Sol :, I= K tan &;, , I1= K tan 30 … …(1) ;, I2= K tan60…..(2), Equ, , : ±4², , õ, , ⇒ õ = : ±4² a, =, , √a, √S, , = √3√3 = 3 A, , 18. Peltier effect is the converse of, a) Joule effect, b) Raman effect, , c) Thomson effect, , d) Seebck effect, , 19. The torque experienced by a rectangular current loop placed perpendicular to a uniform, magnetic field is, a) maximum, b) zero, c) finite minimum, d) infinity, Hints: τ = nBIA sin θ; when the rectangular current loop placed perpendicular to a uniform, magnetic field the torque is zero. Because θ = 00., τ = nBIA sin θ = nBIA (0) = 0, , 20. In which of the following pairs of metals of a thermocouple the e.m.f is maximum?, a) Fe –Cu, b) Cu – Zn, c) Pt – Ag, d) Sb – Bi, Hints: The direction of the current at the hot junction is from the metal occurring earlier in the, series to the one occurring later in the series. The magnitude of thermo emf is larger for metals, appearing farther apart in the series, , www.Padasalai.Net, 21. Which of the following principles used in a thermopile, a) Thomson effect, b) Peltier effect, c) Seebeck effect, , d) Joule’s effect, , 22. Fuse wire is an alloy of, a) Lead and Tin, b) Tin and copper, , c) Lead and copper, , d) Lead and Iron, , 23. Thermopile is used to, a) measure temperature, c) detect thermal radiation, , b) measure current, d) measure pressure, , 24. The magnitude and direction of the magnetic Lorentz force is given by, ^_ 6, ^_), ^_ 6, ^_, ^^_ ;, ^^_), a) g_ = (e, b) g_ =q/(e, c) ^_ = q(¢, 25. Unit of peltier coefficient is, a) ohm, b) mho, , c) Volt, , ^_ h, d)g_ =e, d) ampere, , 26. For a given thermocouple the neural temperature, a) depends upon the temperature of cold junction, b) depends upon the temperature of the hot junction, c) is a constant, d) depends upon the temperature of cold junction and the temperature of the hot junction, , 81, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html, , ^_), 6
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 27. When the number of turns (n) in a galvanometer is doubled, current sensitivity, a) remains constant, b) decreases twice, c) increases twice, d) increases four times, Sol :current sensitivity, / £, , 5õ9 (, , m ¡, P, , & ′, , /, õ, , =, , ⇒ 5 < 9 =2, , m ¡, , /, , P, , ;, , No. of turns =2n, , õ, , 28. An electron is moving with a velocity of 3 106m/s perpendicular to a magnetic field of, induction 0.5 T. The force experienced by the electron is, a) 2.4 10-13N, b) 13.6 10-27N, c) 13.6 10-11N, d) zero, Sol : F = Bqv sin & = Bqv sin 90, , 10-19, , = 0.5, , 1.6, , = 2.4, , 10-13 N, , 29. Fuse wire, a)is an alloy of lead and copper, c) has high resistance, , 3, , 106, , 1, , b) has low resistance, d) has high melting point, , 30. In the experiment to verify Joule’s law when the current passed through the circuit is, doubled keeping resistance (R) constant and time of passage of current (t) constant, the, temperature of the liquid is, a) increases twice, b) increases four times, c) increase sixteen times, d) decreases four times, Sol :, H =I2Rt;, , www.Padasalai.Net, current = 2I;, H′ = (2I)2Rt, , = 4I2Rt= 4H, , 31. AB is a rod of lead. The end A Is heated. A current I is allowed to flow along AB. Now,, due to Thomson effect, in rod AB:, a) heat is absorbed, b) heat is liberated, d) heat is first absorbed and then liberated, b) heat is neither absorbed nor liberated, Hints: Thomson effect of lead is nil, 32. The direction of force on a current carrying conductor placed in a magnetic field is given, by :, a) Fleming’s Left hand Rule, b) Fleming’s Right hand Rule, b) End Rule, d) Right Hand Palm Rule, 33. Which of the following produces large joule heating effect, a) 1 A current through 2 Ω resistor for 3 second, b) 1 A current through 3 Ω resistor for 2 second, c) 2 A current through 1 Ω resistor for 2 second, d) 3 A current through 1 Ω resistor for 1 second, Sol:, H = I2Rt;, a) H = 12 2 3= 6 J, b) H = 12 3 2 = 6 J, c) H = 22 1 2= 8 J, d) H = 32 1 1 = 9 J, , 82, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 34. In a thermocouple, the temperature of the cold junction is 20℃, the temperature of, inversion is 520℃. The neutral temperature is, a) 500℃, b) 54℃, c) 270℃, d) 510℃, Sol :, , Θn =, , =, =, , /4 F/5, , E, , F, , E, , =270℃, , 35. In a thermocouple, the temperature of the cold junction is -30℃, and the neutral, temperature is270℃. Then the temperature inversion is, a) 520℃, b) 540℃, c) 500℃, d) 570℃, /4 F/5, Θn =, Sol:, , ⇒&7 ( 2&m 8 &= ( 2 270 8 30, ( 540 +30, = 570℃, 36. Consider a circular coil of radius 10 cm in an air medium. If 5A current passes through it,, what would be the magnetic induction at its centre?, a) ?, <@#Æ T, b) U, 10E T, c) U, 10# E T, d) U, 10# T, •0 n<, Sol:, B = 2>, , www.Padasalai.Net, «, , =, , E, , \, , =, =U, , 10#E T, , 37. In Joule’s Calorimeter experiment, when the current of 1 ampere is passed through a coil, for a known interval of time ‘t’ , the temperature of water increases from 30℃ to 33℃., When a current of 2 A is passed through the same coil placed in the same quantity of, water and for the same time, the temperature of water increases from 30℃ to :, d)42℃, a)33℃, b)36℃, c)39℃, Sol: Given : I1 = 1A;, &1 = 30℃ ; & 2 = 33℃ ;, I2 = 2A;, &1 = 30℃ ; & 2 = ? ;, , H = I2Rt, H = W ( & 2 – &1 ), W ( &2 – & 1 ) = I12Rt, W . 3 = 12Rt, W. 3 = Rt ………(1), W ( & 2 – & 1 ) = I22Rt, W ( &2 – 30 ) = 22Rt, , 83, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , W ( &2 – 30 ) = 4Rt ……..(2), <, =, , ⇒, , B / #a, W .3, , / #a, 3, , =, , ìë, , ìë, , =4, , & 2 – 30 = 12, , &2 = 12 + 30 = 42 ℃, , 38. A proton and an Particle are Projected with the same velocity normal to a uniform, magnetic field. The ratio of the magnetic Lorentz force experienced by the proton and, particle is, a) 1 : 1, b) 1 : 2, c)2:1, d) 1 : 0, sol: proton has one unit positive charge ( q ), Lorentz force of proton Fp = BqV sin &, , ™ Particle has two unit positive charge ( 2q ), Lorentz force of ™ particle Fα = B(2q)V sin & = 2 BqV sin &, ©), ©@, , =, , A ¯ .%² /, , A ¯ .%² /, , =, , g+ : g, = 1 : 2, , www.Padasalai.Net, 39. a wire of length 1 m is made into a circular loop and it carries a current of 3.14 A . The, magnetic dipole moment of the current loop ( in A m2 ) is :, a) 1, b) 0.5, c) 0.25, d) 0.314, sol:, 2U ( 1, , r=, U, , M =I A = I, =I, , U§ ¨, , = 3.14, , U, , = = 0.25, 40. In a thermocouple, when the temperature of cold junction is increased (but less than, neutral temperature) the temperature of inversion, a) increases, b) decreases, c) does not change, d) first increases then decreases, 41. In a given thermocouple, the neutral temperature :, a) is a constant, b) depends on the temperature of cold junction, c) depends upon the temperature of inversion, d) both (b) and (c)., Hints : For a given thermocouple, the neutral temperature is a constant, , 84, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , CHOOSE THE CORRECT ANSWER FROM THE OPTIONS GIVEN, 42. Which of the following is wrong, according to Joules law of heating effect ______, a) H ™ <2, for a given R, b) H ™ R for given I, c) H V for a given R, d) H ™ ì for a given V, 43. In which of the following, Joule heating effect is undesirable?, a) electric iron, b) electric toaster, c) transformer and dyanamos, d) fuse wire, 44. Which of the following is not a thermo emf effect?, a) Peltier effect, b) Thomson effect, d) Seeback effect, c) Joule effect, 45. In a circuit consisting of two dissimilar metals, an emf is developed, when the junctions, are maintained at ____, a) very high temperature, b) very low temperature, c) same temperatures, d) different temperatures, 46. In a Cu-Fe thermocouple, the direction of the current at the hot junction is ____, a) from Cu to Fe, b) from Fe to Cu, c) either (a) or (b) depending on temperature of hot junction, d) random direction, 47. Position of the metal in the thermoelectric series depends on _____, b) nature of the metal, a) temperature, c) magnitude of thermo emf, d) atomic number of metal, , www.Padasalai.Net, 48. For small temperature difference, the graph showing the variation of thermo emf with, temperature of the hot junction is _____, a) parabola, b) circle, c) straight line, d) hyperbola, 49. For a given thermocouple, the neutral temperature is ____, a) maximum, b) minimum, c) zero, , d) a constant, , 50. For a given thermocouple, the temperature of inversion ____, a) is constant, b) depends upon the temperature of the cold junction, c) is independent of temperature of cold junction, d) depends on the neutral temperature, 51. Peltier co-efficient of a junction of a thermocouple depends on ____, a) the current in the thermocouple, b) time for which the current flows, c) temperature of the junction, d) charge that passes through the junction, 52. In a thermocouple, the temperature of cold junction is 200C, while the neutral, temperature is 3000C. It temperature of inversion is _____, b) 8500C, c) 5080C, d) 8050C, a) 5800C, , 85, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 53. Two wires of equal length are first connected in series, and then in parallel with a voltage, source. The ratio of heat developed in two cases is ____, a) 2:1, b) 1:2, c) 4:1, d) 1:4, 54. If the heating element of an electric toaster has a resistance of 22 ohm and is connected to, an voltage source of 110V, the amount of heat generated in 1 minute is ____, a) 33 KJ, b) 22 KJ, c) 66 KJ, d) 3.3 KJ, 55. In a thermopile, the deflection in the galvanometer is proportional to ___ of the radiation, b) frequency, c) velocity, d) energy, a) intensity, 56. Magnetic induction at a point due to infinitely long straight conductor carrying current, at a distance of ‘a’ from the axis is, a) directly proportional to a, b) directly proportional to a2, c) inversely proportional to a, d) inversely proportional to a2, 57. Magnetic needle of a tangent galvanometer is kept small because, the magnetic field is, (a) very large at the centre, (b) considered to be small and uniform at the centre, (c) such that it is convenient to handle small needle, (d) radial at the centre, 58. A current of √ A produces a deflection of 450 in a tangent galvanometer having 50 turns, and radius 10 cm. The reduction factor of the tangent galvanometer is, a) 1.732 A, b) A, c) 50√3, d) 5√3 A, , www.Padasalai.Net, √a, , 59. Reduction factor of the tangent galvanometer is, , a), , µ, , m, , B, , b), , µm, , DB, , c), , =8;C, , d), , œ@ º, , m B, , µ, , 60. A Current of 2 A flows through 5 turn coil of a tangent galvanometer having a radius of, 12.5 cm. If the deflection of the needle at the centre is 450, the horizontal component of the, earth field at that point is, a) 16UX10-5T, b) 16?X10-6T, c) 16UX10-7T, d) 16UX10-8T, 61. If the reduction factor of a tangent galvanometer is 0.9 A, then the current that produces, a deflection of 300 is, a) 450 mA, b) 520 mA, c) 780 mA, d) 520 A, 62. When a current of 1.5 A flows through a tangent galvanometer, a deflection of 600 is, produced in3 it. The current required to produce a deflection of 300 is, a) 500 mA, b) 250 mA, c) 250 A, d) 50 A, 63. Biot-Savart law expressed in an alternative way is called, a) end rule, b) Gauss law, c) Ampere circuital law, d) Fleming left hand rule, 64. Direction of magnetic field due to circular closed loop is given by, a) Ampere – circuital law, b) right hand rule, c) right hand screw rule, d) right hand palm rule, , 86, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 65. The magnetic polarity of a current carrying solenoid is given by, a) Ampere – circuital law, b) right hand palm rule, c) end rule, d) Biot-Savart rule, , 66. An electron is moving with a velocity of 3X106 m/s perpendicular toa magnetic field of 0.5, T, then the force experienced by the electron is, a) 24 X10-11 N, b)2.4 X10-13 N, c) 13.6 X10-27 N, d) 13.6 X10-11 N, 67. Angular frequency and period of rotation of the charged particle in a magnetic field is, independent of _____ of the particle, a) mass and radius, b) velocity and radius, c) charge and velocity, d) mass and charge, 68. A charged particle of mass 3.2 X10-27 kg and charge 1.6 X10-19 C moves in a circular orbit, under the influence of perpendicular magnetic field of 3.14T, then the period of revolution, of the particle is, a) 10-8 S, b) 2 X10-8 S, c) 3 X10-8 S, d) 4 X10-8 S, 69. An - particle with (e/m) ratio 4.8 X10-11CKg-1 travels in a circular path of radius 0.45 m, in a magnetic field of 1.2 T, then the speed of the - particle is, a) 2.6 X104 m/s, b) 2.6 X10-11 m/s, 6, c) 2.6 X10 m/s, d) 1.3 X107 m/s, 70. Cyclotron cannot accelerate, b) a proton, a) an electron, , c) a deuteron, , d) an ™ 8 s> ïyzb{, , www.Padasalai.Net, 71. Workdone by a Lorentz force is, a) zero when & = 900, c) always zero, , b) zero when & = 450, d) maximum & = 900, , 72. Two parallel straight conductors carrying currents the same direction, a) repel each other, b) attract each other, c) do not experience any force, d) experience a maximum force, 73. A current of 2 A flows through two long straight parallel conductors separated by a, distance of 10 cm. The force per unit length on each conductor is, a) 0.0458 N, b) 8 X10-4 N, c) 8 X10-5 N, d) 8 X10-6 N, 74. Two straight parallel current carrying conductor separated by certain distance carrying, equal current experience a force of 16 N. If the distance between them is doubled and the, current in each conductor is halved then the force between them will be, a) 64 N, b) 16 N, c) 4 N, d) 2 N, 75. When the number of turns in a galvanometer is doubled, then, (a) current sensitivity and voltage sensitivity doubled, (b) current sensitivity is doubled and voltage sensitivity remains unchanged, (c) current sensitivity remains unchanged, (d) voltage sensitivity is doubled, 76. Phosphour – bronze wire is used for suspension in a moving coil galvanometer, because it, has, a) high conductivity, b) high resistivity, c) large couple per unit twist, d) small couple per unit twist, , 87, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 77. The deflection in moving coil galvanometer is reduced to half, when it is shunted with a, resistance of40 ohm, then the resistance of the galvanometer will be, a) 80 ohm, b) 40 ohm, c) 20 ohm, d) 10 ohm, 78. A galvanometer of resistance 50 ohm is shunted with a wire of 10 ohm. The current, through the galvanometer when the current in the circuit is 12 A is, a) 3 A, b) 2 A, c) 5 A, d) 6 A, 79. If the resistance of a moving coil galvanometer is 100 ohm and if it shows a full scale, deflection for 1 mA, then by connecting 900 ohm in series, the range of the voltmeter is, a) 1 V, b) 10 V, c) 100 V, d) 100 mV, 80. The value of gyromagnetic ratio is, a) 8.8 X109 CKg-1, b) 8.8 X1010 CKg-1, c) 8.8 X10-10 CKg-1, d) 8.8 X10-9 CKg-1, , 3 MARK QUESTIONS: ( 1 – QUESTIONS:, , Q.No: 36 ), , PUBLIC ‘3’ MARKS:, 1. Why nichrome is used as heating element ? (OR) What are the characteristics of, heating element used in electric heating device? (J–07,M -10,O-15,M-17), , www.Padasalai.Net, i) It has high specific resistance, ii) It has high melting point, iii) It is not easily oxidized, , 2. What is seebeck effect? (J – 16), Seebeck discovered that in a circuit consisting of two dissimilar metals like, iron and copper, an emf is developed when the junctions are maintained at, different temperatures. Two dissimilar metals connected to form two junctions is, called thermocouple. The emf developed in the circuit is thermo electric emf. The, current through the circuit is called thermoelectric current. This effect is called, thermoelectric effect or Seebeck effect., 3. What is neutral temperature? ( O – 08 ), Keeping the temperature of the cold junction constant, the temperature of the, hot junction is gradually increased. The thermo emf rises to a maximum at a, temperature (& n) called neutral temperature., , &n =, , 88, , J.SHANMUGAVELU, , /5 F/4, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 4. Define: Peltier coefficient and write its unit. ( J – 06, J – 11,M – 12,M-14 ), The amount of heat energy absorbed or evolved at one of the junctions of, a thermocouple when one ampere current flows for one second (one coulomb) is, called Peltier coefficient., It is denoted by π. Its unit is volt., , H = π It, π =H/It, 5. Define Thomson coefficient. ( J-15 ), The amount of heat energy absorbed or evolved when one ampere current, flows for one second (one coulomb) in a metal between two points which differ, in temperature by 1oC is called Thomson coefficient., It is denoted by σ. Its unit is volt per oC., 6. Give any two differences between Peltier effect and Joule’s law of heating., ( M – 06 ), Peltier effect, , Joule’s law of heating, Rate of evolution of heat is directly, of heat is directly proportional to proportional to the square of the, the current H ∝ I., current H ∝ I ., Depends on direction of current. Independent on direction of current., Reversible process, Irreversible process, Takes place only at the junction Takes place throughout the conductor, At one junction heat is evolved, Heat is evolved throughout the, at the other absorbed, conductor, , 1. Rate of evolution or absorption, , www.Padasalai.Net, 2., 3., 4., 5., , 7. State Tangent law. ( M – 11, O-16 ), A magnetic needle suspended at a point where there are two crossed fields, at right angles to each other will come to rest in the direction of the resultant of, the two fields., B = Bh tan θ, 8. Define: ampere’s circuital law. ( M – 09,J-14 ), , ^_.^^^_, The line integral ∮ 6, qb for a closed curve is equal to μ0 times the net, current Io through the area bounded by the curve., ^_.^^^_, qb = μ0I0, ∮6, 89, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 9. What are the limitations of a cyclotron? ( O – 06, J – 10,M-13 ), i) Maintaining a uniform magnetic field over a large area of the Dees is, difficult., ii) At high velocities, relativistic variation of mass of the particle upsets the, resonance condition., iii) At high frequencies, relativistic variation of mass of the electron is appreciable, and hence electrons cannot be accelerated by cyclotron., 10. State Fleming’s left hand rule. ( O – 10, M-15 ), The forefinger, the middle finger and the thumb of the left hand are, stretched in mutually perpendicular directions. If the forefinger points in the, direction of the magnetic field, the middle finger points in the direction of the, current, then the thumb points in the direction of the force on the conductor., 11. Define ampere in terms of force between two long parallel current carrying, conductors ( M – 08 , J – 08, O – 11, O-14 ), Ampere is defined as that constant current which when flowing through two, parallel infinitely long straight conductors of negligible cross section and placed, in air or vacuum at a distance of one metre apart, experience a force of 2 × 10-7, newton per unit length of the conductor., 12. How the current sensitivity of a galvanometer can be increased? ( O – 09 ), , www.Padasalai.Net, The current sensitivity of a galvanometer can be increased by, i) Increasing the number of turns, ii) Increasing the magnetic induction, iii) Increasing the area of the coil, iv) Decreasing the couple per unit twist of the suspension wire., , 13. In a galvanometer, increasing the current sensitivity does not necessarily, increase voltage sensitivity. Explain ( M – 07 ), An interesting point to note is that, increasing the current sensitivity does not, necessarily, increase the voltage sensitivity. When the number of turns (n) is, /, m ¡, doubled, current sensitivity is also doubled ( from the equation (, ). But, õ, P, increasing the number of turns correspondingly increases the resistance (G). Hence, voltage sensitivity remains unchanged., 14. How is a galvanometer converted in to (a) an ammeter and, (b) a voltmeter?(J-09), A galvanometer is converted into an ammeter by connecting a low resistance, in parallel with it. The low resistance connected in parallel with the galvanometer, is called shunt resistance., A galvanometer can be converted into a voltmeter by connecting a high, resistance in series with it., , 90, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , OTHER IMPORTANT ‘3’ MARKS:, 15. State Joule’s law of heating., The heat produced in a conductor is, i) directly proportional to the square of the current for a given R, ii) directly proportional to resistance R for a given I and, 2, iii) directly proportional to the time of passage of current. ( i.e.) H = I Rt, , 16. Write short notes on Fuse wire, Fuse wire is an alloy of lead 37% and tin 63%. It is connected in series in, an electric circuit. It has high resistance and low melting point. When large, current flows through a circuit due to short circuiting, the fuse wire melts due to, heating and hence the circuit becomes open. Therefore, the electric appliances are, saved from damage., 17. What is meant by thermo electric current?, Two dissimilar metals connected to form two junctions is called, thermocouple. The emf developed in the circuit is thermo electric emf., The current through the circuit is called thermoelectric current., , www.Padasalai.Net, 18. Define inversion temperature of a thermocouple., , Keeping the temperature of the cold junction constant, the temperature of the, hot junction is gradually increased. The thermo emf rises to a maximum at a, temperature (& n) called neutral temperature and then gradually decreases and, eventually becomes zero at a particular temperature (θi) called temperature of, inversion., , &7 =2&n -&=, , 19. Distinguish between neutral and inversion temperature of a thermocouple., Neutral temperature, 1. The thermo emf rises to a, maximum at a temperature, 2. For a given thermocouple, the, neutral temperature is a constant, 3., , &n =, , /5 F/4, , Inversion temperature, Thermo emf decreases and eventually, becomes zero, The temperature of inversion depends, upon the temperature of cold junction., , &7 =2&n -&=, , 20. Define Peltier effect., When electric current is passed through a circuit consisting of two dissimilar, metals, heat is evolved at one junction and absorbed at the other junction. This, is called Peltier effect. Peltier effect is the converse of Seebeck effect., , 91, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 21. What is meant by Thomson effect?, When a current flows through unequally heated conductors, heat energy is, absorbed or evolved throughout the body of the metal., 22. What is a thermopile?, Thermopile is a device used to detect thermal radiation. It works on the, principle of Seebeck effect., 23. Show by graph , how the thermo emf of a thermocouple varies with, temperature of hot junction., The graph showing the variation of thermo emf with temperature of the hot, junction , taking the temperature of the cold junction (θC) as origin. For small, difference in temperature between the junctions, the graph is a straight line. For, large difference in temperature, the graph is a parabola., 24. Name few metals which have positive Thomson effect and negative Thomson, effect ?, Positive Thomson effect: Ag, Zn, Cu, Cd, Sb, , www.Padasalai.Net, Negative Thomson effect: Fe, Bi, Co, Ni, Pt, Hg, , 25. State Biot – Savart Law, , i) Directly proportional to the current (I), ii) Directly proportional to the length of the element (dl ), iii) Directly proportional to the sine of the angle between dl and the line joining, , element dl and the point P (sin θ), iv) Inversely proportional to the square of the distance of the point from the, element (, , ), dB =, , ³° .%² D, , 26. State Maxwell’s right hand cork – screw rule., If a right handed cork screw is rotated to advance along the direction of the, current through a conductor, then the direction of rotation of the screw gives the, direction of the magnetic lines of force around the conductor., 27. State right hand palm rule., The coil is held in the right hand so that the fingers point in the direction of, the current in the windings. The extended thumb, points in the direction of the, magnetic field., , 92, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 28. State end rule., When looked from one end, if the current through the solenoid is along, clockwise direction, the nearer end corresponds to south pole and the other end is, north pole., When looked from one end, if the current through the solenoid is along anticlock wise direction, the nearer end corresponds to north pole and the other end, is south pole, 29. What is a solenoid?, A long closely wound helical coil is called a solenoid. The magnetic field, due to the solenoid is the vector sum of the magnetic fields due to current, through individual turns of the solenoid., 30. What is Magnetic Lorentz Force?, The force experienced by a moving charged particle placed in a uniform, magnetic field is called as magnetic Lorentz force., F = BqV sin θ, 31. State Principle of cyclotron, , www.Padasalai.Net, A charged particle moving normal to a magnetic field experiences magnetic, Lorentz force due to which the particle moves in a circular path., , 32. Is electron can be accelerated by a cyclotron? Support your answer with, reason., , Electron cannot be accelerated by a cyclotron. Because electron is negative, charge. Cyclotron is used to accelerate protons , deutrons and, , α - particles., , 33. Define: current sensitivity of a galvanometer., The current sensitivity of a galvanometer is defined as the deflection produced, when unit current passes through the galvanometer., /, m ¡, (, Current sensitivity, õ, P, 34. Define voltage sensitivity of a galvanometer, The voltage sensitivity of a galvanometer is defined as the deflection per unit, voltage., , ∴ Voltage sensitivity, , 93, , /, , f, , J.SHANMUGAVELU, , =, , /, , õE, , (, , m ¡, PE, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 95 : www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 35. How the voltage sensitivity of a galvanometer can be increased?, i., ii., iii., , The voltage sensitivity of a galvanometer can be increased by, Increasing the magnetic induction, Increasing the area of the coil, Decreasing the couple per unit twist of the suspension wire., , 36. Why an ammeter is always connected in series?, Ammeter resistance is very low and this explains why an ammeter should be, connected in series. When connected in series, the ammeter does not appreciably, change the resistance and current in the circuit. Hence an ideal ammeter is one, which has zero resistance, 37. Define magnetic moment of a current loop?, The magnetic moment of a current loop is defined as the product of the, current and the loop area. Its direction is perpendicular to the plane of the loop., Its unit Am2 ., M = IA, , E ß Ú õÀ • i ² ®, E ß Ú õÀ © m k ÷ © • i ² ®, E ß Ú õÀ • i ¯ õu x J ß Ö ª À ø », , www.Padasalai.Net, Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], Contact for subject doubt, Email :
[email protected], Phone No: 9952223467, , JEYAM TUITION CENTRE, [PARAMAKUDI], Contact for tuition, Email :
[email protected], Phone No: 9944888512, , 94, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PUBLIC ‘3’ MARK PROBLEMS:, , 1. Calculate the resistance of the filament of a 100 W, 220 V electric bulb [O-07], , Data : P = 100 W,, Sol, , V = 220 V,, , Power P =, , R=?, , f, , Resistance R =, , ì, f, , +, , =, , R = 484 Ω., 2. A long straight wire carrying a current produces a magnetic induction of, , 4 x 10-6 T at a point 15 cm from the wire. Calculate the current through the, wire. ( J – 12 ), , Given data: b = 15 cm = 15 10-2 m;, , B=4, , 10# T, , www.Padasalai.Net, Sol:, , õ, , B=, I=, , µ, , µ, , R, , E, , I=, , «, , I = 3A, 3. A conductor of length 50 cm carrying a current of 5 A is placed perpendicular, , to a magnetic field of induction 2, 13], , Given data: B = 2 × 10-3 ;, Sol:, , 10-3 T. find the force on the conductor. [O-, , I = 5A;, , b ( 50 z ;, , F=?, , F = BI b sinθ, = 2 × 10-3 × 5 × 5 × 10-1 × sin 900, , ∴ F = 5 × 10-3 N, 95, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PUBLIC ‘5’ MARK PROBLEMS:, , 4. A long straight wire carrying a current produces a magnetic induction of, , 4 10-6 T at appoint 15 cm from the wire. Calculate the current through the, wire. [O-07], , Given data: b = 15 cm = 15 10-2 m;, Sol:, B=, I=, , B=4, , 10# T, , õ, , µ, µ, , E, , I=, , R, «, , I = 3A, , www.Padasalai.Net, 5. A circular coil of 50 turns and radius 25 cm carries a current of 6A. It is, , suspended in a uniform magnetic field of induction 10-3 T. The normal to the, plane of the coil makes an angle of 600 with the field. Calculate the torque of, the coil., , Given data: n = 50 ;, Sol:, , r = 25 10-2 m;, , % = nBIA sin θ, , = 50 10-3 6 U, , (25 10-2)2, , I = 6A;, , B = 10-3 T; θ = 60°, , √a, , = 150 3.14 625 10-7 1.732, = 5.098 105 10-7, , = 5.098 10-2 Nm, 6. A circular coil of radius 20 cm has 100 turns of wire and it carries a current of, , 5A. Find the magnetic induction at a point along its axis at a distance of 20 cm, from the Centre of the coil. [M-06,O-06,M-09, J – 15,M-17] [compulsory], , Given data: a = 20 10-2m = 2 10-1m;, x = 20 10-2m = 2 10-1m, 96, , J.SHANMUGAVELU, , n = 100;, I = 5A;, x = a;, B=?, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 10-4 m2 is suspended inside a, radial magnetic field of induction 10-4 T by a suspension wire of torsional, constant 5 10-10 Nm per degree. Calculate the current required to produce, deflection of 100. [J-06,O-09,M-13,O-15] [compulsory], Given data : n= 500;, A = 6 10-4 m2;, B = 10-4 T ;, -10, &= 10°;, I=?, C = 5 10 Nm/ ° ;, , 7. A rectangular coil of 500 turns and of area 6, , Sol:, , τ = nBIA, , % = Cθ, I=, , m ¡, , I=, , i/, , m ¡, E, , =E, , Ó, , Ó, , =, , 10-3 = 0.166 10-3, , I = 0.166 mA, 8. A moving coil galvanometer of resistance 20 Ω produces full scale deflection, , for a current of 50 mA. How will you convert the galvanometer into, i) an ammeter of range 20 A and ii) a voltmeter of range 120 volt? [M-07,M09, J-13, M-15] [compulsory], , www.Padasalai.Net, Given data: G = 20 Ω;, R= ?, Sol:, i., , S= G ., , õG, , õ#õG, , =, , Ig = 50 10-3 A; I= 20A;, E, , #E, , S, S, , =, , V = 120 V;, , S=?&, , # . E, , S = 0.05 Ω, A shunt of 0.05 Ω should be connected in parallel, ii., , R=, , f, , õG, , =, , E, , –G, , S, , - 20 = 2400-20, , R = 2380 Ω, A resistance of 2380 Ω should be connected in series with, the galvanometer., 98, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 9. A galvanometer has a resistance 40 Ω. It shows full scale deflection for a, , current of 2 mA. How will you convert the galvanometer into voltmeter of, range 0 to 20 V? [O-10,J-16], Ig = 2 10-3 A;, , Given data: G = 40 Ω;, Sol:, , R=, , f, , V = 20 V;, , -G, , õG, , =, , S, , = 10, , – 40, , 103 – 40, , = 10000 – 40, , R = 9960 Ω, 9960 Ω resistance should connected in series, 10. The deflection galvanometer falls from 50 divisions to 10 divisions when 12Ω, , resistance is connected across the galvanometer. Calculate the galvanometer, resistance. [O-12], , www.Padasalai.Net, Given data:, , & = 50 divs;, , I∝ &, , Sol:, , &H ( 10 divs;, , S = 12 Ω;, , G=?, , <H ∝ &H, , In a parallel circuit potential is common., ∴, , G. Ig = S (I-Ig), , G=, , õ# õG, õG, , =, , E #, , G = 48 Ω, , 99, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 11. Two parallel wires each of length 5 m are placed at a distance of 10 cm apart, , in air. They carry equal currents along the same direction and experience a, mutually attractive force of 3.6 10-4 N. Find the current through the, conductors. [O-09,J-10,M-13,O-16] [compulsory], I1 = I2= I; b = 5 m;, , Given data :, Sol:, , a = 10-1m;, , F = 3.6 10-4 N;, , I=?, , õ õ d, , F=, , µ, , «, , =, , õ d, , µ, «, , F=, < =, , < =, , õ d, , µ, , ©µ, , «, , d, Ó, , a., , «, , E, , = 36, , I=6A, , www.Padasalai.Net, 12. Two straight infinitely long parallel wires carrying equal current placed at a, , distance of 20 cm apart experience a mutually attractive force of 4.9, per unit length of the wire. Calculate the current. [O-11], 10-1 m;, , Given data: a = 20cm = 2, õ õ d, , F=, , Sol:, , I1 = I2 = I;, , ©, d, , 10-5 N, , ( 4.9 10-5 N/m;, , µ, , force of per unit length, ©, d, , ©, d, , 4.9, , =, =, , õ õ, µ, , õ, , µ, «, , 10-5 =, I2 =, , .T, , õ, R, , «, , = 4.9 10 = 49, , I=7A, 100, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 0, , 13. In a hydrogen atom electron moves in an orbit of radius 0.5 A making, 16, , 10 revolutions per second. Determine the magnetic moment associated with, orbital motion of the electron. (Given : e = 1.6 10-19 C) [J-08], , Data :, , r = 0.5 Å = 0.5 10-10 m,, , Solution :, , n = 1016 s-1, , Orbital magnetic moment μl = i.A ….....(1), i = I = e.n ……....(2), , A = πr2 ………....(3), substituting equation (2), (3) in (1), , μl = e.n. πr2, = 1.6 × 10-19 × 1016 × 3.14 (0.5 × 10-10)2, = 1.256 × 10-23, , ∴, , μl = 1.256 × 10-23 Am2, , www.Padasalai.Net, 14. A current of 4 A flows through 5 turns coil of a tangent galvanometer having, , a diameter of 30 cm. if the horizontal component of Earth’s magnetic, induction is 4 10-5T. find the deflection produced in the coil., [given μ0 = 4π 10-7 Hm-1] [M-14], , Data : n = 5;, , a = 1.5 × 10–1 m;, Solution :, , d = 3 × 10–1 m;, , I = 4A;, &=?, I=, , tan θ =, , =, , µ J, m, , Bh = 4, , 10–5 T;, , tan &, , mõ, , µ J, «, , .E, , E, , \, , tan θ = 2.093, , ∴, , 101, , θ = 64o 28′, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 104 : www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 17. A uniform magnetic field of induction 0.5 T acts perpendicular to the plane, of the Dees of a cyclotron. Calculate the frequency of the oscillator to, accelerate protons. (mass of proton =1.67 × 10-27 kg) (O-15), Data: B = 0.5 T; mp =1.67 × 10-27 kg;, Sol:, , N(, =, , q = 1.6 × 10-19 C; N ( ?, , p), .E, , a., , ., , «, , . Ï, , 107 = 7.63, , = 0.763, , 106 Hz, , N = 7.63 MHz, 18. A stream of deutrons is projected with a velocity of 104 ms-1 in XY – plane., A uniform magnetic field of induction 10-3 T acts along the Z-axis. Find the, radius of the circular path of the particle. (Mass of deuteron is 3.32 × 10-27, kg and charge of deuteron is 1.6 x 10-19C) (M-17 ) [compulsory], , Data : v = 104 ms–1, B = 10–3T, m = 3.32 × 10–27 kg , e = 1.6 x 10-19C, r = ?, Sol :, , Bqv =, , pË, ì, , www.Padasalai.Net, r=, =, , pË, , a.a, , «, , S, , Ó, , ., , = 2.08 10#, r = 0.208 m, , E ß Ú õÀ • i ² ®, E ß Ú õÀ © m k ÷ © • i ² ®, E ß Ú õÀ • i ¯ õu x J ß Ö ª À ø », Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], For contact, Email :
[email protected], Phone No: 9952223467, 103, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , FIVE MARKS: (1-question: Q.No: 54), 1. State and explain Biot-Savart law. (J-09), 2. Explain the principle and the construction of tangent galvanometer.(diagram, theory not, necessary (O-08), 3. What are the special features of magnetic Lorentz force? (J-07,M-11), 4. Explain the conversion of galvanometer into an ammeter. (M-08,J-12), 5. Explain how will you convert a galvanometer into a voltmeter. (M-10,J-11,M-12,O-13), , TEN MARKS: (1-question: Q.No: 64), 1. State Joule’s law. Explain Joule’s calorimeter experiment to verify Joule’s law of, , heating. (J - 07, J - 12), 2. State Biot – Savart law. Derive an expression for a magnetic induction due to an, , infinitely long straight conductor carrying current. (M -10, M – 06,O-13), 3. Derive an expression for the magnetic induction at a point due to an infinitely long, , straight conductor carrying current. Write the expression for the magnetic induction, when the conductor is placed in a medium of permeability ‘μ’, (J - 06, O – 09, M-15,M-17), , www.Padasalai.Net, 4. Deduce an expression for the magnetic induction, at a point along the axis of a circular, , coil carrying current. (O - 07, M – 08, M – 12, O-14), 5. State Tangent law. Explain in detail the principle, construction and theory of a tangent, , galvanometer. (J - 08), 6. Define ampere’s circuit law. Applying it find the magnetic induction at a point due to a, , long solenoid carrying current. (O - 06, J - 09), 7. Discuss the motion of a charged particle in a uniform magnetic field. Define magnetic, , Lorentz force. (J – 10,O-15), 8. Explain the motion of a charged particle in a uniform magnetic field. Deduce the time, , period of rotation of it. ( M-13,J-14), 9. Explain in detail the principle, construction, working and limitations of a cyclotron with, , a diagram. (M - 07, O-10, O -11, J -13, J - 15), 10. Derive an expression for the force on a current carrying conductor placed in a magnetic, , field. Find the magnitude of the force. (O - 08, M - 09, J - 11, O -12,M-14, O-16), 11. Obtain an expression for the force between two long parallel current carrying conductors, , and also define “ ampere ’’. (M – 11, J-16), , 104, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, Only for slow bloomers, ( 5 & 10 Mark answer ), 1. features of Magnetic Lorentz force:, i), ii), iii), iv), v), , Force F on the charge is zero, if the charge is at rest. ., Force is proportional to the magnitude of the charge (q), Force is proportional to the magnetic induction (B), Force is proportional to the speed of the charge (v), Direction of the force is oppositely directed for charges of opposite sign, , 2. Conversion of galvanometer into an ammeter:, , A galvanometer is converted into an ammeter, by connecting a low resistance in parallel with, it., Iø be the maximum current, , passed through the galvanometer., Galvanometer resistance = G, Shunt resistance = S, Current in the circuit = I, Current through the shunt resistance = Is = (I–Iø ), , www.Padasalai.Net, Ig . G = (I- Ig)S, S= G ., , Q, , #Q, , 3. Conversion of galvanometer into a voltmeter:, , A galvanometer can be converted into a voltmeter by, connecting a high resistance in series with it, , Galvanometer resistance = G, The current produce full scale deflection in the galvanometer = Iø, Range of voltmeter = V, Resistance to be connected in series = R, ¯, , Ig = .ôFR, R=, , ¯, , Q, , –G, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 1. Magnetic induction due to infinitely long straight conductor carrying current, , XY is an infinitely long straight conductor carrying a current I. P is a point at a, distance a from the conductor. AB is a small element of length dl. & is the angle, between the current element I dl and the line joining the element dl and the point P., The magnetic induction at the point P due to the current element Idl is, S qb .%²D, , q6 =, , 2, , 6, , ...(1), , www.Padasalai.Net, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 2. Magnetic induction along the axis of a circular coil carrying Current, , Let us consider a circular coil of radius, ‘a’ with a current I . P is a point along, the axis of the coil at a distance x from, the centre O of the coil., Magnetic induction at P due to the element dl is q6 =, , S, , q6 =, , S, , äd .%²D, , 2, , 6, , Here & = 90°, äd, , 2 6, , dB is resolved into two components :- dB sin ™ along OP and dB cos ™ perpendicular to, OP. dB cos ™ components due to two opposite elements cancel each other whereas dB sin, ™ components get added up., , Total magnetic induction due to entire coil, , B = r q6 Yyn™ = r, S, , S, , qb 4, , 2 6, , 6, , www.Padasalai.Net, B=, B=, , If the coil contains n turns, , B=, , At the centre of the coil,, , B=, , x=0, , 4, , 2 6S, , S 4, , 26S, , r qb, , 2U> =, , mõµ, µ F", , õµ, , µ F", , Sæ, , mõ, , µ, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html, , Sæ
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 3. Cyclotron, , Principle:, Cyclotron works on the principle that a charged particle moving normal to a magnetic field, experiences magnetic lorentz force due to which the particle moves in a circular path., Construction:, It consists of a hollow metal cylinder divided into, two sections D1 and D2 called Dees, enclosed in an, evacuated chamber ., The Dees are kept separated and a source of ions is, placed at the centre in the gap between the Dees., They are placed between the pole pieces of a, strong electromagnet. The magnetic, field acts perpendicular to the plane of the Dees., The Dees are connected to a high frequency, oscillator., Working: When a positive ion of charge q and mass m is accelerated towards the Dee, having a negative potential ., Due to the normal magnetic field, the ion experiences magnetic lorentz force and moves in, a circular path., The ion arrives at the gap between the Dees, the polarity of the Dees gets reversed and, moves into the other Dee with a greater velocity along a circle of greater radius., Thus the particle moves in a spiral path of increasing radius and when it comes near the, edge. The particle with high energy is now allowed to hit the target T., , www.Padasalai.Net, The necessary centripetal force., , Bqv =, T, , =, 6, , The time taken to describe a semi-circle t =, Substituting equation (1) in (2),, , t=, , kT, , A, , 6, , k, , = constant, , π6, , T, , πk, , A, , the time taken by the ion to describe a semi-circle is independent of (i) the radius (r), of the path and (ii) the velocity (v) of the particle, time period T =, Frequency =, , πk, , A, , = constant, , A, , πk, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 4.ELECTROMAGNETIC INDUCTION, AND ALTERNATING CURRENT, PREPARED BY, J.SHANMUGAVELU M.Sc, B.Ed. [P.G. Assist in Physics], , ONE MARK QUESTIONS(4-questions:), BOOK BACK ONE MARKS:, 1. Electromagnetic induction is not used in, a) transformer, b) room heater, , c) AC generator, , d) choke coil, , 2. A coil of area of cross section 0.5 m2 with 10 turns is in a plane which is perpendicular to, an uniform magnetic field of 0.2 Wb/m2.The flux though the coil is, a)100 Wb, b) 10 Wb, c) 1 Wb, d) zero, Sol: the coil is perpendicular to an uniform magnetic field. So θ=00, ;, angle θ is in between direction of magnetic field and normal to the surface of the coil., Φ = nBA cos θ, = 10 0.2 0.5 cos 0 = 1 Wb, , www.Padasalai.Net, 3. Lenz’s law is in accordance with the law of, , a) conservation of charges, c) conservation of momentum, , b) conservation of flux, d) conservation of energy, , 4. The self – inductance of a straight conductor is, , a) zero, b) infinity, c) very large, d) very small, Hints :A coil only opposes any change in current that flows through it by inducing an emf in it., , 5. The unit henry can also be written as, , a) Vs A−1, Sol:, , b) Wb A−1, äë, äõ, a) e =-L ⇒ L = = VsA-1, äë, äõ, b) ϕ = LI ⇒ L =, c) L=VsA-1 =, , eY, , U, õ, , d) all, , c) Ωs, , = Wb A-1, , = Ωs, , 6. An emf of 12 V is induced when the current in the coil changes at the rate of 40 A S–1. The, , coefficient of self induction of the coil is, a) 0.3 H, b) 0.003 H, , c) 30 H, , d) 4.8 H, , Sol:L = -äõ, , =-, , æäë, , = -0.3 H, , (- ve sign indicates that the emf is produced in opposite direction), , 7. A DC of 5A produces the same heating effect as an AC of, , a) 50 A rms current, b) 5 A peak current, Hints: AC current = rms value of DC current, , 109, , J.SHANMUGAVELU, , c) 5A rms current, , [P.G. T. in Physics], , d) none of these, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 8. Transformer works on, , a) AC only, c) both AC and DC, , b) DC only, d) AC more effectively than DC, , 9. The part of the AC generator that passes the current from the coil to the external circuit is, , a) field magnet, , b) split rings, , d) brushes, , c) slip rings, , 10. In an AC circuit the applied emf e = Eosin (ωt + π/2) leads the current i = Io sin (ωt – π/2), , by, a) π/2, Sol:, , 5ωt 7, , U, 2, , 8 §ωt 8, , b) π/4, , U, ¨9, 2, , = 7 =U, , c) π, , d) 0, , 11. Which of the following cannot be stepped up in a transformer?, , a) input current, , b) input voltage, , c) input power, , d) all, , 12. The power loss is less in transmission lines when, , a) voltage is less but current is more, c) voltage is more but current is less, , b) both voltage and current are more, d) both voltage and current are less, , 13. Which of the following devices does not allow d.c. to pass through?, , b) capacitor, , a) resistor, , c) inductor, , d) all the above, , 14. In an ac circuit, , a)the average value of current is zero., c)the average power dissipation is zero, , b) the average value of square of current is zero, d) the rms current is 2time of peak current., , www.Padasalai.Net, PUBLIC ONE MARKS :, 15. The average power consumed over one cycle in an A.C circuit is, , b) Erms Irms cosϕ, , a) Erms Irms, , c) Erms Irms sinϕ, , 16. The angle between area vectors _ and plane of the area A is, , a) U, , b)2 U, , c) ?/=, , d) E0 I0cosϕ, d) zero, , 17. If the flux associated with a coil varies at the rate of 1wb/minute then the induced emf is, , a) 1V, Sol: e = -, , =-, , äU, , b) 1/60, , c) 60V, , d) 0.60V, , äë, , ( -ve sign indicates that the emf is produced in opposite direction), , 18. In a.c circuit with an inductor, , a)Voltage lags current by U/2, c) voltage leads current by U, , 110, , J.SHANMUGAVELU, , b) voltage and current are in phase, d) current lags the voltage by ?/2, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 19. In LCR series a.c circuit, the phase difference between current and voltage is 300 The, , reactance of the circuit is 17.32 Ω. The value of resistance is, a) 30 Ω, b) 10 Ω, c) 17.32Ω, àX #àY, , Sol:, , tan ϕ =, ⇒ R=, , ì, , àX #àY, , =, , ±4² Z, Ï.a, , æ, √a, , =, , a, , .Ïa, , Ï.a, , ±4² a, , Ï.a, , =, , Ï.a, , =, , d) 1.732Ω, , √a, , √a, √a, , =30 Ω, , 20. The generator rule is, , b) Fleming’s right hand rule, d) right hand palm rule, , a) Fleming’s left hand rule, c) Maxwell’s cork screw rule, , 21. The power loss is less in transmission line when, , a) voltage is less but current is more, c) voltage is more but current is less, , b) both voltage and current are more, d) both voltage and current are less, , 22. In a step up transformer, the input voltage is 220 V and the output voltage is 11 kV. The, , ratio of number of turns of primary to secondary is, a) 50 : 1, b) 1: 50, c) 25 : 1, [ß, [, , d) 1 : 25, , www.Padasalai.Net, Sol:, , ⇒, , 220, , 11, , 10, , 3, , (, , (, , tß, t, , 220, , 110 102, , (, , (, , E, , 23. In LCR circuit when XL = XC, the current, , a) is zero, c) leads the voltage, , b) is in phase with the voltage, d) lags behind the voltage, , 24. In an AC circuit with capacitor only, if the frequency of the signal is zero, then the, , capacitive reactance is, b) zero, a) infinity, sol:N ( 0; \P (, , ]P, , c) finite maximum, , = =∞, , d) finite minimum, , 25. In a step up transformer, the output voltage is 11 kV and the input voltage is 220 V The, , ratio of number of turns of secondary to primary is, a)20:1, b)22:1, c)50:1, sol:, , 111, , [Y, [s, , (, , tY, ts, , (, , 11 103, 220, , (, , 11000, 220, , (, , E, , J.SHANMUGAVELU, , [P.G. T. in Physics], , d)1:50, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 26. The reactance offered by 300 mH inductor to an AC supply of frequency 50 Hz is, , b)94.2 Ω, , a)1046 Ω, sol:, , c)9420Ω, , d)104.6Ω, , XL = ωL = 2πν L,, =2, , 3.14, , = 3.14, , 3, , 50, , 10-3, , 300, , 10 = 94.2 Ω, , 27. The r.m.s value of an AC voltage with a peak value of 311 V, , a) 110 V, sol :, , Erms =, , š, , √, , =, , a, , = 0.707, , b) 220 V, , d) 70.7 V, , c) lead, , d) air, , √, , 311 = 220 V, , 28. The core used in audio frequency choke is, , a) iron, , c) 50 V, , b) carbon, , 29. A power of 11,000 W is transmitted at 220 V. The current through the line wire is, , a) 50 A, sol:, , b) 5 A, , c) 500 A, , d) 0.5 A, , P = VI, ⇒I=, , +, , f, , =, , = 50 A, , 30. In a transformer, eddy current loss can be minimized by using, , b) laminated core made of stelloy, d) thick copper wires, , a) laminated core made of Mumetal, c) shell type core, , www.Padasalai.Net, 31. The Q - factor of an a.c circuit containing a resistance R, inductance L, and capacitor C is, , a)Q =, , _, , ì, P, , b) Q =, , <, , _, , ¥, , c) Q =, , ì, , _, , P, , 1 `, d) Q = _, X î, , 32. In a three phase AC generator the three coils are fastened rigidly together and are, , displaced from each other by an angle, a) 900, b) 1800, 33. In RLC circuit, at resonance, , a) current is minimum, c) circuit is purely inductive, , d) 3600, , b) impedance is maximum, d) current is in phase with the voltage, , 34. In LCR series circuit at resonance, , a) impedance (Z) is maximum, c) impedance (Z) is equal to R, , c) 1200, , b) current is minimum, q N =1/√LC, , 35. A coil of the area of cross section 0.5 m2with 10 turns is in a plane which is parallel to a, , uniform magnetic field of 0.2 wb/m2. The magnetic flux through the coil is, a) 100 Wb, b) 10 Wb, c) 1 Wb, d) zero, sol: the coil is parallel to an uniform magnetic field. So θ = 900 ;, angle θ is in between direction of magnetic field and normal to the surface of the coil., Φ = nBA cos θ, = 10 0.2 0.5 cos 90 = 0, , 112, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 36. The rms value of AC flowing through a resistor is 5 A. Its peak value is, , a) 3.536 A, sol:, , Irms =, , õ, , c) 7.07 A, , b) 70.7 A, , d) 7 A, , √, , ⇒ < = Irms x √2, =5, , 1.414 = 7.07A, , 37. The effective value of alternating current is, , b) I0/√=, , a) I0/2, , c) I0√2, , d) 2I0, , 38. In an AC circuit average power consumed is 200 W and the apparent power is 300 W. The, , power factor is, a) 1.5, , b) 0.66, , KL: Pav ( Apparent power x power factor, , power factor (, , Û4T, , Üùù46 ²± ù-d 6, , (, , c)0.33, , d) 1, , = 0.66, , a, , 39. A rectangular coil is uniformly rotated in a uniform magnetic field such that the axis of, , rotation is perpendicular to the direction of the magnetic field. When the plane of the coil, is perpendicular to the magnetic field, a) (i) magnetic flux is zero (ii) induced emf is zero, b) (i) magnetic flux is maximum (ii) induced emf is maximum, c) (i) magnetic flux is maximum (ii) induced emf is zero, d) (i) magnetic flux is zero (ii) induced emf is maximum, Sol:The coil is perpendicular to the direction of the magnetic field . so θ = ωt = 00, Flux ɸ = NBA cos θ = NBA cos 0, = NBA (1) = NBA → >ey f, emf e = E0 sinωt = E0 sin0 = 0, , www.Padasalai.Net, 40. In a series LCR circuit, at resonance, , a) XL = XC, , b) XL>XC, , c) XL<XC, , d) ω=, , P, , 41. In a.c circuit voltage leads the current by a phase of ?/2 , then the circuit has, , a) only an inductor L b) only a capacitor (C) c) Only resistor (R) d) L, C and R in series, 2, Sol: voltage (e) = E0 sinωt ; current ( i) = Io . sin (ωt – )., Hence in inductor circuit the voltage leads the current by a phase of U/2, , ν0. The inductance is doubled. The capacitance, also doubled. Now the resonant frequency of the circuit is, a) 2 ν0, b) ν0/2, c) ν0 /4, d) ν0 /√2, , 42. The resonant frequency of RLC circuit is, , Sol:, , νo =, , 2√ P, , ;, , [L & C are doubled ], , ′, , νo =, , 2√, , P, , =, , 2‘ P, , = νo/2, , 43. When the frequency of AC increases , the capacitive reactance offered by capacitor, , connected in the circuit, a) increases, b) decreases, c) remains the same, Hints:, XC = ωX = 2πνX;, The capacitive reactance is inversely proportional to the frequency, , 113, , J.SHANMUGAVELU, , [P.G. T. in Physics], , d) becomes zero, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 44. The instantaneous emf and current equations of an a.c. circuit are respectively, , e = 200 sin (ωt+?/Â ) and i = 10 sin ωt .The average power consumed over one complete, cycle is:, a) 2000 W, b) 1000 W, c) 500 W, d) 707 W, sol:, phase difference ϕ = 60;, , Pav =, , š, , õ, , √, , √, , Pav = (200, , cos ϕ = (E0, , 0.707), , = 141, , 7.07, , 0.707), , (10, , (I0 x 0.707) cos 60, , 0.707), , = 498.435 ≃ 500 W, , 45. In a series RLC circuit, the instantaneous values of current and emf are, , i = I0 sin(ωt - ?/3) and e = E0 sin ωt respectively. The phase difference between the, current and voltage is :, a) zero, b) 1800, c) 600, d) 450, sol:phase difference = xωt 8 ωt 8 U/3 | = U/3 =180/3 = 60°, , 46. For a D.C circuit the value of capacitive reactance is, , b) infinity, c) U/2, d) U, XC = ωX = 2πνX ; ν is the frequency of the a.c. supply., , a) zero, sol:, , In a d.c. circuit ν = 0, , ∴ XC = ∞, 47. In an A.C circuit the applied emf e = E0 sin (ωt + ? /2) leads the current, , www.Padasalai.Net, i = I0 sin(ωt - ? /2) by, a) U/2, , b) U/4, , sol:5§ωt 7 ¨ 8 §ωt 8 ¨9 = 7, , c) ?, , =U, , d) 0, , 48. A DC of 5A produces the same heating effect of as an AC (alternating current) of, , a) 50 Arms current, , b) 5 A peak current c) 15 Arms current, , d) none of these, , 49. If an emf of 25 V is induced when the current in the coil changes at the rate of 100As-1,, , then the coefficient of self induction of the coil is, a) 0.3 H, b) 0.25 H, c) 2.5 H, #, Sol:, L = äõ, , =, , d) 0.25 mH, , æäë, E, , = 0.25 H, , 50. Eddy current in a transformer can be minimized by using a laminated core made of, , a) Stelloy, c) Soft iron, , b) mumetal, d) Silicon steel, , 51. In LCR series a.c circuit, the phase difference between current and voltage is 600 The, , reactance of the circuit is 17.32 Ω. The value of resistance is, a) 30 Ω, b) 17.32Ω, c) 10 Ω, Sol:, , tan ϕ =, , àX #àY, , ⇒ R=, , àX #àY, , =, , 114, , d) 1.732Ω, , ì, , ±4² Z, Ï.a, , ±4², , J.SHANMUGAVELU, , =, , Ï.a, √a, , =, , Ï.a, , .Ïa, , = 10Ω, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 52. In an A.C circuit , the instantaneous values of emf and current are respectively e = 200 sin, ?, , ?, , §i¸ 8 ¨; I = 10 sin §i¸ 7 ¨. The phase relation between current and voltage is :, Â, j, ?, , a) voltage lags behind current by a phase angle of Â, ?, , b) current leads voltage by a phase angle of j, , ?, , c) current leads voltage by a phase angle of =, ?, , d) voltage leads current by a phase angle of =, Sol: Phase difference = maximum – minimum, , = 5§ωt 7 ¨ 8 §ωt 8 a ¨9, = ωt 7, =, , F, , =, , 8 ωt 7, a, , a, , =, , 7, , a, , =, , 53. The Q - factor of series resonant circuit is, , a) Q =, , b) Q = ì _, , P, , P, , c) Q =, , √ P, , d) Q = _¥, <, , 54. If the frequency of AC circuit connected with an inductor of inductance 0.03 H only is 50, , Hz, then inductive reactance is:, a) 3.14 Ω, b) 9.42 Ω, Sol:, XL = L k ( ` 2UN, , = 0.03, , c) 3 Ω, 3.14, , 2, , d) 6.28 Ω, , 50, , www.Padasalai.Net, = 9.42Ω, , 55. The instantaneous emf and current equations of an RLC series circuit are e = 200 sin §i¸ 8 j¨, i = 20 sin §i¸ 7 ¨ The average power consumed per cycle is :, j, a) zero, b) 2000 W, c) 1000 W, Pav = Erms Irms cos ϕ, Sol:, , ?, , ?, , =, , =, =, , š, , √, , √, , õ, , √, √, , d) 500 W, , cos ϕ, , cos 60, = 1000 W, , 56. A rectangular coil of wire is placed in a uniform magnetic field such that the plane of the, , coil is parallel to the magnetic field. The magnetic flux linked with the coil and the emf, induced are respectively, a) zero and zero, b) zero and maximum, c)maximum and zero, d) maximum and zero, sol: The coil is parallel to the direction of the magnetic field . so θ = ωt = 900, Flux ɸ = NBA cos θ = NBA cos 90, = NBA (0) = 0, Hints: coil is does not rotate ( placed ) in a magnetic field . So magnetic flux linked with the, coil is does not change., emf e = 0, , 115, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 57. The instantaneous current in an AC circuit containing a pure inductor is i = I0 sin ωt. The, , instantaneous emf is :, ?, a) e = E0 sin §m 7 = ¨, b) e = E0 sin §ωt 8 ¨, a) e = E0 sin ωt 8 U, b) e = E0 sin ωt 7 U, Hints : in an a.c. circuit containing a pure inductor the current i lags behind the voltage e by the, , phase angle of . Conversely the voltage across L leads the current by the phase angle of ., So e = E0 sin §ωt 7 ¨, 58. In RLC series AC circuit at resonance, , a) resistance is zero, c) impedance is maximum, , b) net reactance is zero, d) voltage leads the current by a phase angle, , 59. An LCR series circuit is connected to 240 V A.C. supply. At resonance, the values of VR′ VL, and VC are respectively :, a) 80 V, 80 V and 80 V, b) 120 V, 60 V and 60 V, c) 240 V, 120 V and 120 V, d) 180 V, 40 V and 40 V., , Hints : The applied voltage at resonance is the potential drop across R, because the potential, drop across L is equal to the drop across C and they are 180o out of phase. Therefore they, cancel out and only potential drop across R will exist., , Mathematically , V = _eì 7 e 8 eP, At resonance e ( eP, , V = _eì, , www.Padasalai.Net, ( eì, , Voltage supplied by the alternating source of emf = voltage across resistor, , CHOOSE THE CORRECT ANSWER FROM THE OPTIONS GIVEN, 60. The reverse effect of Oersted experiment was demonstrated by ___, , a) Faraday, , b) Ohm, , c) Henry, , d) Lenz, -1, , 61. An emf of 12 volt is induced when the current in the coil changes at the rate of 40 As ., , The co-efficient of self induction of the coil is ____, a) 0.3 H, b) 0.003 H, c) 30H, , 62. Transformer works on ____, , a) AC only, c) both AC and DC, , d) 4.8 H, , b) DC only, d) Ac more effectively than DC, , 63. A fuse wire has a current rating of 5 A. Then the peak value of the current in the fuse wire, , is __, a) 0.7A, , c) 7.07 A, , b) 1 A, , d) 70.7 A, , 64. The power loss is less in transmission lines when ___, , a) voltage is less but current is more, c) voltage is more but current is less, , 116, , J.SHANMUGAVELU, , b) both voltage and current are more, d) both voltage and3 current are less, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 65. A generator produces an emf given by e=141 sin 88t. The frequency and rms value of, , voltage are _____, a) 50 Hz and 99.7 V, c) 14 Hz and 99.7 V, , b) 7 Hz and 49.5 V, d) 50 Hz and 49.5V, , 66. The emf in an AC containing only inductance will ___, , ?, , b) be ahead of current by =, d) always be out of phase., , a) lag behind current by, c) have current in phase with the applied voltage, , 67. An ideal transformer has a power input of 10kW. The secondary current is 25 A. If the, , ratio of number of turns in the primary and the secondary coils is 5:1, then the potential, difference applied to the primary is ___, a) 100V, b) 200 V, c) 2000 V, d) 1500 V, 68. Which of the following devices does not allow d.c.to pass through?, , b) capacitor, , a) resistor, , c) inductor, , d) all the above, , 69. In a LCR circuit, when Xl = Xc m____, , a) Current is minmum, impedance is maximum, b) current is maximum, impedance is maximum, c) current is maximum, impedance is minimum, d) current is minimum, impedance is minimum., , 70. An aeroplane having a wingspan of 35 m flies at a speed of 100 m/s., , If the vertical, component of earth’s magnetic field is 4X10-4 T, then the induced emf across the wingspan, is, a) 28 V, b) 2.8 V, c) 14 V, d) 1.4 V, , www.Padasalai.Net, 71. The co-efficient of mutual induction between a pair of coils depends upon ___, , a) size and shape of the coil, c) proximity of the coil, , b) number of turns of the coil, d) all the above, , 72. The generator rule is ____, , a) Fleming’s left hand rule, c) Maxwell’s right hand cork screw rule, , b) Fleming’s right hand rule, d) Ampere’s swimming rule, , 73. A field of induction 20 T acts at right angles to a coil of area 20 m2 with 50 turns. The flux, , linked with the coil is, a) 2000 Wb, b) 20000 Wb, , c) 0 Wb, , d) 200 Wb, , 74. The frequency of A.C for guided rocket is ___, , a) 50 Hz, , b) 400 KHz, , c) 400 Hz, , d) 100 KHz to 100 MHz, , 75. The number of lines of force crossing unit area normally is ____, , a) Magnetic flux, , b) magnetic induction, , c) induced emf, , d) total flux, , 76. If in an LCR circuit, XL= 500n, Xc= 326.8n , R = 100 n, then o= ________, , a) 600, , b) 300, , c) 450, , d) 900, , 77. Lenz’s law is in accordance with law of conservation of _____, , a) charges, , 117, , b) momentum, , J.SHANMUGAVELU, , c) mass, , [P.G. T. in Physics], , d) energy, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 78. Q factor has values lying between ___ for normal frequencies, , a) 0 to 10, , b) 10 to 50, , d) 10 to 100, , c) 50 to 100, , 79. Cores of chokes used in low frequency AC circuits are made of ___, , a) stelloy, , c) iron, , b)mumetal, , d) aluminum, , 80. In an A.C.circuit the applied emf e=E0sin(i¸ 7 ) leads the current I=I0sin(i¸ 8 ) by, ?, , ?, , =, , ____, a), , =, , c) ?, , b), , d) 0, , 81. At what rate must the current change in a 65 mH coil to have a 1 volt self induced emf?, , a) 25 As-1, , b) 17 As-1, , d) 15.4 As-1, , c) 25.4 As-1, , 82. Power loss due to Joule – heating is also called as ____, , a) copper loss, , b) Eddy current loss, , c) flux leakage, , d) Hysteresis loss, , 83. The equation of a 25 cycle current sine wave having rms value of 30 A is _____, , a) 30 sin 157 t, , c) 30√= sin 157 t, , b) 30 sin 150 t, , d) 30 sin 160 t, , 84. The power factor of a choke coil having inductance ‘L ‘ and resistance ‘r’ is given by, , 7k `, , a) √, , b), , √, , Fp, , c), , d), , ‘ = Fi= =, , 7k `, , 85. The Q factor of an a.c circuit containing a resistance R, inductance L and a capacitor C is, , www.Padasalai.Net, ____, , a) q (, , b) q ( ì _, , P, , √ P, , z r ( _¥, , d) q (, , <, , √ ì, , 86. The part of the AC generator that passes the current from the coil to the external circuit is, , ___, a) field magnet, , b) slip rings, , c) split rings, , d) brushes, , 87. The peak voltage and peak current in a circuit containing resistor alone are 220 V and 1 A, , respectively then the power in the circuit is, a) 110 W, b) 11 W, c) 110 kW, , d) 0, , 88. The energy stored in a coil of inductance 5H and resistance 20 n, when the emf applied to, , the coil is 100 volt is, a) 62.5 J, b) 125 J, , c) 12.5 J, , d) 15.6 J, , 89. _____ transmits large amount of power with low cost and high efficiency., , a) two phase alternator, c) single phase alternator, , b) three phase alternator, d) none of the above, , 90. AC frequency of 100 kHz to 100 MHz is required for _____, , a) satellite purpose, c) transmission of audio and video signals, , 118, , J.SHANMUGAVELU, , b) domestic purpose, d) high transmission, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 91. Power dissipation in an AC circuit in which voltage and current are given by E=300, , sin(i¸ 7 = ) and 1=6 sin i¸ is ____, a) 0 Watt, b) 750 Watt, ?, , c) 375 Watt, , d) 1500 Watt, , 92. How much current is drawn by the primary of a transformer connected to a 220 V supply,, , when it delivers power to a 110 V and 550 W refrigerator?, a) 55 A, b) 2.5 A, c) 04. A, , d) 44 A, , 93. In low power AC dynaoes magnetic field is provided by ___, , a) permanent magnets, c) horse-shoe magnets, , b) electromagnets, d) cylindrical magnets, , 94. The average power of an ac is also called ___ of the circuit., , a) true power, c) resonant power, , b) instantaneous power, d) RMS power, , 95. As the coil rotates with an angular velocity (i sº 8º uniform magnetic field, the emf, , induced is maximum when ___, ?, a) kï ( 0, b) i¸ (, =, , c) kï (, , d) kï (, , a, , 96. In a three phase AC generator, the three coils are inclined at an angle of ___, , a) 450, , c) 1200, , b) 900, , d) 1800, , 97. For an ideal transformer, efficiency t su ____, , www.Padasalai.Net, a) greater than one, , b) less than one, , 98. Induction motors are used in ____, , a) Grinders, , b) Generator, , c) equal to one, , d) infinity, , d) Fans, , c) Refrigirators, , 99. The co-efficient of Mutual inductance of a pair of coils is 4 mH. If the current in one of the, , coils change from 0.6 A to 0.61 A in 0.02 seconds, then induced is, a) 20•V In the same coil, b) 2 mV in the other coil, c) 20 •V in the other coil, d) 20 mV in the same coil, In a step-down transformer, the following condition satisfied, a) Es>Ep, b) K<1, c) Ip>Is, , 100., , d) Np<Ns, , A wire cuts across a flux of 0.2X10-2 weber 0.12 second. What is the emf induced in the, wire?, a) 0.06 V, b) 0.02 V, c) 0.0167 V, d) 0.24 V, , 101., , Choke coils are commonly seen in _____, a) Incandescent bulbs b) fluorescent tubes c) stabilizer circuits, , 102., , The unit of self inductance is ____, #µpè, B v #ëw mÍ, a), b) µpè, ëw mÍ, , 103., , B v, , c), , B v, , #ëw mÍ, , µpè, , A circuit will have flat resonance. If the Q value is _____, a) high, b) infinits, c) low, , d) radio, , d), , B v, , #ëw mÍ, , Ëxdë, , 104., , 119, , J.SHANMUGAVELU, , [P.G. T. in Physics], , d) Zero, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 400 MW power produced at 15,000 V at Neyveli power station is stepped upto ___, before transmission., a) 22,000V, b) 230,000 V, c) 110,000 V, d) 20,000 V, , 105., , In a step-own transformer, the input voltage is 22 kV and the output voltage is 550 V., The ratio of number of turns in the primary to that in the secondary is, a) 1 : 40, b) 40 : 1, c) 1 : 20, d) 20 : 1, , 106., , Flux loss can be minimized by ___, a) using wires of low resistance, c) using shell type of core, , 107., , b) using laminated core made of mumetal, d) using laminated core made of stelloy, , The co-efficient of mutual induction between two long solenoids S1 and S2, Whose core, is filled with a magnetic material of permeability œ is, o o ¡, o o, o o ¡, o o ¡, a) M=, b), M=, c), M=, d), M=, d, d, d, d, , 108., , ‘3’ MARK QUESTIONS: (2 – Questions: Q.No: 37 & 38), PUBLIC ‘3’ MARKS:, , www.Padasalai.Net, 1. What is electromagnetic induction? ( M – 08,O-16 ), , The phenomenon of producing an induced emf due to the changes in the, magnetic flux associated with a closed circuit is known as electromagnetic, induction., , 2. State Faraday’s laws of electromagnetic induction. (J– 06 ,J– 07 ,O – 07,O-14 ), , First law: Whenever the amount of magnetic flux linked with a closed circuit, changes, an emf is induced in the circuit. The induced emf lasts so long as the, change in magnetic flux continues., Second law: The magnitude of emf induced in a closed circuit is directly, proportional to the rate of change of magnetic flux linked with the circuit., , eα, , äU, äë, , 3. State Lenz’s law. ( O – 08, J - 10 ,M-14), , Lenz’s law states that the induced current produced in a circuit always flows, in such a direction that it opposes the change or cause that produces it., , e=–, , 120, , J.SHANMUGAVELU, , [P.G. T. in Physics], , oäU, äë, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 4. State Fleming’s right hand (generator) rule.(M–07,M–09,O–09,M–10,J–11,M-17), , The forefinger, the middle finger and the thumb of the right hand are held in, the three mutually perpendicular directions. If the forefinger points along the, direction of the magnetic field and the thumb is along the direction of motion of, the conductor, then the middle finger points in the direction of the induced, current. This rule is also called generator rule., 5. Define: self inductance (or) coefficient of self – induction of a coil and give its, , unit. ( J – 09,O-15 ), The coefficient of self induction of a coil is numerically equal to the, opposing emf induced in the coil when the rate of change of current through the, coil is unity. The unit of self inductance is henry (H)., , e=–L, , äõ, , äë, , 6. Define the unit of self inductance. ( J – 12 ), , The unit of self inductance is henry (H), One henry is defined as the self-inductance of a coil in which a change in, current of one ampere per second produces an opposing emf of one volt., 7. What are the methods of producing induced emf in a circuit?(M–06,J–06,O-, , www.Padasalai.Net, 10,M–11,M -12,J-16), , The induced emf can be produced by changing, 1. The magnetic induction (B), 2. Area enclosed by the coil (A), 3. The orientation of the coil (& ) with respect to the magnetic field., , 8. Define efficiency of a transformer? (S- 13, J – 15,J-16 ), , Efficiency of a transformer is defined as the ratio of output power to the, input power., - ±ù ± ù-d 6, y=, , %²ù ± ù-d 6, , The efficiency y = 1 (ie. 100%), only for an ideal transformer where there is, no power loss. But practically there are numerous factors leading to energy loss, in a transformer and hence the efficiency is always less than one., 9. Define: rms (effective) value of AC. ( J – 07, O – 09,M-13,J-14 ), , The rms value of alternating current is defined as that value of the steady, current, which when passed through a resistor for a given time, will generate the, same amount of heat as generated by an alternating current when passed through, the same resistor for the same time., , 121, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 10. A capacitor blocks d.c. but allows a.c. Why ?( O – 11 ), , Capacitive reactance XC =, , pP, , =, , ]P, , Where N is the frequency of the a.c. supply., In a d.c. circuit, N = 0, , ∴ XC = ∞, , Thus a capacitor offers infinite resistance to d.c. and blocks the d.c., For an a.c., the capacitive reactance varies inversely as the frequency of a.c. and, also inversely as the capacitance of the capacitor., 11. Why a DC ammeter cannot read AC ?( O – 07 ), , The average value of the alternating current or emf over one complete cycle, is zero. A d.c ammeter or voltmeter connected in an a.c circuit will does not, show any deflection. Hence a d.c ammeter cannot read a.c., , 12. Define: Q- factor (or) quality factor. ( O – 06, J – 11, M- 15 ), , www.Padasalai.Net, The Q factor of a series resonant circuit is defined as the ratio of the voltage, across a coil or capacitor to the applied voltage., i.e. Q =, , T-°±4ø 4,6-.. / -6 i, 4ùù°% ³ T-°±4ø, , 13. Give the differences between AF choke and RF choke. ( J – 08 ), , 1., 2., 3., 4., , 122, , Audio – frequency(A.F) chokes, Used in low frequency a.c. circuit, An iron core is used, The inductance is high, Used in fluorescent tubes, , J.SHANMUGAVELU, , Radio frequency (R. F) chokes, Used in high frequency a.c. circuit, Air core is used, The inductance is low, Used in wireless receiver circuits, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , OTHER IMPORTANT ‘3’ MARKS:, 14. Define magnetic flux., , The magnetic flux (z) linked with a surface held in a magnetic field (B) is, defined as the number of magnetic lines of force crossing a closed area (A)., ^_ . _ = BA cos θ, z=6, , 15. Define self induction., , The property of a coil which enables to produce an opposing induced emf in, it when the current in the coil changes is called self induction., 16. Define mutual induction., , Whenever there is a change in the magnetic flux linked with a coil, there is, also a change of flux linked with the neighbouring coil, producing an induced, emf in the second coil. This phenomenon of producing an induced emf in a coil, due to the change in current in the other coil is known as mutual induction., 17. Define: Mutual inductance (or) coefficient of mutual – induction., , The coefficient of mutual induction of two coils is numerically equal to the, emf induced in one coil when the rate of change of current through the other, coil is unity. The unit of coefficient of mutual induction is henry., , www.Padasalai.Net, W, )æ, , M = - äõ, , äë, , 18. What are the factors depend on which coefficient of mutual induction?, , i) Size and shape of the coils, number of turns and permeability of material on, which the coils are wound., ii) Proximity of the coils, 19. What is meant by Eddy current?, , Foucault in the year 1895 observed that when a mass of metal moves in a, magnetic field or when the magnetic field through a stationary mass of metal is, altered, induced current is produced in the metal. This induced current flows in, the metal in the form of closed loops resembling ‘eddies’ or whirl pool. Hence, this current is called eddy current. The direction of the eddy current is given by, Lenz’s law., 20. State the principle of ac single phase generator?, , It is based on the principle of electromagnetic induction, according to which, an emf is induced in a coil when it is rotated in a uniform magnetic field., , 123, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 21. What is called transformer? State its principle, , Transformer is an electrical device used for converting low alternating, voltage into high alternating voltage and vice versa. It transfers electric power, from one circuit to another. The transformer is based on the principle of, electromagnetic induction., 22. What is meant by alternating current?, , A rotating coil in a magnetic field, induces an alternating emf and hence an, alternating current. Since the emf induced in the coil varies in magnitude and, direction periodically, it is called an alternating emf., 23. Mention the differences between step up and step down transformer., , Step up transformer, 1. It increases the voltage by, 2., , 3., 4., , decreasing the current, The number of turns in the, secondary is greater than that in, the primary, Transformer ratio k >1, Output (secondary) voltage is, greater than that of input (primary), voltage, Input (primary) current is greater, than that of Output (secondary), current, , Step down transformer, It decreases the voltage by increasing, the current, The number of turns in the secondary, is less than that in the primary, Transformer ratio k <1, Output (secondary) voltage is less, than that of Input (primary) voltage, , www.Padasalai.Net, 5., , input (primary) current is less than, that of Output (secondary) current, , 24. What is inductive reactance?, , The resistance offered by an inductor is called the inductive reactance., XL = k` ( 2UN L. Its unit is ohm., , 25. What is capacitive reactance?, , The resistance offered by an capacitor is called the capacitive reactance., XC =, , pP, , =, , ]P, , . Its unit is ohm., , 26. Define resonant frequency in RLC circuit., , The particular frequency νo at which the impedance of the circuit becomes, minimum and therefore the current becomes maximum is called resonant, frequency of the circuit., , N (, , 124, , J.SHANMUGAVELU, , √ P, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 27. What happens to the value of current in RLC series circuit if frequency of, , the source is increased?, , As the frequency increases the inductive reactance XL = k` ( 2UN`, also, , increases. But the capacitive reactance XC =, =, decreases. Hence the, pP, ]P, reactance of the RLC circuit (XL – XC) increases. So the impedance, Z = ‘î 7 \ 8 \P increases. Thus the value of current in RLC series circuit is, decreases, 28. What is acceptor circuit ? Give the uses of it., , The series resonant circuit is often called an ‘acceptor’ circuit. By offering, minimum impedance to current at the resonant frequency it is able to select or, accept most readily this particular frequency among many frequencies. In radio, receivers the resonant frequency of the circuit is tuned to the frequency of the, signal desired to be detected. This is usually done by varying the capacitance of, a capacitor., 29. Define power factor?, , ßµË (, , š õ, , Cos z, , www.Padasalai.Net, =, , š, , √, , =t, , ., , õ, , √, , Cos z, , pÍ < pÍ, , Cos z, , Pav = apparent power × power factor, , where Apparent power = Erms Irms and power factor = cos z, , The average power of an ac circuit is also called the true power of the circuit., 30. Write short notes on choke coil?, , A choke coil is an inductance coil of very small resistance used for, controlling current in an a.c. circuit. If a resistance is used to control current,, there is wastage of power due to Joule heating effect in the resistance. On the, other hand there is no dissipation of power when a current flows through a pure, inductor., , 125, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , PUBLIC ‘3’ MARK PROBLEMS:, 1. An emf of 5 V is induced when the current in the coil changes at the rate of, , 100 As-1. Find the coefficient of self-induction of the coil. [M-10], äõ, , Given data: e = 5V ;, Solution:, , äë, , = 100 As-1; L = ?, , Induced emf e = - L, , äõ, , äë, , #, , Coefficient of self induction L = äõ, , æäë, , =, , #E, , L = - 0.05 H, -1, , 2. If the rate change of current of 2 As induces an emf of 10 mV in a solenoid,, , what is self- inductance of the solenoid?, , www.Padasalai.Net, äõ, , Given data: e = 10 mV ;, Sol:, , äë, , = 2 As-1;, , L=?, , #, , Self induction L = äõ, , æäë, , =, , #, , S, , = - 5 10#a H, , L = 5 mH, , 126, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 3. Calculate the mutual inductance between two coils when a current of, , 4 A changing to 8 A. in 0.5 s, in one coil, induces an emf of 50 mV in the, other coil. [M-06,M-09, J - 15], , Given data: e = 50 mV ;, Sol:, , Induced emf, , dI = 8 - 4 = 4A ; t = 0.5 s; M = ?, , e=-M, M=-, , e, , äë, , {Q, §{ ¨, , M=-e, M=-, , äõ, , äë, äõ, , |E, , –, , S}, , .E, , = −6.25 × 10−3H, , M = 6.25 mH ., 4. An air craft having a wingspan of 20.48 m flies due north at a speed of, , 40 ms-1. If the vertical component of earth’s magnetic field at the place is, 2 10-5 T, calculate the induced emf between the ends of the wings. [J-06,M08,J-10, O-10,M-11], , www.Padasalai.Net, Data :, , l = 20.48 m;, , v = 40 ms−1;, , B = 2 × 10−5T;, , e=?, , e=− Bbv, , Solution :, , = − 2 × 10−5 × 20.48 × 40, , e = − 0.0164 volt, Negative sign indicates that the emf is produced in opposite direction., 5. An aircraft having a wing span of 10 m flies at a speed of 720 kmph. If the, , vertical component of the earth’s magnetic field is 3 × 10 -5T, calculate the, emf induced between the ends of the wings. ( O – 06 ), , Data :, , l = 10 m;, , Solution :, , v=, , Ï, , a, , =, , Ï, , E, , ms−1;, , B = 3 × 10−5T;, , e=?, , e=−Bbv, = − 3 × 10−5× 10 × 720 × 5 / 18, , e = − 0.06 volt, Negative sign indicates that the emf is produced in opposite direction, , 127, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 6. In an ideal transformer has transformation ration 1:20. If the input power and, , the primary voltage are 600 mW and 6 V respectively, find the primary and, secondary currents. [O-08], , Given data: EP = 6 V;, Solution:, , EpIp = 600 mW = 600, , Primary current Ip =, , š) õ), , 10-3W, +, , ( since I = f ), , š), , S, , =, , Ip = 100 × 10 -3A, õ), õÎ, , =, , Secondary current Is = Ip × 20, = 2000 × 10 -3 A, , Is = 2 A, , www.Padasalai.Net, 7. 11 kW power is transmitted at 22000 V through a wire of resistance 2 Ω., , Calculate the power loss (J-14), , Given data: P = 11000 W;, Sol:, , V = 22000 V ;, , R=2Ω, , +, , I= =, f, , = 0.5 A, Power loss = I2 R, = (0.5)2, , 2 = 0.25, , 2, , Power loss = 0.5 W, , 128, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 8. A solenoid of length 1 m and 0.05 m diameter has 500 turns. If a current of 2, , A passes through the coil, calculate the co-efficient of self-induction of the, coil. [M-13], , Data : l = 1 m;, Sol:, , L=, , d = 0.05 m;, o ¡, d, , o, , =, , =, , «, , r = 0.025 m;, , N = 500 ;, , I = 2A ; L = ?, , d, , E, , a., , ., , E, , = 0.616 10-3, , L = 0.616 mH, 2, , 9. A coil of area of cross section 0.5 m with 10 turns is in a plane perpendicular, , to a uniform magnetic field of 0.2 Wb/m2. Calculate the flux through the coil., [M-07], , Given data: B = 0.2 Wb/m2;, Sol:, , A = 0.5 m2;, , θ = 0°, , Magnetic flux ɸ = BA cos θ, = 10 × 0.5 × 0.2 × cos 0, , www.Padasalai.Net, ɸ = 1 Wb., , 10. A capacitor of capacitance 2 μF is in an ac circuit of frequency 1000 Hz. If the, , rms value of the applied emf is 10 V, find the effective current flowing, through the circuit. ( J – 08,O-15 ), , Given data: N = 1000 Hz;, , 10# F; t, , C=2, , Solution: Capacitive reactance XC =, =, =, , pÍ =, , 10 V;, , pP, , <, , pÍ, , ]P, a., , R, , = 79.6 Ω, Irms =, =, , šŸ~W, à5, , ÏT., , Irm = 0.126 A, 129, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html, , (?
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 14. Magnetic field through a coil having 200 turns and cross sectional area, , 0.04 m2 changes from 0.1 wbm-2 to 0.04 wbm-2 in 0.02 sec. find the induced, emf.[M-14], , Data : N = 200, A = 0.04 m2,, t = 0.02 s, e = ?, Solution :, , äɸ, , e=-, , =-, , äë, , ä, , äë, , ä, , =-, , äë, , B1 = 0.1 wb m−2,, , (ɸ, , (NBA) = - NA ., , = - NA., e = -200, , B2 = 0.04 wb m−2,, , ä, , äë, , #, , äë, , 10-2, , 4, , ., , ., , # ., , e = 24 V, , www.Padasalai.Net, 15. A train runs at a speed of 180 km/hr on a railway track with the two rails, , insulated from each other and the ground and connected to a millivoltmeter. If, the vertical component of earth’s magnetic field is 0.2 × 10-4 Wb/m2 and the, distance of separation between the rails is 1m. what would be the reading in, the voltmeter ?(O-14), , Data : v = 180 km/hr =, b= 1m; e = ?, Sol:, , a, , = 50 ms-1; B = 0.2 10−4 Wb/m2;, , e = -Bb v, = - 0.2 10−4 1 50, = -10-3, , e = -1 mV, , 131, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 2, , 16. An a.c generator consists of a coil of 10,000 turns and of area 100 cm . The, , coil rotates at an angular speed of 140 rpm in a uniform magnetic field of, 3.6 10-2 T. Find the maximum value of the emf induced. [M – 15,M-17], A = 102 cm2 = 10-2 m2;, , Given data: N = 10,000;, N = 140 rpm =, , Sol:, , rps;, , B = 3.6, , 10-2 T;, , E0 = ?, , E0 = NBAω, = NBA 2UN, , = 104 10-2 3.6 10-2 2U, , Ï, a, , E0 = 52.75 V, , 17. Calculate the power loss in the form of heat when a power of 11,000 W is, , transmitted at 220 V. (O-16), , www.Padasalai.Net, Given data: P = 11000 W;, , Sol:, , V = 220 V ;, , +, , I= =, f, , I = 50 A, R=, , f, , +, , =, = 4.4 Ω, , Power loss = I2 R, = 502, , 4.4 = 11000, , Power loss = 11 kW, , 132, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , PUBLIC ‘5’ MARK PROBLEMS:, 2, , 18. An a.c generator consists of a coil of 10,000 turns and of area 100 cm . The, , coil rotates at an angular speed of 140 rpm in a uniform magnetic field of, 3.6 10-2 T. Find the maximum value of the emf induced. [J-09], , Given data: N = 10,000;, rps;, N = 140 rpm =, Sol:, , A = 102 cm2 = 10-2 m2;, B = 3.6 10-2 T;, E0 = ?, , E0 = NBAω, = NBA 2UN, , = 104 10-2 3.6 10-2 2U, , Ï, a, , E0 = 52.75 V, , PUBLIC ‘5’ MARKS: (1-questions: Q. No: 55), 1. State Faraday’s laws and Lenz law in electromagnetic induction.(J-11), , www.Padasalai.Net, 2. Derive an expression for the self - inductance of a long solenoid. (M-14 , J-16,M-17), 3. Obtain an expression for the energy associated with an inductor(O-13), , 4. Obtain an expression for the mutual inductance between two long solenoids., (J-08,M-12), 5. Obtain an expression for the coefficient of mutual induction between two long, solenoids., 6. Explain how an emf can be produced by changing the area enclosed by a coil., (J-07,O-07,M-09,O-12,J-13, J-15,O-15), 7. Explain the applications of eddy currents. (M-07,O-08,M-10,J-14), 8. Explain the energy losses in a transformer? How are they minimised?, (J-06,O-06,O-09,J-10,O-10,M-11,M-13 O-14,M-15), 9. Obtain an expression for current flowing in a circuit containing resistance only to which, alternating emf is applied. Find the phase relation between the current and the voltage, Draw the corresponding graph. (O-11,J-12), 10. Obtain the phase relation between the current and the voltage in an AC circuit, containing an inductor only. Draw the corresponding graph. (M-06,M-08), , 133, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 135 : www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PUBLIC ‘10’ MARKS: (1 –question: Q. No: 65), , 1. Discuss with theory the method of inducing e.m.f in a coil by changing its orientation, , with respect to the direction of the magnetic field., (J - 08, O- 09, J -10, O -11, M - 11, M-13, O-14, M-15), 2. Describe principle, construction, and working of a single phase A C generator., , (M - 07, M - 08, J - 07, O – 07, O - 10, J - 11, M -11, J - 12, J-13,M-14), 3. What are eddy current? Explain their applications. How are they minimized?(M -9), 4. Principle, construction, theory of working of a transformer. Define its efficiency., , Mention the energy losses. (M - 06, M – 12,O-13), 5. Obtain the phase relation between voltage and current in an A.C circuit containing a, , pure inductance. Draw the necessary graph. (O - 08), , (OR), , Obtain an expression for the current in an AC circuit containing a pure inductance only., Find the phase relationship between voltage and current. Draw the necessary graph, ( O-16, 6. In an ac circuit containing a capacitor, the instantaneous emf is e = Eosinωt. Obtain the, , expression for instantaneous current. Explain the phase relation between emf and current, by graph. (O - 06), , www.Padasalai.Net, 7. A source of alternating emf is connected to a series combination of a resistor R, inductor, , L, and a capacitor C. Obtain with the help of vector ( voltage phasor) diagram and, impedance diagram an expression for i) effective voltage ii) the impedance iii) the, phase relationship between current and voltage, [J - 06, J – 09, O -12,J-14, J-15,O-15,M-17], , E ß Ú õÀ • i ² ®, E ß Ú õÀ © m k ÷ © • i ² ®, E ß Ú õÀ • i ¯ õu x J ß Ö ª À ø », Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], , For contact, Email :
[email protected], Phone No: 9952223467, , 134, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, Only for slow bloomers, ( 5 & 10 Mark answer ), 1. Emf induced by changing the area enclosed by the coil:, , L1M1 is a sliding conductor of length l resting on, the arms PQ and RS., A uniform magnetic field ‘B’ acts perpendicular to, the plane of the conductor., When L1M1 is moved through a distance dx in time, dt, Due to the change in area L2L1M1M2, there is a, change in the flux linked with the conductor., Therefore, an induced emf is produced., Change in area dA = b dx, dϕ = B.dA = B b dx, e=8, , ³Z, ³±, , =8, , A b ³0, ³±, , e = -B b v, , www.Padasalai.Net, 2. Energy losses in a transformer:, , Energy losses, Hysteresis loss, , Copper loss, , Eddy current, loss, Flux loss, , Causes, Repeated magnetisation and, demagnetisation of the iron, core caused by the alternating, input current, Current flowing through the, primary and secondary, windings lead to Joule heating, effect, Varying magnetic flux leads to, the wastage of energy in the, form of heat, Flux produced in the primary, coil is not completely linked, with the secondary coil due to, leakage., , To Minimise, Alloys like mumetal and, silicon steel, , Thick wires with low, resistance, Laminated core made of, stelloy, an alloy of steel, Shell type core, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , JEYAM TUITION CENTRE PARAMAKUDI, , www.TrbTnpsc.com, , 2017-2018, , 2. AC generator (Dynamo) – Single phase, , Principle: Electro magnetic induction, Diagram :, , Essential parts:, , Armature : large number of loops or turns of insulated copper wire wound over a, laminated soft iron core., Field magnets :Permanent magnets in the case of low power dynamos and electro, magnets in the case of high power dynamos., Slip rings : The metal rings R1,R2 are connected to the ends of the armature and, they rotate along with armature., Brushes : B1,B2 are the two flexible carbon brushes which pass on the current, from armature to external power ., , www.Padasalai.Net, Working:, The direction of the induced current is given by Fleming’s right hand rule., , First half cycle, Second half cycle, AB moves down, CD moves down, CD moves up, AB moves up, current flows in the coil along current flows in the coil along, DCBA., ABCD., current in outer circuit B1 → B2 current in outer circuit B2 → B1, e = Eo sin ωt, Eo = NBAω, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 3. Eddy currents - methods to reduce , applications, When a mass of metal moves in a magnetic field or when the magnetic field, through a stationary mass of metal is altered, induced current is produced in the, metal. This induced current flows in the metal hence this current is called eddy, current., , Methods to reduce, i), using thin laminated sheets instead of solid metal., ii), holes drilled in the plates, Dead beat galvanometer:, When current is passed through a galvanometer the coil oscillates., Eddy currents are set up in the metallic frame, which opposes further, oscillations of the coil., The oscillations of the coil die out instantaneously, the galvanometer is called, dead beat galvanometer., , Induction furnace :, The material to be melted is placed in a varying magnetic field of high, frequency., A strong eddy current is developed inside the metal. Due to the heating effect, of the current, the metal melts., , www.Padasalai.Net, Induction motors :, Eddy currents are produced in a metallic cylinder called rotor, when it is, placed in a rotating magnetic field., The eddy current initially tries to decrease the relative motion between the, cylinder and the rotating magnetic field and metallic cylinder is set in to, rotation., These motors are used in fans., , Electromagnetic brakes :, The drum rotates along with the wheel when the train is in motion., When the brake is applied, a strong magnetic field is developed and eddy, currents are produced in the drum ., The eddy current oppose the motion of the drum and the train comes to rest., Speedometer :, A magnet rotates inside an aluminium cylinder (drum) according to the speed, of the vehicle., Eddy currents produced in the drum makes it deflected through a certain angle, A pointer attached to the drum moves over a calibrated scale which indicates, the speed of the vehicle., , 138, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 4. Transformer :, principle :- Electromagnetic induction, construction: A transformer consists of primary and secondary coils wound on a soft, iron core. To minimise eddy currents a laminated iron core is used., , Explanation : A varying alternating voltage is given to primary coils. The, magnetic flux changes in the primary coils. Magnetic flux in the secondary coil, changes . An emf is induced in the secondary coils., EP and ES be the induced emf in the primary and secondary coils and NP and NS be the, number of turns in the primary and secondary coils respectively, šÎ, , oÎ, , (, , š), , o), , For an ideal transformer , the input power = output power, Ep Ip = Es Is, šÎ, , (, , šÎ, , (, , š), š), , õ), , õÎ, , õ), , õÎ, , (, , oÎ, , o), , = k (transformer ratio), , www.Padasalai.Net, For step up transformer k > 1 , t > t+ and < < <+, for step down transformer k < 1, , Efficiency, , η=, , Power losses:, i., , 139, , - ±ù ± ù-d 6, %²ù ± ù-d 6, , Hysteresis loss ii) copper loss iii) eddy current loss iv) flux loss, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 5.ELECTROMAGNETIC WAVES, AND WAVE OPTICS, PREPARED BY, J.SHANMUGAVELU M.Sc, B.Ed. [P.G. Assist in Physics], , ONE MARK QUESTIONS: (4-question), BOOK BACK QUESTIONS:, 4. In an electromagnetic wave, a)power is equally transferred along the electric and magnetic fields, b)power is transmitted in a direction perpendicular to both the fields, c)power is transmitted along electric field, d)power is transmitted along magnetic field, 5. Electromagnetic waves are, a) transverse, c) may be longitudinal or transverse, , b) longitudinal, d) neither longitudinal nor transverse, , www.Padasalai.Net, 6. Refractive index of glass is 1.5. Time taken for light to pass through a glass plate of, thickness 10cm is, a) 2 × 10–8s, b) 2 × 10–10s, c) 5 × 10–8s, d) 5 × 10–10s, Sol :, , μ =, , ⇒ Xp =, , P•, , P~, , P•, S, , =, , a, , .E, , ', , = 2 × 108 ms-1;, , Time =, , =, , ±´%,÷² .., P~, , ', , =5 × 10-10s, , ^^_ and magnetic, 7. In an electromagnetic wave the phase difference between electric field `, ^_ is, field ^;, a) π/4, , b)π/2, , 8. Atomic spectrum should be, a) pure line spectrum, c) absorption line spectrum, , 141, , J.SHANMUGAVELU, , c)π, , d) zero, , b) emission band spectrum, d) absorption band spectrum., , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 9. When a drop of water is introduced between the glass plate and plano convex lens in, Newton’s rings system, the ring system, a) contracts, b) expands, c) remains same, d) first expands, then contracts, 10. A beam of monochromatic light enters from vacuum into a medium of refractive index µ., The ratio of the wavelengths of the incident and refracted waves is, b) 1 : µ, c) µ 2: 1, d) 1 : µ 2, a) µ : 1, sol:, , μ=, ⇒, , S, , c•, , c~, , = c•, c, , ~, , ⇒ ѵ : Ñp = μ : 1, 11. If the wavelength of the light is reduced to one fourth, then the amount of scattering is, a) increased by 16 times, b) decreased by 16 times, c) increased by 256 times, d) decreased by 256 times, sol:, , A∝, A1∝, , cÓ, , ;, , 5cæ 9, , Ó, , ⇒A1∝ 256 cÓ = 256 A1, , www.Padasalai.Net, 12. In Newton’s ring experiment the radii of the m th and (m + 4)th dark rings are, respectively √Æmm and √€mm. What is the value of m?, a) 2, b) 4, c) 8, d) 10, îÑ … … … … … . . 1 ;, KL:, p (, pF, , ⇒, , ~, , ~ Ó, , (, , =, , 7 4 îÑ … … … 2, p, , E, , ⇒ =, , pF, , Ï, , p, , pF, , ⇒5 (m+4) = 7m, 5m + 20 =7m, 2m = 20 ;, m = 10, , 13. The path difference between two monochromatic light waves of wavelength 4000 Å is, 2 × 10−7m. The phase difference between them is, a) π, b) 2π, c) 3, d) π/2, sol:, , Φ=, , =, , 142, , c, , •, , «, «, , =U, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 14. In Young’s experiment, the third bright band for wavelength of light 6000 Å coincides, with the fourth bright band for another source in the same arrangement. The wave length, of the another source is, a) 4500 Å, b) 6000 Å, c) 5000 Å, d) 4000 Å, sol:, , X=, , ‚, ä, , n Ñ;, , X3 =, ‚, ä, , X4 =, , ‚, ä, , ‚, ä, , ƒ, , 3 × Ñ …….(1) ;, 4 × Ñ ……..(2);, , Now X3 = X4, , 3×Ñ =q4×Ñ, , ⇒3Ñ ( 4Ñ, ⇒ Ñ (3, , c, , (, , Å, , a, , = 4500Å, , 15. A light of wavelength 6000 Å is incident normally on a grating 0.005 m wide with 2500, lines. Then the maximum order is, b) 2, c) 1, d) 4, a) 3, Sol :, Sin θ= Nmλ, θ, ⇒ m = Sin, [λ, , =, , Õ E, , æ ., , Ú%² T, , E×, , = 3.33 ≈ 3, , www.Padasalai.Net, =, , E, , ., , E, , =, \, , E, , Ó, , E, , Ó, , 16. A diffraction pattern is obtained using a beam of red light. What happens if the red light, is replaced by blue light?, a) bands disappear, b) no change, c) diffraction pattern becomes narrower and crowded together, d) diffraction pattern becomes broader and farther apart, 17. The refractive index of the medium, for the polarising angle 60ois, a) 1.732, b) 1.414, c) 1.5, d) 1.468, sol:, μ = tan yè, ⇒ μ = tan 60 = √3 =1.732, , PUBLIC QUESTIONS, 18. The existence of electromagnetic waves was confirmed by, a)Hertz, b)Maxwell, c) Huygens, , d) Plank, , 19. When a ray of light is incident on a glass surface at polarising angle of 57.5°, the angle, between the incident ray and the reflected ray is, a) 57.5°, b) 32.5°, c)115°, d) 90°, sol: angle between the incident ray and the reflected ray = i+r;, (i = r), , i + i = 57.5°+57.5° = 115°, , 143, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 20. Unpolarised light passes through a tourmaline crystal. The emergent light is analysed by, an analyser. When the analyser is rotated through 90°, the intensity of light, a) remains uniformly bright, b) remains uniformly dark, d) varies between maximum to zero, c) varies between maximum to minimum, Hints: the emergent light is polarised light, 21. Velocity of the electromagnetic waves through vacuum is, <, a)√•ₒNₒ, b) œₒ ₒ, c)‘•ₒ/Nₒ, √, , 22. In a plane transmission grating, the unit of grating element is, a) no unit, b) metre, c) metre-1, , d)‘Nₒ /•ₒ, d) degree, , 23. A ray of light is incident on a glass plate at its polarising angle. The angle between the, Incident ray and the reflected ray is, a) 57.5°, b) 32.5°, c)90°, d)115°, sol: angle between the incident ray and the reflected ray = i+r;, (i = r), , i + i = 57.5°+57.5° = 115°, 24. Which one of the following is not an electromagnetic wave?, a) X – rays, b) γ – rays, c) UV – rays, , d) β – rays, , 25. If C is the velocity of light in vacuum, the velocity of light in a medium with refractive, index μ is, a) μ C, b) C/μ, c)μ /C, d) 1/μC, P•, , www.Padasalai.Net, μ=, , sol:, , P~, , ⇒ cm=, , P•, μ, , =, , =, , μ, , 26. A ray of light passes from a denser medium into a rarer medium. For an angle of, incidence of 45°, the refracted ray grazes the surface of separation of the two media. The, refractive index of the denser medium is, a) 3/2, b) 1/√2, c) √2, d) 2, 7m 7, , sol, μ=, 7m, light passes from a denser medium into a rarer medium, μ, , =, , 7m, , 7m 7, , .%² T, , = .%², =, , æ, √, , E, , =√2, , 27. Of the following, which one is a uniaxial crystal?, a) Mica, b) Aragonite, c) Topaz, , d) Quartz, , 28. The radiations used in physiotherapy are, a) Ultra violet, b) infra red, , d)microwaves, , 144, , J.SHANMUGAVELU, , c) radio waves, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 29. In Newton’s rings experiment, light of wavelength 5890 Å is used. The order of the dark, ring produced where the thickness of the air film is 0.589 μm is, a) 2, b) 3, c) 4, d) 5, Sol:, 2t = nλ, , ⇒n=, =, =, , ±, , c, , R, , .E T, , E T, , E T, , =2, , E T, , 30. Of the following, optically active material is, a) Sodium chloride, b) calcium chloride c) sodium, 31. Electric filament lamp gives rise to, a) Line spectrum, c) band spectrum, , d) chlorine, , b) continuous spectrum, d) line absorption spectrum, , 32. In Young’s double slit experiment, the separation between the slits is halved, and the, distance between the slits and the screen is doubled. Then the fringes width is, a) Unchanged, b) halved, c) doubled, d) quadrupled, ä, ‚, Sol : Initial fringe width β = Ñ, now [q £ = ; ƒ £ = 2D ], ‚ˆ, ä, , Ñ, , www.Padasalai.Net, Final fringe width, , β' =, , =, , äˆ, , ‚, , q, 2, , Ñ =, , ‚, , ä, , Ñ =4‰, , 33. The phenomenon of light used in the formation of Newton’s rings is, a) Diffraction, b) interference, c) refraction, d) polarisation, 34. An example for uniaxial crystal is, a) Tourmaline, b) Mica, , c) Topaz, , d) Selenite, , 35. In Raman effect, the spectral line with lower frequency than the incident frequency is, a) Fraunhofer line, b) Rayleigh line, c) Stokes’ line, d) Anti stokes’ line, 36. The optical rotation does not depend on, a) Concentration of the solution, c) the temperature of the solution, , b) frequency of the light used, d) intensity of the light used, , 37. Which of the following gives rise to continuous emission spectrum ?, a) Electric filament lamp, b) sodium vapour lamp, c) gases in the discharge tube, d) calcium salt in Bunsen flame, 38. The transverse nature of light waves is demonstrated by the phenomenon, a) interference, b) diffraction, c) polarization, d) reflection, , 145, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 39. If the velocity of light in a medium is 2.25 × 108 ms-1 then the refractive index of the, medium is, a) 1.5, b) 0.5, c) 1.33, d) 1.73, , μ=, , sol:, , =, , P•, , P~, , a, , 2.25, , ', , =1.33, , 108, , 40. The polarising angle for water is 53°4′if the light is incident at the angle on the surface of, water, the angle of refraction in water is, a) 53°4′, b) 26°30′, c)30°4′, d) 36°56′, sol:, r =90 - yè, , = 90 - 53°4′ = 36°56′, 41. In Raman effect, if the scattered photon gains energy, it gives rise to, a) Stokes’ line, b) Anti stokes’ line, c) Stokes’ and Anti stokes’ line, d) Rayleigh line, 42. In case of Fraunhofer diffraction, the wave front undergoing diffraction is, a) Spherical wave front, b) Cylindrical wave front, d) Plane wave front, c) elliptical wave front, 43. A ray of light is incident on a glass surface such that the reflected ray completely plane, polarized. The angle between the reflected ray and the refracted ray is, a) 57.5°, b) 32.5°, c)90°, d)115°, Hints: the reflected and refracted rays are at right angles to each other - Brewster, , www.Padasalai.Net, 44. Soap bubbles exhibit brilliant colours in sunlight due to, a)scattering of light, b)diffraction of light, c)polarization of light, d) interference of light, 45. The radii of Newton’s dark rings are in the ratio, a) 1:2:3, b) √1 ∶ √2 ∶ √3, c) √1 ∶ √3 ∶ √5, Sol: m = √nRÑ ;, ∝, n, ;, n, =, 1,2,3......, √, m, , = √1;, :, , :, , = √2;, a, , a=, , √3, , = √1: √2: √3, , 46. In the grating formula sin θ = Nm λ, the unit of N is, a) metre, b) metre-1, c) No unit, sol:, , d) 1 : 4 : 9, , N=, (or) N =, , .%² /, pc, , µFv, , =, =, , c, , =, =, , p, p, , d) (metre)2, , = m-1, = m-1, , 47. The ratio of the radii of the 4th and 9thdark rings in Newton’s rings experiment is, a) 4 : 9, b) 2 : 3, c) 16 : 81, d) √2: √3, Sol:, , ⇒, 146, , √nRÑ, , m=, Ó, , =, , √, , √T, , =, , a, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 48. Which of the following is not an optically active material?, a) Quartz, b) Sugar crystals, c) Turpentine oil, , 2017-2018, d) Calcium chloride, , 49. In Raman effect, the incident photon makes collision with an excited molecule of the, substance. The scattered photon gives rise to, a) Stokes’ line, b) Anti Stokes’ line, c) Rayleigh line, d) Zeeman line, 50. Refractive index of glass is 1.5. The velocity of light in glass is, a) 2 × 108ms-1, b) 4.5 × 108ms-1, c)3 × 108ms-1, Sol :, , μ =, ⇒ Xp =, , P•, , P~, P•, S, , =, , a, , .E, , ', , d) 1.33 × 108ms-1, , = 2 × 108 ms-1;, , 51. The dark lines found in the solar spectrum are called, a) Raman lines, b) Fraunhofer lines, c) Stokes lines, , d) anti – stokes lines, , 52. A ray of light travelling in a rarer medium and reflected at the surface of a denser, Medium automatically undergoes a, a) Phase change π/2, b) phase change 2π, d) path difference λ/2, c) path difference λ, 53. Waves from two coherent sources interfere with each other. At a point where the trough, of one wave superposes with the trough of the other wave, the intensity of light is, b) minimum, c) zero, d) no change, a) maximum, Hints: crest meets crest, and trough meets trough→ maximum;, Trough meets crest→ minimum, , www.Padasalai.Net, 54. Nature of wave front corresponding to extraordinary ray inside a calcite crystal is, a) plane, b) spherical, c) elliptical, d) cylindrical, , 55. A ray of light is incident on a plane glass surface at an angle of 57˚30’. The angle between, the reflected ray and the refracted ray is, a) 32˚ 30’, b) 90˚, c) 115˚, d) 57˚ 30’, Hints: the reflected and refracted rays are at right angles to each other - Brewster, 56. In a pile of plates arrangement, the angle between the incident light and the reflected, plane polarized light is, a) 32.5˚, b) 57. 5˚, c) 90˚, d) 115˚, sol: polarizing angle of glass ip= 57. 5 and ip = r;, , angle between the incident light and the reflected plane polarized light= ip + r = 115°, 57. An example of uniaxial crystal is, a) Selenite, b) Mica, , c) Topaz, , d) Calcite, , 58. Angle between the electric component and magnetic component of an electromagnetic, wave is, a)0, b) π/4, c) π/2, d) π, 59. If i is the angle of incidence, the angle between the incident wave front and the normal to, the reflecting surface is, a) i, b) 90ᵒ – i, c) 90ᵒ + i, d) i – 90ᵒ, , 147, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 60. Refractive index of a material for a polarizing angle of 55˚is, a) 1.4281, b) 1.7321, c) 1.4141, sol:, μ = tan yè, , 2017-2018, d) 1.5051, , ⇒ μ = tan 55, =1.4281, 61. In a Nicol prism, the ordinary ray is prevented from coming out of Canada balsam by the, phenomenon of, a) reflection, b) polarization, c) diffraction, d)total internal reflection, 62. In a plane transmission grating the width of the ruling is 12000 Å and the width of the, slit is 8000 Å , the grating element is, a)20•m, b) 2œm, c) 1•m, d) 10•m, sol:, grating element e = a +b, , =12000, , 10-10 +8000 × 10-10, , =20000 × 10-10 = 2 × 10-6 = 2 •m, 63. A nicol prism is based on the principle of, a)refraction, b)reflection, , c)double refraction, , d)diffraction, , 64. Of the following which one is biaxial, a)tourmaline, b) ice, , c)calcite, , d) mica, , www.Padasalai.Net, 65. In young’s double slit experiment , bandwidth β contains, a) a bright band only, b) a dark band only, d) both bright band and dark band, c) either a bright band or a dark band, , 66. According to focault and michelson experiment the velocity of light in a rarer medium is, a)greater than in a denser medium, b)lesser than in a denser medium, c)equal to that in a denser medium, d)either greater or lesser than in a denser medium, 67. Which of the following is a biaxial crystal ?, a) calcite, b) quartz, , c) tourmaline, , d) topaz, , 68. Which of the following is a uniaxial crystal?, a) Tourmaline, b) Mica, , c) Topaz, , d)selenite, , 69. The scattering of sunlight by gas molecules in the earth’s atmosphere is, a)Raman’s effect, b)™ - scattering, c)Tyndal scattering, d)Rayleigh scattering, 70. A light of wavelength 4000 Å after travelling a distance of 2 μm produces a phase change, of :, a) zero, b) 3U, c), d) a, Sol :, , Φ=, , c, , •, , =, , R, , «, , = 10 U = 5. 2 U, = 5. 0 = 0, , 148, , J.SHANMUGAVELU, , [ 2U ( 0 |, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, usº s, , 71. In snell’s law of refraction œ ( usº , œ is :, a) directly proportional to sin i, b) inversely proportional to sin r, d) independent of (a) and (b), c) both (a) and (b), Hints: The refractive index of the medium is constant for particular material., So independent of i and r., 72. A wave of length ′‹′ corresponds to a phase of ‘2?′ calculate the phase when a distance of, ′Œ′ corresponds to a phase of ′ɸ′, =?, c, a) ɸ ( ‹, Œ, b) ɸ (, •, c) ɸ ( •, Ñ, d) ɸ ( c •, 73. A ray of light travelling in air is incident on a denser surface at an angle of 60°. If the, velocity of light in the denser medium is 2 108 ms-1 , the angle of refraction inside the, denser medium is :, <, a) 30°, b)sin -1 (0.75), c) sin -1 § ¨, d) sin -1 (0.6666), P•, , μ =, , Sol :, , =, , P~, , ', , a, , ', , .%² %, , √Â, , a, , = ;, , μ ( .%² 6, ⇒ sin r (, =, , .%² %, S, , √a, , .%² °, aæ, , (, , =, , www.Padasalai.Net, a, , √a, , <, , r = sin -1 §√¨, , 74. A ray of light is incident normally on a glass surface of refractive index 1.5. the angle of refraction, is :, , b)sin -1 (0.666), , a) 30°, , c) zero, , d) sin, , -1, , (0.75), , 75. In newton’s ring experiment, when a wave length of the light λ and a plano canvex lens of radius, of curvature of 50 cm is used, the radius of the 10 th dark ring is√ mm. then with the same wave, length, a plano canvex lens of radius of curvature of 2m is used, the radius of the 10 th dark ring, is:, a) 3mm, b) 2√ mm, c) 3√3 mm, d) 4√3 mm, 2, Sol: rn = nRλ, l, , n=, For R = 50, , n=, , ìª, , a, , 10# m = 0.5 m, R, , 0.5, , (2) = (1) ⇒, , ª, , ………(1), l, , ª, , (, , m, , (, m, , 149, , For R = 2 m, , n=, , l, , ª, , ………(2), , R, , a, 0.5, , ª, R, , a, 0.5, , (4, , 3, , 10#, , = 2 √3 mm, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 76. In Raman effect, wave length of incident light is 5890 Å. The wave length of stokes and, anti stokes lines are respectively, a) 5885 Å and 5880 Å, b) 5895 Å and 5900 Å, c) 5885 Å and 5895 Å, d) 5895 Å and 5885 Å, Hints :, Stokes line :The lines having frequencies lower than the incident frequency. In other words,, the lines having wavelengths higher than the incident wavelength., Anti- stokes line: The lines having frequencies higher than the incident frequency. In other, words, the lines having wavelengths lower than the incident wavelength., 77. Which one of the following is not an electromagnetic waves ?, a) X – rays, c) Ultra violet rays ( UV rays), , b) γ – rays, , d) β – rays, , 78. In Raman effect, wave length of incident light is 5890 Å. The wave length of stokes and, anti stokes lines are respectively, a) 5880 Å and 5900 Å, b) 5900 Å and 5880 Å, c) 5900 Å and 5910 Å, d) 5870 Å and 5880 Å, Hints :, Stokes line :The lines having frequencies lower than the incident frequency. In other words,, the lines having wavelengths higher than the incident wavelength., Anti- stokes line: The lines having frequencies higher than the incident frequency. In other, words, the lines having wavelengths lower than the incident wavelength., 79. Light from a source is analysed by an analyser. When the analyser is rotated, the intensity of, , www.Padasalai.Net, the emergent light :, , a) Does not vary, , b) Remains uniformly dark, c) Varies between maximum and zero, d) Varies between maximum and minimum, , Hints : Light from the source is unpolarised light, , Choose the correct answer from the options given, 80. Electromagnetic waves are, a) mechanical waves, c) transverse in nature, , b) similar to sound waves, d) longitudinal in nature, , 81. If an electromagnetic wave is propagating along x-direction and electric field variation is, along y-direction then the magnetic field variation will be, a) along x-direction, b) inclined at an angle of 450 with x-direction, d) along z direction, c) along y direction, 82. If the energy of the electromagnetic wave is ‘E’ then the energy associated with electric, field vector, š, `, a) E, b) 2E, c), d) =, 83. The velocity of electromagnetic waves in vacuum or free space is, <, a), b) • N, c) œ, Ž, , 150, , ‘ @ @, , J.SHANMUGAVELU, , [P.G. T. in Physics], , d) ‘• N, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 84. Electromagnetic waves are not deflected in electric and magnetic fields, because, a) they travel with very high velocity, b) they are chargeless waves, c) they travel even in vacuum, d) they are transverse in nature, 85. Existance of electromagnetic waves was confirmed experimentally by, a) Maxwell, b) Henry, c) Hertz, d) Huygen, 86. In Hertz experimental arrangement, the two metal plates A and B are placed with a, separation of, a) 6m, b) 60 mm, c) 6 cm, d) 60 cm, 87. Frequency of electromagnetic waves produced by Hertz arrangement was about, a) 5X107 Hz, b) 50X107 Hz, c) 8X1014 Hz, d) 4X1014 Hz, 88. The frequency of oscillation of charges between the plates A and B in Hertz experimental, setup is given by, , a) 2 U√`X, , b), , _, , P, , <, , c) = ?√, , d), , ¥, , _, , P, , 89. Frequency range of electromagnetic spectrum is, a) 10 Hz-105 Hz, b) 103 Hz-1022 Hz, 22, c) 10 Hz-10 Hz, d) 10 Hz-10-5 Hz, 90. Physical properties of electromagnetic waves are determined by their, a) wavelength, b) sources, c) method of excitation, d) all these, , www.Padasalai.Net, 91. The overlapping in certain parts of the electromagnetic spectrum reveals that the, particular wave, a) can be produced by only one method, b) has very high energy values, c) can be produced by different methods, d) has very low energy values, 92. Frequency band of radio waves used in cellular phones is, a) high frequency, b) low frequency, c) ultra high frequency, d) very low frequency, 93. Frequency range of AM band of radio waves is from, a) 54 MHz to 890 MHz, b) 530 KHz to 108 MHz, c) 530 KHz to 1710 KHz, d) 88 MHz to 108 MHz, 94. Radio waves used in television communication is ranging from, a) 54 MHz to 890 MHz, b) 530 KHz to 108 MHz, c) 530 KHz 1710 KHz, d) 88 MHz to 108 MHz, 95. Radiation used to destroy bacteria and for sterilizing surgical instruments are, a) radio waves, b) X-ray, c) ultra violet radiation, d) gamma rays, 96. Microwaves are used in, a) radio communication systems, c) Radar communication systems, , 151, , J.SHANMUGAVELU, , b) television communication systems, d) radio & TV communications, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 97. Infrared absorption spectrum is used to study, a) Crystal structure, b) molecular structure, c) atomic structure, d) electronic configuration, 98. Wavelength range of visible portion of electromagnetic spectrum is from, a) 6X10-10 m-4X10-7 m, b) 4X10-7 m-8X10-7 m, 14, -4, 14, c) 8X10 m X10 m, d) 10-3m-0.3 m, 99. Infrared lamps are used in, a) weather forecasting, c) physiotherapy, , b) Infra red photography, d) sterilizing surgical instruments, , 100. Radiation used in the detection of forged documents and finger prints in forensic, laboratories is, a) IR rays, b) UV rays, c) gamma rays, d) X-rays, 101. UV rays are used to find, a) crystal structure, c) structure of atoms, , b) nuclear structure, d) molecular structure, , 102. Match the radiations with their applications, A) IR rays, - a) molecular structure, B) UV rays, - b) crystal structure, C) X-rays, - c) nuclear structure, D) •-rays, - d) structure of atoms, b) A(a) B(d) C(b) D(c), a) A(a) B(b) C(c) D(d), c) A(c)B(b) C(d) D(a), d) A(a)B(c) C(d) D(b), , www.Padasalai.Net, 103. When the light emitted by the source is directly examined by a spectrometer, the, spectrum obtained is, a) continuous spectrum, b) emission spectrum, c)band spectrum, d) absorption spectrum, 104. Wavelength of two sodium lines (D1 and D2) are, a) 8590A and 8596 A, b) 5893A and 5890A, c) 5896A and 5890A, d) 6958A and 6950 A, 105. Spectrum produced by incandescent solid at high temperature is, a) continuous emission spectrum, b) line emission spe4ctrum, c) continuous absorption spectrum, d) line absorption spectrum, 106. Spectrum produced by electric filament lamp, a) depends on temperature of the source only, b) is independent of temperature of the source, c) depends on characteristic of the source, d) depends on method of excitation, , 152, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 107. The spectrum, consisting of unbroken luminous bands of all wavelengths containing, all colours from red to violet is given by, a) mercury in mercury vapour lamp, b) Calcium and barium salts in Bunsen flame, c) carbon arc lamp, d) CO2 gas in molecular state in discharge tube, 108. Free excited atoms emit ____, a) continuous emission, c) band emission, , b) line emission, d) line absorption, , 109. _____ spectrum is characteristic of the emitting substance and is used to identify the, gas., a) Continuous emission, b) Line emission, c) Continuous absorption, d) Line absorption, 110. Ammonia and nitrogen in molecular state in the discharge tube gives ___ spectrum, a)continuous emission, b) line emission, c) band emission, d) line absorption, 111. Spectrum, which is the characteristic of the absorbing substance is, a) line emission spectrum, b) band emission spectrum, c) absorption spectrum, d) emission spectrum, 112. When the temperature of the solid is increased, the spectrum spread from, a) red to green, b) blue to green, c) red to blue, d) violet to red, , www.Padasalai.Net, 113. If a white light is allowed to pass through the solution of blood or Chlorophyll the, resulting spectrum is, a) line absorption, b) band absorption, c) continuous absorption, d) line emission, 114. The spectrum used for making dye is, a) line absorption, c) continuous absorption, , b) band absorption, d) line emission, , 115. Dark lines appearing in the solar spectrum are called, a) Raman lines, b) Tyndall lines, c) Fraunhofer lines, d) Rayleigh lines, 116. Temperature of Sun’s outer layer is about, a) 14X106 K, b) 14X107 k, d) 6000 K, c) 600 K, 117. Lifetime of atoms of the substance exhibiting the Phenomenon of fluorescence is, a) more than 10-5 S, b) equal to 10-5 S, -5, c) less than 10 S, d) equal to 103 S, 118. Delayed fluorescence is known as, a) luminescence, c) phosphorescence, , 153, , J.SHANMUGAVELU, , b) bio-luminescence, d) reflection, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 119. According to corpuscular theory, the difference in colours is due to different ____ of, the corpuscles, a) shapes, b) sizes, c) velocities, d) energies, 120. According to corpuscular theory, when the corpuscles approach a surface between two, media, if they are attracted it gives to the phenomenon of, a) reflection, b) scattering, c) refraction, d)interference, 121. “Velocity of light in denser medium is greater than the velocity of light in rarer, medium”—this statement is TRUE in, a) corpuscular theory, b) wave theory, c) electromagnetic theory, d) quantum theory, 122. Experimental results of Foucault on velocity of light do not support, a) corpuscular theory, b) wave theory, c) electromagnetic theory, d) quantum theory, 123. “No material medium is necessary for the propagation of light waves” this statement is, TRUE’ according to, a) corpuscular theory, b) wave theory, c) electromagnetic theory, d) quantum theory, 124. Energy associated with each photon is given by, •, •, b) E= Ë, c) E=Ë, a) E=hv, , d) E=hv2, , www.Padasalai.Net, 125. The value of planck’s constant is ____, a) 6.625X1034 Js, c) 66.25X1034 Js, , b) 6.626X10-34 Js, d) 6.025X1023 Js, , 126. Light waves behaves as, a) particle in both high and low energy ranges, b) particle in low energy range, but as wave in high energy range, c) wave in both high and low energy ranges, d) wave in low energy range, but as particle in high energy range, 127. Strength of scattering depends on, a) wavelength of the light, c) both (a) and (b), , b) size of the particles, d) velocity of light, , 128., Absorption of light by the molecules, followed by its re-radiation in different, directions is called, a) reflection, b) multiple reflection, c) scattering, d) dispersion, 129., According to Rayleigh scattering law, the amount of scattering is ____, a) directly proportional to Ñ, b) inversely proportional to ‹>, c) directly proportional to Ñ, d) inversely proportional to Ñ, 130., Blue colour of the sky is due to scattering of light by, a) colloidal particles, b) atmosphere, c) molecules of liquid, d) Raman effect, , 154, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 131., When light passes through a colloidal solution, its path is visible inside the solution., This is due to, a) Rayleigh scattering, b) Raman effect, c) Tyndall scattering, d) Scattering of light by atmosphere, 132., In Raman effect, lines of shorter wavelengths are called, a) Stokes lines, b) antistokes lines, c) Raman lines, d) Rayleigh lines, 133., In stokes lines, energy of scattered photon is, a) equal to energy of incident photon, b) lesser than the energy of incident photon, c) greater than the energy of incident photon, d) zero, 134., Raman shift is, a) independent of the frequency of incident light, b) characteristic of the substance, c) independent of characteristic of the substance, d) both (a) and (b), 135., a), b), c), d), , Which of the following is true?, Intensity of stokes lines is always greater than that of antistokes line, Intensity of Stokes lines is always lesser than that of antistokes lines, antistokes lines are of lower frequency than stokes lines, stokes lines are of higher frequency than the antistokes lines, , www.Padasalai.Net, 136. A linear source of light at finite distance in an isotropic medium emits a ____, a) spherical, b) cylindrical, c) circular, d) plane, , 137., If the refractive index of second medium with respect to the first medium is greater, than one, then it implies that, a) first medium is rarer and the second medium is denser, b) first medium is denser and the second medium is rarer., c) velocity of light in first medium is less than that in the second medium, d) velocity of light is same in both media, , 138. If ‘0 is the frequency of incident radiation and νs is the Frequency of scattered, radiation of given molecular sample, then Raman shift or Raman frequency (’“ is given, by, a) ’“ ( “@ 8 “u, b) Δ• ( •Í 8 •, c) Δ• ( •Í 7 •, d) Δ• ( 2 • 8 •Í, 139., If the path difference between two monochromatic waves is Œ,the phase difference, must be, c, =?, ℷ, a) 2UÑ•, b), •, c) ‹ Œ, d) •, 140., a), b), c), d), , 155, , In the interference pattern, the energy is, created at position of maximum, destroyed at the position of minima, conserved but it is redistributed, none of the above, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 141., Colours in thin films is due to, a) dispersion of light, c) interference of light, , b) scattering of light, d) reflection of light, , 142., A ray of light travelling in a rarer medium, gets reflected at the surface of a denser, medium. The automatic path change produced is, ac, ‹, c, a) Ñ, b), c) =, d), 143., In thin films, the condition for getting bright fringe due to interference of the reflected, light is, ‹, a) 2œt cosr=(2n-1) =, b) 2•t=n Ñ, , c) 2t=n Ñ, , d) 2•ï=(2n+1), , c, , 144., In Newton’s ring experiment, the radius of the nth dark ring is proportional to, a) n, b) n2, c) √º, d) m, √, , 145., Bending of light waves around the edges of an obstacle is known as, a) reflection, b) diffraction, c) refraction, d) polarization, 146., In case of Fraunhofer diffraction, the incident wavefront is, a) spherical, b) cylindrical, c) elliptical, , d) plane, , 147., In Frensnel’s diffraction, the shape of the incident wavefront is, a) spherical, b) cylindrical, c) plane, d) (a) or (b), , www.Padasalai.Net, 148., The points on the successive slits separated by a distance equal to grating element are, called as, a) identical points, b) grating points, c) corresponding points, d) equal points, 149. Transverse nature of electromagnetic waves was confirmed by the phenomenon of, a) interference, b) diffraction, c)polarization, d) reflection, 150., In the propagation of light waves, the angle between the plane of vibration and plane, of polarization is, a) 00, b) 900, c) 450, d) 1800, 151., In the propagation of light waves, the angle between the direction of propagation and, plane of polarization is, a) 0, b) 900, c) 450, d) 1800, 152., In case of partially polarized light, when the analyser is rotated through 900, the, intensity of light beam varies from, a) maximum to zero, b) zero to maximum, c) maximum to minimum, d) remains same, 153. The polarizing angle for glass is, a) 57.50, b) 52.50, , 156, , J.SHANMUGAVELU, , c) 32.50, , [P.G. T. in Physics], , d) 37.50, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 154., According to Brewster’s law, a) œ ( tanip, c) • (cot ip, , b) ip=tan •, d) ip=1/tan •, , 155. In the arrangement of pile of plates,the glass plates are inclined at an angle of ____, with the axis of the tube, a)57.50, b) 52.50, c) 32.50, d) 37.50, 156. ___________is an example for uniaxial crystal, a) Mica, b) Topaz, c) Selenite, , d) Quartz, , 157. Of the following optically active material is, a) sodium chloride, b) Calcium chloride, c) sodium, d) Chlorine, 158. Instrument used to determine the optical rotation produced by the substance is _____, a) interfero meter, b) Jamin’s photometer, d) polarimeter, c) polariscope, 159. Atomic spectrum should be a, a) pure line spectrum, c) line absorption spectrum, , b) emission band spectrum, d)band absorption spectrum, , 160. In Raman effect, Raman shift in frequency is always, a) positive, b) negative, c) negative for Stoke’s lines and positive for Antistoke’s lines, d) positive for Stoke’s line and negative for Anti-stoke’s lines, , www.Padasalai.Net, 161. In Raman effect,if frequencies of incident radiation and Stoke’s lines are respectively, 6.198X1015 Hz and 6.602X1015 Hz then the value of Raman shift is, a) -0.038X1015 Hz, b) 0.038X1015 Hz, 15, c) 3.8X10 Hz, d) -3.8X1015 Hz, 162. The ratio of scattering powers of two wavelengths 400 nm and 6000A is, a) 81:16, b) 16;81, c) 81:64, d) 64:81, 163. If A denotes the amount of scattering, the wavelength of light is proportional, a) inversely to A4, b) directly to A4, c) inversely to, , <, >, , d) directly to, , Ó, , 164. If the velocity of light in a medium is 2.25X108 m/s, Then the refractive index of the, medium will be, a) 1.5, b) 0.5, c) 1.33, d) 1.73, 165. If the wavelength of light wave in vacuum is 6.4X10-7 m, Then wavelength of light, wave in water of refractive index 1.33 will be, a) 4X10-7 m, b) 4.8X10-7 m, c) 2.64X10-7 m, d) 5X10-7 m, 166. The refractive indices for glass and water respectively, is 1.5 and 1.33,then the ratio of, velocity of light in glass and water is, a) 4:3, b) 3:4, c) 8:9, d) 9:8, , 157, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 167. The time taken by the light to travel a distance of 200 m in a medium of refractive, index 1.5 is, a) 2X10-8 S, b) 10-6 S, c) 10-8 S, d) 3X108 S, 168. If the path difference between two monochromatic light waves of wavelength 4000 A is, 2X10-7 m. The phase difference between them will be, a, a) ?, b) 2 U, c), d), 169. In young’s double slit experiment, the third order bright band for wavelength of light, 6000A coincides with fourth order bright fringe for another source in the same, arrangement. The wavelength of another source will be, a) 4500A, b) 6000A, c) 5000A, d) 4000A, 170. In young’s double slit experiment, lights of wavelength 5.48X10-7m and 6.85X10-8 m, are used, in turn keeping D and d constant. The ratio of respective bandwidths in the two, cases will be, a) 1:4, b) 1:8, c) 8:1, d) 4:1, 171. In young’s double slit experiment, if the distance between the coherent sources is, increased twice that of initial value then the new bandwidth will be, a) increased 4 times, b) increased 2 times, d) decreased 2 times, c) decreased 4 times, 172. In young’s double slit experiment, if the distance between slits is halved and the, distance between the slits and the screen is doubled, the new fringe width will be, a) remains same, b) halved, d) quadrupled, c) doubled, , www.Padasalai.Net, 173. In young’s double slit experiment, sodium light is employed and interference fringes, are obtained in which the bandwidth of 3rd bright fringe is 2.2 mm. What will be the, bandwidth of 2nd dark fringe?, a) 2.2 mm, b) 1.1 mm, c) 4.4 mm, d) 3.3 mm, , 174. Newton’s rings were obtained with a light of wavelength 5460A the thickness of the air, film where 2nd dark ring is formed is, a) 5.46X10-7 m, b) 3.276X10-6 m, -6, c) 32.76X10 m, d) 54.6X10-7m, 175. In Newton’s ring experiment, the ratio of the radii of 4th ring and 9th ring is, a) 4:9, b) 2:3, c) 16:81, d) √2: √3, <, , 176. In a plane transmission grating experiment, the wavelength of light is, times of, √=, grating element used.The angle of diffraction for the first order maximum will be, a) 300, b) 450, c) 600, d) 900, 177. The distance between the two corresponding points in a grating is 2X10-4 cm. The, number of lines per meter width of the grating will be, a) 20000, b) 2000, c) 500000, d) 50000, , 158, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , ‘3’ MARK QUESTIONS:(2- Questions: Q. No: 39,40), PUBLIC ‘3’ MARKS:, 1. Mention any three characteristics of electromagnetic waves ( J-15), i) Electromagnetic waves are produced by accelerated charges., ii) They do not require any material medium for propagation., ^_ occur simultaneously., ^_ and 6, iii) Variation of maxima and minima in both t, iv) They travel in vacuum or free space with a velocity 3 × 108 m s−1 given by, the relation C =, , ‘, , v) The energy in an electromagnetic wave is equally divided between electric, and magnetic field vectors., vi) The electromagnetic waves being chargeless, are not deflected by electric and, magnetic fields., 2. Write any three uses of Infra Red rays. ( O – 08,M-17 ), i) Infrared lamps are used in physiotherapy., ii) Infrared photographs are used in weather forecasting., iii) As infrared radiations are not absorbed by air, thick fog, mist etc, they are, used to take photograph of long distance objects., iv) Infra red absorption spectrum is used to study the molecular structure., , www.Padasalai.Net, 3. Write any three uses of ultra-violet radiations?(M-14,J-16), i) They are used to destroy the bacteria and for sterilizing surgical instruments., ii) These radiations are used in detection of forged documents, finger prints in, forensic laboratories., iii) They are used to preserve the food items., iv) They help to find the structure of atoms, 4. What are emission and absorption spectra? ( M – 06,O-16 ), Emission spectrum: When the light emitted directly from a source is examined, with a spectrometer, the emission spectrum is obtained. Every source has its own, characteristic emission spectrum., Absorption spectrum: When the light emitted from a source is made to pass, through an absorbing material and then examined with a spectrometer, absorption, spectrum is obtained., , 159, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 5. What is band emission spectrum? Give an example. ( O – 06 ), It consists of a number of bright bands with a sharp edge at one end but, fading out at the other end. Band spectra are obtained from molecules., It is the characteristic of the molecule., Example: Calcium or Barium salts in a Bunsen flame and gases like, carbon – di – oxide , ammonia and nitrogen in molecular state in the discharge, tube give band spectra., 6. What are Fraunhofer lines? (M-13), If the solar spectrum is closely examined it is found that it consists of large, number of dark lines. These dark lines in the solar spectrum are called, Fraunhofer lines. Solar spectrum is an example of line absorption spectrum., 7. Why does the sky appear blue in colour ?( J – 06,0-13,J-14 ), According to Rayleigh’s scattering law, the shorter wavelengths are scattered, much more than the longer wavelengths. The blue appearance of sky is due to, scattering of sunlight by the atmosphere. Blue light is scattered to a greater, extent than red light. This scattered radiation causes the sky to appear blue., , www.Padasalai.Net, 8. What is Tyndal scattering? ( M – 07, J – 09, J – 10, O - 10 ), , When light passes through a colloidal solution its path is visible inside the, solution. This is because, the light is scattered by the particles of solution. The, scattering of light by the colloidal particles is called Tyndal scattering., 9. State Huygen’s principle. ( O – 09, O – 11,M – 12,O-15 ), i) Every point on a given wave front may be considered as a source of, secondary wavelets which spread out with the speed of light in that medium, and, ii) The new wavefront is the forward envelope of the secondary wavelets at that, instant., 10. State the conditions to achieve total internal reflection. ( O – 08,M-14 ), For total internal reflection to take place, i) Light must travel from a denser medium to a rarer medium and, ii) The angle of incidence inside the denser medium must be greater than the, critical angle. i >C, , 160, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 11. List the conditions for sustained interference?(O-14), The interference pattern in which the positions of maximum and minimum, intensity of light remain fixed with time, is called sustained or permanent, interference pattern., i) The two sources should be coherent, ii) Two sources should be very narrow, iii) The sources should lie very close to each other to form distinct and broad, fringes., 12. Why the centre of Newton’s rings pattern appear dark? ( M – 09,O-16 ), The thickness of the air film at the point of contact of lens L with glass, plate P is zero. Hence, there is no path difference between the interfering waves., So, it should appear bright. But the wave reflected from the denser glass plate, has suffered a phase change of π while the wave reflected at the spherical, surface of the lens has not suffered any phase change. Hence the point O, appears dark., 13. Distinguish between Fresnel and Fraunhofer diffractions? ( M -10, J – 12 ), Fresnel diffraction, 1. The source and the screen are at, Finite distances from the obstacle, producing diffraction, 2. The wave front undergoing, diffraction is either spherical or, cylindrical, 3. No lens is used, , Fraunhofer diffraction, The source and the screen are at, infinite distances from the obstacle, producing diffraction, The wave front undergoing, Diffraction is plane wave front, , www.Padasalai.Net, convex lens is used, , 14. Distinguish between interference and diffraction fringes. ( O – 07 ), Interference, 1. Fringes are equally spaced, 2. Bright fringes are of same intensity, 3. Comparing with diffraction, it has, , Diffraction, Fringes are unequally spaced, Intensity falls rapidly, It has less number of fringes, , large number of fringes, 15. State Brewster’s law. ( M – 11 ), The tangent of the polarising angle is numerically equal to the refractive index, of the medium., ( i.e.) tan ip= μ, 16. Define: optic axis of a crystal. ( J – 07, J - 10 ), Inside a double refracting crystal there is a particular direction in which both, the rays travel with same velocity. This direction is called optic axis. The, refractive index is same for both rays and there is no double refraction along this, direction., , 161, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 17. What are uniaxial and biaxial crystal. Give an example for each ? (O-15), Crystals like calcite, quartz, ice and tourmaline having only one optic axis are, called uniaxial crystals., Crystals like mica, topaz, selenite and aragonite having two optic axes are, called biaxial crystals., 18. On what factors does the amount of optical rotation depend?(J–08,J-1, M 15), The amount of optical rotation depends on :, i) Thickness of crystal, ii) Density of the crystal or concentration in the case of solutions., iii) Wavelength of light used, iv) The temperature of the solutions, , 19. Define: specific rotation? ( M – 08, M – 10,0-13,J-14 ), Specific rotation for a given wavelength of light at a given temperature is, defined as the rotation produced by one – decimeter length of the liquid column, containing 1 gram of the active material in 1 cc of the solution., /, S=, d=, , www.Padasalai.Net, OTHER IMPORTANT ‘3’ MARKS:, 20. What are electromagnetic waves?, In an electromagnetic wave electric and magnetic field vectors are at right, angles to each other and both are at right angles to the direction of propagation., They possess the wave character and propagate through free space without any, material medium. These waves are transverse in nature, 21. What is meant by continuous emission spectrum?, It consists of unbroken luminous bands of all wavelengths containing all the, colours from violet to red. These spectra depend only on the temperature of the, source and is independent of the characteristic of the source., Incandescent solids, liquids, Carbon arc, electric filament lamps etc, give, continuous spectra., , 162, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 22. What is fluorescence?, When an atomic or molecular system is excited into higher energy state by, absorption of energy, it returns back to lower energy state in a time less than, −, 10 5 second and the system is found to glow brightly by emitting radiation of, longer wavelength. When ultra violet light is incident on certain substances, they, emit visible light. It may be noted that fluorescence exists as long as the, fluorescing substance remain exposed to incident ultraviolet light and re-emission, of light stops as soon as incident light is cut off., 23. What is phosphorescence?, There are some substances in which the molecules are excited by the, absorption of incident ultraviolet light, and they do not return immediately to, their original state. The emission of light continues even after the exciting, radiation is removed. This type of delayed fluorescence is called phosphorescence., 24. What are two possible modes of propagation of energy?, i) By stream of material particles moving with a finite velocity, ii) By wave motion, wherein the matter through which the wave propagates does, not move along the direction of the wave., , www.Padasalai.Net, 25. Write a short note on electromagnetic theory, , Maxwell showed that light was an electromagnetic wave, conveying, electromagnetic energy and not mechanical energy as believed by Huygens,, Fresnel and others. He showed that the variation of electric and magnetic, intensities had precisely the same characteristics as a transverse wave motion. He, also showed that no medium was necessary for the propagation of electromagnetic, waves., 26. State Rayleigh scattering law?, The amount of scattering is inversely proportional to the fourth power of the, wavelength. This is known as Rayleigh scattering law., 27. What is Raman effect?, The monochromatic light is scattered when it is allowed to pass through a, substance. The scattered light contains some additional frequencies other than that, of incident frequency. This is known as Raman effect., , 163, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 28. What are the uses of Raman spectrum?, i) It is widely used in almost all branches of science., ii) Raman Spectra of different substances enable to classify them according to, their molecular structure., iii) In industry, Raman Spectroscopy is being applied to study the properties of, materials., iv) It is used to analyse the chemical constitution, 29. The sun looks almost reddish at sunset and sunrise. Why ?, At sunrise and sunset the rays from the sun have to travel a larger part of, the atmosphere than at noon. Therefore most of the blue light is scattered away, and only the red light which is least scattered reaches the observer. Hence, sun, appears reddish at sunrise and sunset., , 30. Define wave front. And mention the different types of wave front?, The wave front at any instant is defined as the locus of all the particles of, the medium which are in the same state of vibration., i) Spherical wave front, ii) Plane wave front, iii) Cylindrical wave front, , www.Padasalai.Net, 31. State superposition principle?, , When two or more waves simultaneously pass through the same medium, each, wave acts on every particle of the medium, as if the other waves are not, present. The resultant displacement of any particle is the vector addition of the, displacements due to the individual waves., 32. What are coherent sources?, Two sources are said to be coherent if they emit light waves of the same, wave length and start with same phase or have a constant phase difference., 33. Two independent sources cannot be coherent source. Why?, Two independent monochromatic sources, emit waves of same wave length., But the waves are not in phase. So they are not coherent. This is because, atoms, cannot emit light waves in same phase and these sources are said to be, incoherent sources., , 164, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 34. What is meant by interference of light?, The waves from two coherent sources travel in the same medium and, superpose at various points, the intensity of light will be alternatively maximum, and minimum i.e. bright and dark bands which are referred as interference, fringes. The redistribution of intensity of light on account of the superposition of, two waves is called interference., 35. Distinguish between constructive and destructive interference., Constructive interference, 1. At points where the crest of one, , wave meets the crest of the other, wave or the trough of one wave, meets the trough of the other wave, 2. The waves are in phase, 3. The displacement is maximum, 4. These points appear bright, , Destructive interference, At points where the crest of one, wave meets the trough of the other, wave, The waves are in opposite phase, The displacement is minimum, These points appear dark, , 36. Define band width., The distance between any two consecutive bright or dark bands is called, band width., , www.Padasalai.Net, Band width, β =, , ‚, ä, , Ñ, , 37. Mention the Condition for obtaining clear and broad interference bands?, i) The screen should be as far away from the source as possible., ii) The wavelength of light used must be larger., iii) The two coherent sources must be as close as possible., 38. On what factors does the colours in thin films depend on?, Colours are due to interference between light waves reflected from the top, and the bottom surfaces of thin films. When white light is incident on a thin, film, the film appears coloured and the colour depends upon the thickness of the, film and also the angle of incidence of the light., 39. Write the applications of Newton’s rings., i) Using the method of Newton’s rings, the wavelength of a given, monochromatic source of light can be determined., ii) Using Newton’s rings, the refractive index of a liquid can calculated, , 165, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 40. Distinguish between corpuscle and photons., 1., 2., 3., 4., , Corpuscle, Tiny and massless and perfectly, elastic particle, They travel in straight line, with, particle nature, Different colours are due to, variations in size of corpuscles, Light energy is the kinetic energy, of the corpuscles, , Photons, Small pockets of energy, They show the dual nature. Both, particle and wave nature., Different colours are due to, different amount of energy, Different colours are due to the, frequency of the photons. Their, energy is equal to E=hυ, , 41. What is meant by polarisation?, Light waves coming out of tourmaline crystal A have their vibrations in, only one direction, perpendicular to the direction of propagation. These waves are, said to be polarised. Since the vibrations are restricted to only one plane parallel, to the axis of the crystal, the light is said to be plane polarised. The, phenomenon of restricting the vibrations into a particular plane is known as, polarisation., 42. What are polariser and analyser ?, , www.Padasalai.Net, A device which produces plane polarised light is called a polariser., A device which is used to examine, whether light is plane polarised or not is an, analyser., A polariser can serve as an analyser and vice versa., , 43. Define plane of vibration and plane of polarisation., The plane containing the optic axis in which the vibrations occur is known, as plane of vibration. The plane which is at right angles to the plane of vibration, is called the plane of propagation. Plane of polarisation does not contain, vibrations in it., 44. Define plane polarised and partially polarised light, A ray of light is allowed to pass through an analyser. If the intensity of the, emergent light does not vary, when the analyser is rotated then the incident light, is unpolarised; If the intensity of light varies between maximum and zero, when, the analyser is rotated through 90° then the incident light is plane polarized . If, the intensity of light varies between maximum and minimum (not zero) when the, analyzer is rotated through 90°, then the incident light is partially plane polarized., , 166, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 45. Define angle of polarisation., When the light is allowed to be incident at a particular angle, the reflected, beam is completely plane polarised. The angle of incidence at which the reflected, beam is completely plane polarised is called the polarising angle (ip)., 46. What is double refraction ?, When a ray of unpolarised light is incident on a calcite crystal, two, refracted rays are produced. This phenomenon is called double refraction . Hence,, two images of a single object are formed. This phenomenon is exhibited by, several other crystals like quartz, mica etc., 47. Distinguish between ordinary and extra ordinary rays?, Ordinary ray, 1. It obey the laws of refraction, 2. It travels with same velocity in all, , directions, 3. It produces spherical wave front, , Extra ordinary, It do not obey the laws of refraction, It travels with different velocities, along different directions, It produces elliptical wave front, , 48. Define optic axis of a crystal., , www.Padasalai.Net, The direction inside the crystal in which both the ordinary and extra ordinary, rays travel with the same velocity is called optic axis., , 49. What are polaroids? Give the uses of polaroids., A Polaroid is a material which polarises light. The phenomenon of selective, absorption is made use of in the construction of polaroids., Uses:, i) Polaroids are widely used as polarising sun glasses., ii) They are used to eliminate the head light glare in motor cars., iii) They are used to improve colour contrasts in old oil paintings., 50. What are optically active substances? Give an example, When a plane polarised light is made to pass through certain substances, the, plane of polarisation of the emergent light is not the same as that of incident, light, but it has been rotated through some angle. This phenomenon is known as, optical activity. The substances which rotate the plane of polarisation are said to, be optically active., Examples : quartz, sugar crystals, turpentine oil, sodium chloride etc., , 167, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 51. Distinguish between dextro – rotatory and laevo – rotatory substances, Dextro – rotatory, 1. which rotate the plane of, polarisation in the clock wise, direction on looking towards the, source., , Laevo – rotatory, Which rotate the plane of polarisation, in the anti clock wise direction on, looking towards the source., , 52. Distinguish between polarised and unpolarised light., Polarised light, 1. The vibrations in only one, direction, perpendicular to the, plane of the paper, 2. The light comes from the polariser, is a polarised light., , Unpolarised light, The vibrations in all directions, The light comes from any source may, be unpolarised light., , PUBLIC ‘3’ MARK PROBLEMS:, 1. An LC resonant circuit contains a capacitor 400 pF and an inductor100 μH. It, is sent into oscillations coupled to an antenna. Calculate the wavelength of the, radiated electromagnetic wave.[M-13], , www.Padasalai.Net, C = 4 10-10 F;, , Given data:, Sol:, , N=, λ=, , L = 10-4 H;, , λ=?, , √ P, , P, ], , = C 2U√`X, , = 3 108 2 3.14 √4, , 10#, , 10#, , = 376.8 m, , λ =377 m, , 168, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 2. In Young’s experiment, the width of the fringes obtained with light of, wavelength 6000 Å is 2 mm. Calculate the fringe width if the entire apparatus, is immersed in a liquid of refractive index 1.33.[J-06,M-11, O-14], , Data : λ= 6000 Å = 6 × 10−7 m;, β’= ?, Solution :, , β’=, , ‚cˆ, ä, , ‚c, , =, , ä, , =, , β= 2mm = 2 × 10−3 m ; μ= 1.33;, —, , 5Yynz{ • (, , c, , cˆ, , 9, , β'= 2 10 – 3/ 1.33, 10-3m (or), , = 1.5, , β' = 1.5 mm, 3. A light of wavelength 6000 Å falls normally on a thin air film, 6 dark fringes, are seen between two points. Calculate the thickness of the air film.[ M-06, J, – 08, J – 09, J – 11 ], , Given data: λ = 6000 Å;, , n = 6;, , θ = 900;, , 6000, , 10 -10/ 2, , t=?, , www.Padasalai.Net, Sol:, , 2μt = nλ, , Thickness of the film t =, , mc, , =6, , 10-6m, , t = 1.8, , 4. A light of wavelength 5890 Å falls normally on a thin air film, 6 dark fringes, are seen between two points. Calculate the thickness of the air film. .[O-07], , Given data: λ = 5890 Å;, Sol:, , n = 6;, 2μt = nλ, , Thickness of the film t =, , mc, , =6, , 5890, , t = 1.767, , 169, , t=?, , J.SHANMUGAVELU, , 10 -10/ 2, , 10-6m., , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 5. Two slits 0.3 mm apart are illuminated by light of wavelength 4500 Å. The, screen is placed at 1 m distance from the slits. Find the separation between the, second bright fringe on both sides of the central maximum.[O-06,J-08,09,11], , Data : d = 0.3 mm = 0.3 × 10−3 m ;, D=1m;, n=2;, 2x = ?, Sol:, , x=, 2x =, , λ= 4500 Å = 4.5 × 10−7 m,, , ‚ mc, ä, , ‚ mc, ä, , =, , .a, , .E, , «, S, , ∴ 2x = 6 × 10−3m, 2x = 6 mm, 6. In young’s double slit experiment the distance between the slits is 1.99 mm., the distance between slit and screen is 1 m. If the band width is 0.35 mm,, calculate the wave length of light used.[M-12], , www.Padasalai.Net, Given data: β = 35 × 10 -5m;, ‚c, Sol:, Bandwidth β =, , d = 1.99 × 10 -3m;, , D = 1m; λ = ?, , ä, , λ =, , —ä, ‚, , = 35 × 10 -5× 1.99 × 10 -3/ 1, = 69.65 × 10 -8m, , λ = 6965 Å, 7. In Newton’s rings experiment the diameter of certain order of dark ring is, measured to be double that of second ring. What is the order of the ring?[M07,J-07, O – 11], , Data :dn= 2d2 ; n = ?, Sol:, , dn2= 4nRλ, d22 = 8Rλ, , 170, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 10. A 300 mm long tube containing 60 cc of sugar solution produces a rotation of, 9o when placed in a polarimeter. If the specific rotation is 60o, calculate the, quantity of sugar contained in the solution.[M-06,09,J-13, J-15,M-17], , Data :l = 300 mm = 30 cm = 3 decimeter θ= 9o; S = 60o ; v = 60 cc m = ?, Solution :, , /, , d =, /, , S=, =, , d, , p/Ë, , /Ë, , m=, , dÍ, , T, , =a, , m=3g, PUBLIC ‘5’ MARK PROBLEMS:, 11. A soap film of refractive index 1.33, is illuminated by white light incident at, an angle 30o. The reflected light is examined by spectroscope in which dark, band corresponding to the wavelength 5893Å is found. Calculate the smallest, thickness of the film.[O-07] [Compulsory], , www.Padasalai.Net, Data : • = 1.33; i = 30o; λ = 5893 Å = 5.893× 10–7 m, n = 1 (Smallest thickness); t = ?, , • =, , Solution :, , Sin r =, , .%² 7, , .%² 7, , .%², , =, , .%² a °, , =, , .aa, , .E, , .aa, , = 0.3759, , Cos r = √1 8 0.3759 = 0.9267, 2 •t cos r = nÑ, , t=, , =, t=, , c, , ,-., «, , E. Ta, .aa, , E. Ta, ., , .T, , Ï, , «, , E, , t = 2.390 <@#€ m, , 172, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 12. A soap film of refractive index 1.34, is illuminated by white light incident at, an angle 30o. The reflected light is examined by spectroscope in which dark, band corresponding to the wavelength 5893Å is found. Calculate the smallest, thickness of the film.[O-08] [Compulsory], , Data : • = 1.34; i = 30o; Ñ = 5893 Å = 5.893× 10–7 m, , n = 1 (Smallest thickness); t = ?, .%² 7, , • =, , Solution :, , .%², , .%² 7, , Sin r =, , .%² a °, , =, , .a, , .E, , = 0.3731, .a, Cos r = √1 8 0.3731 = 0.9277, =, , 2 •t cos r = nÑ, c, , t=, , ,-., «, , E. Ta, , www.Padasalai.Net, =, , t=, , .a, , .T ÏÏ, , «, , E. Ta, ., , <@#€ m, , t = 2.370, , 13. A soap film of refractive index 1.33, is illuminated by white light incident at, an angle 30o. The reflected light is examined by spectroscope in which dark, band corresponding to the wavelength 6000Å is found. Calculate the smallest, thickness of the film.[O-13], , Data : • = 1.33; i = 30o; Ñ = 6000 Å = 6 × 10–7 m, n = 1 (Smallest thickness); t = ?, , Solution :, , • =, , .%², , Sin r =, , =, , 173, , .%² 7, .%² 7, .E, , .aa, , J.SHANMUGAVELU, , =, , .%² a °, .aa, , = 0.3759, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , Cos r = √1 8 0.3759 = 0.9267, 2 •t cos r = nÑ, c, , t=, , ,-., «, , =, , .aa, , .T, , Ï, , «, , t=, , ., , E, , t = 2.434 <@#€ m, , 14. A monochromatic light of wavelength 589 nm is incident on a water surface, having refractive index 1.33. Find the velocity, frequency and wavelength of, light in water.[M-11] [Compulsory], λ = 589 10-9 m;, , Given data :, Sol:, , μ = 1.33, , P•, , Velocity Cw =, , www.Padasalai.Net, =, , a, , ', , .aa, , = 2.255 108, , Cw = 2.26 108 ms-1, 53. Frequency N =, , P•, c, , =, , 54. Wavelength Ñw =, , c•, , ', , a, , E T, , ‘ = 5.093, , =, , E T, , =, , E T, , .aa, , ‹w = 4429 Å, , J.SHANMUGAVELU, , Ó, , <@<> Hz, , = 442.9, , 174, , a, , 10#T, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 15. A monochromatic light of wavelength 5893 Å is incident on a water surface, having refractive index 1.33. Find the velocity, frequency and wavelength of, light in water.[O-08] [Compulsory], Given data :, , λ = 5893 10-10 m;, , Sol:, i., , μ = 1.33, , P•, , Velocity Cw =, , ', , a, , =, , .aa, , = 2.255 108, , Cw = 2.26 108 ms-1, ii., , Frequency N =, =, , P•, c, , a, , E Ta, , ', , =, , a, , Ó, , E Ta, , www.Padasalai.Net, ‘ = 5.09, , iii., , Wavelength Ñw =, =, , <@<> Hz, , c•, , E Ta, , .aa, , = 4430, ‹w = 4430 Å, , 175, , J.SHANMUGAVELU, , 10#, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 16. In Young’s experiment a light of frequency 6 × 1014 Hz is used. Distance, between the centres of adjacent fringes is 0.75 mm. Calculate the distance, between the slits, if the screen is 1.5 m away.[O-07,O-14 , J-15], [Compulsory], Given data:, Sol:, , ‰ = 0.75, , ν = 6 × 1014 Hz;, c = νλ, , λ=, , P, ], , =, , 10#a m;, , D = 1.5 m;, , d=?, , ', , a, , Ó, , λ = 5 10-7 m, β=, , c‚, , d=, d=, , ä, , c‚, —, , E, , «, , .E, S, , .ÏE, , www.Padasalai.Net, = 10, , 10# = 10-3m, , d = 1 mm, , 17. A parallel beam of monochromatic light is allowed to incident normally on a, plane transmission grating having 5000 lines per centimetre. A second order, spectral line is found to be diffracted at an angle 30o. Find the wavelength of, the light.[M-08,M-10, J-15] [Compulsory], Data : N = 5000 lines / cm = 5000 × 102 lines / m = 5 105 lines/m, m = 2 ; & = 30o ; Ñ = ?, Solution :, , sin & = Nm Ñ, , Ñ =, Ñ=, , .%² /, op, , .%² a °, , E, , \, , =, , E, , .E, , \, , λ = 5 10-7, λ = 5000 Å, , 176, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 18. In a Newton’s rings experiment the diameter of the 20th dark ring was found to, be 5.82 mm and that of the 10th ring 3.36 mm. If the radius of the planoconvex lens is 1 m. Calculate the wavelength of light used.[M-10,M-14,O-15], [Compulsory], d20 = 5.82 mm; r20 = 2.91 10#a m, d10 = 3.36 mm; r10 = 1.68 10#a m;, m = [(m+n) –n] =10; R = 1m, , Given data:, sol:, Ñ=, , ~ l, , # l, pì, , S, , S, , .ET, , E., , .T # ., , S, , S, , . a, R, , E, , S, , # ., , .T F ., , =, =, , S, , .T, , =, , www.Padasalai.Net, =, , = 0.5645 10-6 =5645 10-10, , ‹ = 5645 Å, Alternate method :, sol:, , Ñ=, , =, =, =, =, , ‚~ l #‚l, pì, S, , E., , E., , S, , Fa.a, S, , T., , .E, , ., R, , S, , # a.a, , E., , #a.a, , S, , S, , = 0.5645 10-6 = 5645 10-10, , ‹ = 5645 Å, , 177, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 19. A plane transmission grating has 5000 lines/cm. Calculate the angular, separation in second order spectrum of red line 7070 Å and blue line 5000, Å.[J-13], , Given data: N = 5000 lines / cm = 5000 × 102 lines / m = 5 105 lines/m, m = 2 ; Ñì = 7070 Å; Ñ = 5000 Å;, Solution :, , sin & = Nm Ñ, sin &ì = 5 105, , 2, , 10-10, , 7070, , 10-4 = 0.7070, , = 7070, , 104, , 10-4, , sin &ì = 0.707, , &ì = 45°, sin &, , = 5 105, , 2, , 10-4 = 0.5, , = 5000, , 10-10, , 5000, , 104, , 10-4, , sin & = 0.5, , & = 30°, , www.Padasalai.Net, angular separation (&ì - & ) = 45° 8 30°, , Z - Z; = 15°, , 20. A plano – convex lens of radius 3 m is placed on an optically flat glass plate, and is illuminated by monochromatic light. The radius of the 8th dark ring is, 3.6 mm. Calculate the wavelength of light used.[M-11] [Compulsory], Data :R = 3m ; n = 8 ; r8 = 3.6 mm = 3.6 × 10−3 m ; λ= ?, Solution : rn= √nîÑ, rn2= nRλ, λ=, =, , l, , mì, Õa., , a, , S×, , λ = 5400 × 10−10m (or) 5400 Å, 178, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , PUBLIC ‘10’ MARKS: (1 – question : Q. No: 66), 1. What is emission and absorption spectra? Explain the different types of emission and, , absorption spectra with examples? (J - 09, M - 10, M - 12, J - 12, O -12, J-13,M-15), 2. What is Raman scattering? Explain Raman scattering of light with the help of energy, , level diagram. (M - 07, O – 07, M - 08, J - 11, M – 13,O-15, J-16), 3. State Huygens’s principle on the basis of wave theory Prove the laws of reflection., , (O – 08, J-15), 4. On the basis of wave theory, explain total internal reflection. Write the conditions for, , the total internal reflection to take place. (M - 06, J – 06,O-13,O-16), 5. What is known as interference? Derive an expression for band width of interference, , fringes in Young’s double slit experiment., (O - 06, O - 11, O - 10, J - 07, J - 10, M - 09, M -11,M-14,J-14,M-17), 6. Discuss the theory of interference in thin transparent film due to reflected light and, , obtain the condition for the intensity to be maximum and minimum. (J - 08, O - 09), , Only for slow bloomers, , www.Padasalai.Net, ( 5 & 10 mark answer ), , 1. Radius of the nth dark ring:, Consider the vertical section SOP of the plano, convex lens its centre of curvature C, R be the, radius of curvature of the plano convex lens and O, be the point of contact with the surface. t be the, thickness of the air film., Then ST = AO = PQ = t, Let r² be the radius of the nth dark ring Then SA = AP = r², By the law of segments SA. AP = OA. AN, , 2t (, , 6š, , ô, , 2t = nλ, , r² ( √nRλ, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 2. Brewster’s law: The tangent of the polarising angle is numerically equal to the, refractive index of the medium., ip + 90° + r = 180°, , .%² %Ù, , r = 90° - ip, , .%² 6, .%² %Ù, , 90° 8%Ù, , .%² %Ù, , ,-. %Ù, , tan iù, , ( μ, ( μ, ( μ, , =μ, , 3. Pile of plates :, It consists of a number of glass plates placed one over, the other., The plates are inclined at an angle of 32.5° to the axis, of the tube, A beam of monochromatic light is allowed to fall on, the pile of plates along the axis of the tube. So, the, angle of incidence will be 57.5° which is the, polarising angle for glass., The vibrations perpendicular to the plane of incidence are reflected at each surface and, those parallel to it are transmitted., The larger the number of surfaces, the greater is the intensity of the reflected plane, polarised light., The pile of plates is used as a polarizer and an analyser., , www.Padasalai.Net, 4. Nicol prism, , Construction: Calcite crystal whose length is, three times its breadth. It is cut along the diagonal, so that their face angles are 72° and 108°. And, joined together by a layer of Canada balsam, a, transparent cement ., For ordinary light µ = 1.658 and, for extra-ordinary light µ = 1.486., for Canada balsam µ = 1.550 for both rays, Working: A unpolarised light is splits up into ordinary ray (O), and extraordinary ray (E) inside the nicol prism. The ordinary ray is totally internally, reflected at the layer of Canada balsam and is prevented., The extraordinary ray alone is transmitted through the crystal which is plane polarised. The, nicol prism serves as a polariser and also an analyser., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 5. Types of spectra:, Emission spectra, When the light emitted directly from a, source is examined with a spectrometer,, the emission spectrum is obtained, Continuous spectrum, It consists of unbroken luminous bands, of all wavelengths containing all the, colours from violet to red. These spectra, depend only on the temperature of the, source, Example: Incandescent solids, liquids, Line spectrum, Line spectra are sharp lines of definite, wavelengths. It is the characteristic of, the emitting substance, Example: Atoms in the gaseous state,, Band Spectrum, It consists of a number of bright bands, with a sharp edge at one end but fading, out at the other end., Example: Barium salts in a bunsen, flame, , 2017-2018, , Absorption Spectra, When the light emitted from a source is, made to pass through an absorbing, material and then examined with a, spectrometer, the obtained spectrum is, called absorption spectrum., Continuous absorption spectrum, A pure green glass plate when placed in, the path of white light,, absorbs everything except green and, gives continuous absorption, spectrum., Line absorption spectrum, When light from the carbon arc is made, to pass through sodium vapour and then, examined by a spectrometer,, Band absorption spectrum, If white light is allowed to pass through, iodine vapour dark bands on continuous, bright background are obtained, , www.Padasalai.Net, 6. Raman effect:, The monochromatic light is scattered when it is allowed to pass through a substance., The scattered light contains some additional frequencies other than that of incident, frequency, Stoke’s lines: If a photon strikes an atom in a liquid, part of the energy may be used, to excite the atom and the rest is scattered. It have lower frequency, Anti−stokes lines: If a photon strikes an atom or a molecule in a liquid, which is in, an excited state, the scattered photon gains energy. It have higher frequency, Rayleigh line : when a light photon strikes atoms or molecules, photons may be, scattered elastically. Then the photons neither gain nor lose energy. It have, unmodified frequency., Raman Shift or Raman frequency Δν = νο − νs., The Raman shift does not depend upon the frequency of the incident light but it is, the characteristic of the substance producing Raman effect., For Stoke’s lines, Δν is positive and for Anti–stoke’s lines Δν is negative., The intensity of Stoke’s line is always greater than Anti−stoke’s Line., , 183, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 7. Total internal reflection :, Let XY be a plane surface which separates a rarer medium (air) and a denser, medium. Let the velocity of the wavefront in these media be Ca and Cm respectively., A plane wavefront AB passes from denser medium to rarer medium., , www.Padasalai.Net, =~, , i), , =•, , .%² 7, , .%², , =, , P ⁄¡P, ¡‚⁄¡P, , =, , P, , ¡‚, , (, , =~ ë, =• ë, , =, , =~, =•, , < 1; i < r, the refracted wavefront is deflected away from the surface XY., , AD < AC : For small values of i, BC will be small and so AD > BC but less than, AC, ¡‚, , sin = ¡P < 1 ; r < 900 For each value of i, a refracted wavefront is possible, ii), , AD = AC : As i increases r also increases. When AD = AC,, sin r = 1 (or) r = 900. i.e a refracted wavefront is just possible., The angle of incidence at which the angle of refraction is 900 is called the critical, angle C., , iii), , AD > AC : When AD > AC, sin r > 1. This is not possible ., Therefore no refracted wave front is possible., Total internal reflection: When the angle of incidence increases beyond the, critical angle. The incident wavefront is totally reflected into the denser medium, itself., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 8. Young’s double slit experiment:, d - distance between two coherent, sources A and B., XY - parallel to AB at a distance D, from the coherent sources., C - mid point of AB., O - point on the screen equidistant, from A and B., P - point at a distance x from O,, Waves from A and B meet at P in, phase or out of phase depending upon, the path difference between two, waves., , www.Padasalai.Net, Bright fringe:, condition for constructive interference is the path difference, "ä, ‚, , = nλ, , x =, dark fringe :, , ‚, ä, , nλ, , condition for destructive interference is the path difference, "ä, ‚, , c, , = 2(n-1), ‚, , x = 2(n-1), ä, , c, , Band width (β), The distance between any two consecutive bright or dark bands is called bandwidth., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 6.ATOMIC PHYSICS, PREPARED BY, J.SHANMUGAVELU M.Sc, B.Ed. [P.G. Assist in Physics], , ONE MARK QUESTIONS: (4 - questions), BOOK BACK QUESTIONS:, 1. The cathode rays are, a) a stream of electrons, c) a stream of uncharged particles, , b) a stream of positive ions, d) the same as canal rays, , 2. A narrow electron beam passes undeviated through an electric field E = 3 × 104 V/m and, an overlapping magnetic field B = 2×10−3Wb/m2. The electron motion, electric field and, magnetic field are mutually perpendicular. The speed of the electron is, a) 60 ms−1, b) 10.3 × 107ms−1, c) 1.5 × 107ms−1, d) 0.67 × 10−7ms−1, , v=, , Sol:, , =, 3., , ö, , A, , a, , Ó, , 107 ms-1, , = 1.5, , S, , According to Bohr’s postulates, which of the following quantities take discrete values?, a) kinetic energy, b) potential energy, c) angular momentum, d) momentum, , www.Padasalai.Net, 4. The ratio of the radii of the first three Bohr orbit is,, a) 1 : 1/2 : 1/3, b) 1 : 2 : 3, c) 1 : 4 : 9, , rn =, , Sol:, , ² ´ 3, 2k, , ⇒ rn∝ n, , d) 1 : 8 : 27, , 2, , ⇒ r1: r2:r3= 12:22:32, =1 : 4 : 9, 5. The first excitation potential energy or the minimum energy required to excite the atom, from ground state of hydrogen atom is,, a) 13.6 eV, b) 10.2eV, c) 3.4 eV, d) 1.89 eV, Sol: The first excitation potential of the atom is the energy required to shift the electron from, n=1 to n= 2., E1=, , 813.6, 12, , = -13.6 ;, , En =, E2 =, , # a., ², , # a., , ;, = -3.4, , excitation potential energy = final energy(E2) – initial energy(E1)= -3.4 –(-13.6) = 10.2, 6. According to Rutherford atom model, the spectral lines emitted by an atom is,, a) line spectrum, b) continuous spectrum, c) continuous absorption spectrum, d) band spectrum, , 186, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 14. In hydrogen atom, which of the following transitions produce a spectral line of maximum, frequency, a) 2 →1, b) 6 →2, c) 4 →3, d) 5 →2, Sol : E = hν = h, , i, ª, , [Energy is directly proportional to the frequency and inversely proportional to the wavelength., Hence energy is maximum frequency also maximum but wavelength is minimum and vice, En =, , versa], , # a., ², , a) 2 →1 ⇒ E2 –E1 ( -3.4 eV - -13.6 eV ( 10.2 eV, [6 →5 produce maximum energy and frequency but produce maximum wavelength], b) 6 →2 ⇒ E6 –E2 ( -0.37 eV - -3.4 eV ( 3.03 eV, c) 4 →3 ⇒ E4 –E3 ( -0.85 eV - -1.51 eV ( 0.66 eV, d) 5 →2 ⇒ E5 –E2 ( -0.54 eV - -3.4 eV ( 2.86 eV, Alternate method :, ª, , = R5, , ², , 8² 9;, i, , a, , a, , 8 9 = R ; ⇒ν =ª = R, , a) = R5, ª, , C = 0.75R, , C, , www.Padasalai.Net, b) = R5, , 8, , 9=, , c) = R5, , 8, , 9=, , 8, , 9=, , ª, , ª, , a, , d) = R5, ª, , E, , a, , Ï, , i, , R ; ⇒ν = =, ª, , i, , R ; ⇒ν = =, ª, , i, , R ; ⇒ν = =, ª, , a, , Ï, , R, , C = 0.222R, , R, , C = 0.0486R, , R, , C = 0.21R, , C, , C, , C, , ∴ (2 →1) produce maximum frequency, 15. After pumping process in laser,, a) the number of atoms in the ground state is greater than the number of atoms in the excited, state., b)the number of atoms in the excited state is greater than the number of atoms in the, ground state., c)the number of atoms in the ground state is equal to the number atoms in the excited state., d)No atoms are available in the excited state., 16. The chromium ions doped in the ruby rod, a) absorbs red light, b) absorbs green light, , 189, , J.SHANMUGAVELU, , c) absorbs blue light d) emits green light, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 24. If a and b are semi – major and semi – minor axes of the ellipse respectively and is the, orbital quantum number, then the expression to find the possible elliptical orbits is, ¡, , a) =, ¿, , F<, , b) 4 =, , À, , #, , 4, , c) =, , ², , F, , d), , ², , 4, , =, , #, , ², , 25. A crystal diffracts monochromatic X – Rays, If the angle of diffraction for the second, order is 90°, then that for the first order will be, a) 60°, b) 45°, c)30°, d)15°, Sol: 2d sin θ = n λ; and (d = λ);, , for n=2, , 2d sin 900 = 2 λ … . 1 ;, , for n=1, , 2d sin θ = 1 λ … . 2, , (2) ⇒ sin θ =, , ª, , ª, , =, ³, , ª, , (, , =300, , Alternating method, , for n=2 2d sin 900 = 2 λ…….(1);, 2d sin θ = 1 λ…….(2);, , for n= 1, , =, , .%² D, , .%² T, , =, , ⇒ sin θ (, , .%² T, , = ⇒ θ ( 30, , 26. If R is Rydberg’s constant, the minimum wavelength of hydrogen spectrum is, b) R/4, c)4/R, d) R, a) 1/R, , www.Padasalai.Net, Sol:ν, ¢=, , ª, , 8, , =R5, , ², , =R5, , 8, , λ( (, , ô, , £, , 9, , ², , 9 = R;, , â, , 27. The unit of Rydberg’s constant is, a) m, b) no unit, , d) m-1, , c) m2, , 28. For the first order X – ray diffraction, the wavelength of the X – ray is equal to the lattice, spacing at a glancing angle of, a) 15°, b) 60°, c)45°, d) 30°, Sol: 2d sin θ = n λ; and (d = λ);, , ⇒ 2d sin θ = 1d ⇒ sin θ = = 300, 29. A Coolidge tube operates at 18600 V. The maximum frequency of X – radiation emitted, from it is, a) 4.5 × 1018Hz, b) 45 × 1018Hz, c)4.05 × 1018Hz, d) 45.5 × 1018Hz, Sol:, , λmin =, , ¯, , νmax = ª, , i, , œ•š, , Å=, =, , a, , a, , ', , 10-10;, a, , = 4.5, , 1018 Hz, , Frequency can also be calculated as using h νmax = eV ⇒ νmax =, , 191, , J.SHANMUGAVELU, , [P.G. T. in Physics], , ¯, , ´, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 30. If the minimum wavelength of X – ray produced from a Coolidge tube is 0.062 nm , then, the potential difference between the cathode and target material is, c) 2 × 105 V, a) 2000 V, b) 20,000 V, d) 6.2 × 103V, Sol:, , λmin=, ⇒V=, , ¯, ªk%², , =, , Å, Å =, , ., , = 200, , S, , 102 =20000 V, , 31. The spectral series of hydrogen atom in UV region are called, a) Balmer series, b) Lymen series, c) Paschen series, Hints:, Series, Lyman, Balmer, Paschen, Brackett, Pfund, , d) Pfund series, , Region, Ultraviolet, Visible, Infrared, Infrared, Infrared, , 32. Maser materials are, a) Diamagnetic ions, c) ferromagnetic ions, , b) paramagnetic ions, d) non – magnetic ions, , 33. Number of waves per unit length is known as, a) Wavelength, b) wave number, c) bandwidth, , www.Padasalai.Net, d) frequency, , 34. The three dimensional image of an object can be formed by, a) Atomic spectroscopy, b) holography, c) molecular spectroscopy, d) MASER, , 35. In a discharge tube, the source of positive rays (canal rays) is, a) Cathode, b) anode, c) gas atoms present in the discharge tube, d) fluorescent screen, 36. The minimum wavelength of X – rays produced in an X – ray tube at 1000 KV is, c) 1.24 Å, a) 0.0124 Å, b) 0.124 Å, d) 0.00124 Å, Sol:, , λmin =, , ¯, , Å, , =, , S, , Å = 0.0124 Å, , 37. The ionisation potential of hydrogen atom is, b) − 13.6 eV, a) 13.6eV, , c) 13.6 V, , d) – 13.6 V, , 38. The value of Rydberg’s constant is, a) 1.094 × 10-7m1, b) 1.094 × 10-7m-1, , c) 1.094 × 107m-1, , d) 1.094 × 17m1, , 39. When an electric field is applied to an atom each of the spectral lines split into several, lines. This effect is known as, a) Zeeman effect, b) Stark effect, c) Raman effect, d) Seebeck effect, Hints:Electric field → stark effect , Magnetic field → Zeeman effect, , 192, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 40. The direction of viscous force in Millikon’s oil drop experiment is, a) always downwards, b) always upwards, c) opposite to the direction of motion of the oil drop d) either upwards or downwards, 41. In Sommerfeld atom model, for principal quantum number n = 3, which of the following, sub shells represents circular orbit?, a) 3s, b) 3p, c)3d, d) None of these, , Hints: 3d is circular orbit, remaining(3s,3p) is elliptical orbits, 42. In Millikon’s experiment, the plates are kept at a distance of 16 mm and are maintained, at a potential difference of 10000 V. The electric intensity is, a)62.5 V/m, b) 6.25 105 V/m, c)6.25 103 V/m, d) 1.6 105 V/m, V, , E=d=, , Sol:, , S, \, , =, , = 6.25, , 10E V/m, , 43. If R is Rydberg constant, the shortest wavelength of Paschen series is, a) R/9, b) 9/R, c) 16/R, d) 25/R, , www.Padasalai.Net, Sol: = R 5, , 8, , =R5, , 8, , ª, , 44., , þ, , ü, , ², , a, , ², , â, , 9, , ô, , 9 = T;, , λ(, , T, , ô, , of cathode ray particle, , a depends upon the nature of the cathode, b) depends upon the nature of the anode, c) depends upon the nature of the gas atoms present inside the discharge tube, d) is independent of all of these ., þ, , =¤–, , Hints: ( = = ( y- displacement of spot of light, E- electric field, K- constant, l- length of, ü, ¥L ¦, the plate ,B-magnetic field), 45. If c is the velocity ν the frequency and λ the wavelength of a radiation, then its frequency, is defined as, a) The number of waves in a distance of one metre, b) The number of waves in a distance of λ, c) The number of waves in a distance of c, d) The number of waves produced in a period of T second, 46. The wave number of a spectral line of hydrogen atom is equal to Rydberg’s constant. The, line is, a) first line of Lymen series, b) series limit of Lymen series, c) first line of Pfund series, d) series limit of Pfund series, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 54. If the potential difference between cathode and target of Coolidge tube is 1.24, then the minimum wavelength of continuous x ray is, a) 10Å, b) 1 Å, c) 0.1 Å, d) 0.01 Å, Sol:, , λmin =, , Å=, , ¯, , \, , ., , =, , 105 V, , Å, , Å=, , Å = 0.1 Å, , 55. A Coolidge tube operates at 24800 V. the minimum wave length of X – ray radiation, emitted from Coolidge tube is, a)6 1018m, b)3 1018m, c)0.6 10-10m, d)0.5 1010m, , λmin=, , Sol:, , Å, , ¯, , 10#, , =, , = 0.5, , 10# m, , 56. The ratio of areas enclosed by first three orbits of hydrogen atom is, a)1:2:3, b)1:8:27, c)1:4:9, d)1:16:81, Sol:, ( Area A ( πr ;, , r² (, , ² ´ 3, 2k, , ⇒ r² ∝ n ;, , r² ( n so r² ( xn |, , i. e, , = xn | : xn | : xna |, , πr : πr : πra = r : r : ra, , (π 8 is constant for all orbits, , = x1 | : x2 | : x3 | = 1:16:81, , www.Padasalai.Net, 57. In Thomson experiment, cathode rays moving with a velocity ‘v’ enter perpendicular to, an electric field of intensity ‘E’. the deflection produced by the cathode rays is directly, proportional to, a)v, b)v-1, c)v2, d)v-2, Hints : deflection y =, , y =, y∝, , š, , p, , §f ¨, , d, , š, , d, , p f, , f, , (or) v-2, , ∴ deflection y is inversely proportional to V2 (or) directly proportional to V-2, 58. The direction of electric field in Millikan’s oil drop experiment acts:, a) downwards, b) upwards, c) first upwards then downwards, d)first downwards, then upwards, Hints: The direction of electric field is positive to negative because the direction of electric, lines of force is positive to negative., In millikan’s oil drop experiment, positive terminal is located at upper side and negative, terminal is located at lower side. So the direction of electric field is positive to negative, (i.e) downwards., , 196, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 59. A beam of cathode rays moves from left to right in a plane of the paper and it enters into a, uniform magnetic field acting perpendicular to the plane of the paper and inwards. Now ,, the cathode rays are deflected :, a) downwards, b) upwards, c) in a direction perpendicular to the plane of the paper and inwards, d) in a direction perpendicular to the plane of the paper and outwards, Hints: the direction of current from right to left (opposite to the direction of cathode rays), Applying Fleming left hand rule, Direction of magnetic field – perpendicular to the plane of the paper and inwards, Direction of current, – right to left, Direction of force, – downward direction, 60. Wave number is :, a) number of waves produced in one second, b) number of waves in one meter length, c) number of waves in 3 x 108 m length, d) number of waves in λ meter length, 61. The colour of light emitted by ruby laser :, a) Green light, b) Red light, c) Yellow light, , d) white light, , 62. Arrange electron (e), proton(p) and deutron(d) in the increasing order of their specific, charge:, a) e, p, d, b) d, p, e, c) p, e, d, d) d, e, p, , www.Padasalai.Net, Hints : Electron, proton and deutron has one unit (1.6 10# T X charge., , But me = 9.11 10#a ’©, mp = 1.66 10# Ï ’©, md = 3.32 10# Ï kg., If mass is greater the value of, , p, , is smaller and vice versa., , Increasing order of their specific charge: d, p, e, Mathematically ,, For electron,, For proton ,, For proton ,, , 197, , p, , p, , p, , = T., =, =, , ., , ., , S, , ., , ., , a.a, , J.SHANMUGAVELU, , «, , «, , = 1.7592, , 10 CKg-1, , = 9.6385, , 10Ï CKg-1, , = 4.8192, , 10Ï CKg-1, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , CHOOSE THE CORRECT ANSWER FROM THE OPTIONS GIVEN, 63. In hydrogen atom, which of the following transition prduce spectral line of maximum, wavelength ______, a) 2 → 1, b) 4 → 1, c) 6 → 5, d) 5 → 2, 64. The chromium ions doped in ruby rod ____, a) absorbs red light, b) absorbs green light, c) absorbs blue light, d) emits green light, 65. The value of Rydberg’s constant is ______, a) 1.094X10-7 m-1, c) 1.094X107m1, , b) 1.094X10-7 m-1, d) 1.094X10-7 m-1, , 66. The quantity which takes discrete values, according to Bohrs postulates is ____, a)Kinetic energy, b) Potential energy, c) angular momentum, d) momentum, 67. The first excitation potential energy or the minimum energy required to excite the atom, from ground state of hydrogen atom is,, a) 13.6eV, b) 10.2eV, c) 3.4 eV, d) 1.89eV, 68. The life time of atoms for laser in excited state is ____, a) 10-8s, b) 10-3s, c) 108s, , d) 103s, , www.Padasalai.Net, 69. When an electron jumps from M shell to the vacant K shell, it contributes _____, a) ª— line, b) ª, line, c) `, line, d) `— line, 70. Positive column in a discharge tube is produced at a pressure of, a) 110 mm of Hg, b) 100 mm of Hg, c) 10 mm of Hg, d) 0.1 mm of Hg, , 71. The wavelength of radiations asorbed by chromium ions in Ruby laser is ______, a) 5800 A, b) 5400 A, c) 5400 A, d) 5500 A, 72. Sommerfield model explains the, a) Zeeman effect, b) distribution and arrangement of electrons in atom, c) intensities of spectral lines, d) back ground of fine structure of spectral lines., 73. For a given operating voltage the minimum wavelength of X-rays is ____, a) the same for all metals, b) not same for all metals, c) zero for some metals, d) high for certain metals, 74. Based on Thomson atom model, the frequency of spectralline obtained from Hydrodgen, atom is, a) 1300 A, b) 4861 A, c) 6363 A, d) 1500 A, 75. Laue used ____ crystals to demonstrate the diffraction of X-rays, a) rock salt, b) ZnS, c) quartz, , 198, , J.SHANMUGAVELU, , [P.G. T. in Physics], , d) CaCO3, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 76. Bohr quantization condition is, •, •, a) mvr=, b) mv=, , c) mvr=, , m•, , 2017-2018, d) mvr =, , m•, , 77. Mosley’s law led to the discovery of Chemical element ______, a) Helium, b) Iodine, c) Rhenium, , d) Radon, , 78. The life time of metastable state is ______, a) 10-5s, b) 10-3s, , d) 10-8s, , c) 10-4s, , 79. When X-rays fall on certain metals, they liberate _______, a) positrons, b) electrons, c) photons, , d) photoelectrons, , 80. In an X-ray tube, the intensity of emitted X-ray beam is increased by ____, a) increasing the filament current, b) decreasing the filament current, c) increasing the target potential, d) decreasing the target potential, 81. In He-Ne laser, the ratio of helium and neon is ____, a) 4:1, b) 1:4, c) 1:2, , d) 2:1, , 82. When an electron jumps from any outer orbits to the first orbit, then the emitted spectral, line is ____, a) Lymen series, b) Balmer series, c) Paschen series, d) Pfund series, , www.Padasalai.Net, 83. Striations are produced in a discharge tube _____ Hg pressure, a)1 mm, b) 10 mm, c) 0.01 mm, , d) 0.1 mm, , 84. The spacing between the atoms arranged in three dimensional space in crystal is of the, order of _____, a) 10-10m, b) 10-10cm, c) 10-8m, d) 10-8mm, 85. The continuous X-ray spectra consists of radiations of ______, a) well defined wavelengths, b) very low wavelengths, c) very high wavelengths, d) all possible wavelengths, 86. The size of an atom from Rutherford experiment is _____, a) 10-10m, b) 10-16m, c) 10-14m, , d) 10-12m, , 87. The energy of meta stable of Ne in He-Ne laser is ____, a) 20.66 eV, b) 1.89 eV, c) 10.2eV, , d) 3.4 eV, , 88. Achieving more atoms in the excited state than in the ground state is ____, a) population inversion, b) normal population, c) stimulated emission, d) spontaneous emission, 89. Bohr’s model fails because it explain ______, a) only the continuous spectrum, b) only the line spectrum, c) spectral lines of hydrogen atom and hydrogen like atom, d) the spectral lines of all atoms., , 199, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 90. The light from a LASER source is monochromatic because all the photons, a) are in phase, b) have same energy, c) have same amplitude, d) are in the same direction, 91. Einstein’s photoelectric effect and Bohr’s theory of hydrogen spectral lines confirmed, ______, a) energy of matter, b) dual nature of radiant energy, c) radiant energy, d) matter waves, 92. In Bohr atom model the energy of electron in the nth orbit is ____, a., # a., a) m eV, b) m eV, c) -13.6eV, 93. In an hydrogen atom, value of bohr values of bohr radius is, a) 0.53X10-8 m, b) 53 A, c) 0.53 A, , d) -1.36eV, , d) 530 nm, , 94. The spectral lines of hydrogen in UV region are called _______, a) Balmer’s series, b) Lymen series, c) Paschen series, d) Pfund series, 95. A crystal diffracts monochromatic X-rays. If the angle of diffraction for first order is 300,, then that for second order will be _____, a) 900, b) 150, c) 600, d) 450, 96. For the Principal quantum number 3, the possible l values are ________, a) 3,2,1, b) 2,1,0, c) 1,0,-1, d) 0,-1,-2, , www.Padasalai.Net, 97. The glancing angle of monochromatic X-ray of wavelength 1A0 is 300. The lattice space, between the scond order reflection is ____, a) 2X10-10 m, b) 2X10-9 m, -10, d) 2X10-9 cm, c) 2X10 cm, , 98. In a Bragg’s spectrometer the glancing angle for the fourth order of spectrum of X-ray, found to be 300. What will be the glancing angle for the occurrence of first order, spectrum?, a) sin-1(0.25), b) sin-1 (0.217), c) sin-1 (0.5), d) sin-1 (0.125), 99. The radius of second orbit of hydrogen atom is ________, a) 0.53A, b) 2.12A, c) 1.06A, , d) 4.24A, , 100. The ionization potential of the hydrogen atom is 13.6eV. The energy of the atom in n=2, state is, a) -13 eV, b) -3.4 eV, c) 3.4 eV, d) 13.6 eV, 101. The ratio of the specific charge of an electron to that of a proton is ____, a) 1:2, b) 1:1, c) 2:1, d) 1;4, 102., An electron moves through an electric field of intensity 9X103V/m. If the mass of the, electron is 9.1X10-31 kg, then the acceleration of electron is _____, a) 1.71X10-15 m/s2, b) 1.6X10-15 m, c) 1.6X1015 m/s2, d) 1.58X10-14 m/s2, , 200, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 202 : www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 103. An electron with speed of 2.5X107 m/s suffers a deflection in a magnetic field of, induction 2X10-3 T then the electric field that would give the same deflection is ________, a) 5X10-3 V/m, b) 1.25X103 V/m, 2, d) 5X104 V/m, c) 12.5X10 V/m, 104. For a first order X-ray diffraction, the wavelength of X-ray equal to the interplanar, distance at a glancing angle of ______, a) 450, b) 300, c) 600, d) 150, 105. Radius of first orbit of hydrogen atom is 0.53 A0 then the radius of third orbit is ____, a) 59 A, b) 4.774 A, c) 1.06 A, d) 2.12 A, 106. The minimum wavelength an X-ray coming out of X-ray tube under a potential, difference of 1000 volt is _____, a) 12.4 A, b) 12400 A, c) 1240 A, d) 0.0124 A, 107. Given that the charge on an electron is 1.6X10-19C,What is the energy gained by the, cathode ray particles when a voltage of 800 volts is applied between the electrodes of a, cathode ray tube_____, a) 2X10-21 J, b) 8X10-17 J, c) 1.28X10-18 J, d) 1.28X10-16 J, 108. The charge on an oil drop is 12.82X10-19 C, then the number of elementary charges are, ____, a) 6, b) 2, c) 7, d) 8, , www.Padasalai.Net, 109. In Millikan’s oil drop experiment, two plates separated by 5 cm in air are at a, potential of 5 V, then the electric field is _______, a) 1 V/m, b) 10 V/m, c) 100 V/m, d) 2 V/m, 110. 1 MeV=___________, a) 1.602X10-19 J, c) 1.602X10-13 J, , b) 1.602X10-16 J, d) 1.602X10-31 J, , Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], Contact for subject doubt, Email :
[email protected], Phone No: 9952223467, , JEYAM TUITION CENTRE, [PARAMAKUDI], Contact for tuition, Email :
[email protected], Phone No: 9944888512, , 201, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , ‘3’ MARK QUESTIONS:(2- Questions : Q. No: 41, 42), PUBLIC ‘3’ MARKS:, 1., , Write an three properties of cathode rays? (O-15), i) They travel in straight lines, ii) Cathode rays ionize the gas through which they pass., iii) Cathode rays affect the photographic plates., iv) The cathode rays are deflected from their straight line path by both electric and, magnetic fields. The direction of deflection shows that they are negatively charged, particles., , 2., , What is the principle of Millikan’s oil drop method? ( J – 06, M – 12, J-15 ), This method is based on the study of the motion of uncharged oil drop, under free fall due to gravity and charged oil drop in a uniform electric field., By adjusting uniform electric field suitably, a charged oil drop can be made to, move up or down or even kept balanced in the field of view for sufficiently, long time and a series of observations can be made., , 3., , Explain any one of the drawbacks of Rutherford atom model. ( O – 08 ), According to classical electromagnetic theory, the accelerating electron must, radiate energy at a frequency proportional to the angular velocity of the electron., Therefore, as the electron spiral towards the nucleus, the angular velocity tends to, become infinity and hence the frequency of the emitted energy will tend to, infinity. This will result in a continuous spectrum with all possible wavelengths., , www.Padasalai.Net, 4., , Define: ionisation potential of on atom. ( M – 07, O – 10, M-15 ), The ionization potential is that accelerating potential which makes the, impinging electron acquire sufficient energy to knock out an electron from the, atom and thereby ionise the atom., 13.6 V is the ionisation potential of hydrogen atom., , 5., , What is ionisation potential energy? ( J – 09 ), For hydrogen atom, the energy required to remove an electron from first orbit, to its outermost orbit (n=∞) is 13.6-0 = 13.6eV. This energy is known as the, ionisation potential energy for hydrogen atom., , 6., , Give the drawbacks of Sommerfeld’s atom model. ( O – 11 ), i) It could not explain the distribution and arrangement of electrons in atoms., ii) Sommerfeld’s model was unable to explain the spectra of alkali metals such, , as sodium, potassium etc., iii) It could not explain Zeeman and Stark effect., iv) This model does not give any explanation for the intensities of the spectral, lines., , 202, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 7., , Give the differences between hard X-rays and soft X-rays.(M-14), , 1., 2., 3., 4., , 8., , 2017-2018, , Soft X rays, Wavelength is 4 Å or above, Have lesser frequency and lesser, energy, Have low penetrating power, They are produced at low potential, , Hard X rays, Wavelength is in the order of 1 Å, Have high frequency and high energy, Have high penetrating power, They are produced at high potential, , Write down two important facts of Laue experiment on X-ray diffraction., (O–07,J–08,M-13), The Laue experiment has established following two important facts :, i) X–rays are electromagnetic waves of extremely short wave length., ii) The atoms in a crystal are arranged in a regular three dimensional lattice., , 9., , State Moseley’s law. ( M – 07, M – 08, M – 09, M – 11,J-14 ), The frequency of the spectral line in the characteristics X-ray spectrum is, directly proportional to the square of the atomic number (Z) of the element, considered., , √ν = a(Z − b), , www.Padasalai.Net, ναZ2, , or, , Where a and b are constants depending upon the particular spectral line., , 10., , Give the applications of Moseley’s law.(O-14, J-15), i) Any discrepancy in the order of the elements in the periodic table can be, , removed by this law by arranging the elements according to the atomic, numbers and not according to the atomic weights., ii) Led to the discovery of new elements like hafnium (72), technetium (43),, rhenium (75) etc., iii) helpful in determining the atomic number of rare earths, thereby fixing their, position in the periodic table., , 11. Write, , any three medical applications of X-rays. ( M-17), , i) X–rays are being widely used for detecting fractures, tumours, the presence of, foreign matter like bullet etc., in the human body., ii) X–rays are also used for the diagnosis of tuberculosis, stones in kidneys, gall bladder, etc., iii) Many types of skin diseases, malignant sores, cancer and tumours have been cured, by controlled exposure of X-rays of suitable quality., iv) Hard X–rays are used to destroy tumours very deep inside the body., , 203, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 12., , What are the characteristics of laser beam?(O – 06,J–09, M – 10 , J– 10, J – 12 ), The laser beam, i) Is monochromatic., ii) Is coherent, with the waves, all exactly in phase with one another,, iii) Does not diverge at all and, iv) Is extremely intense., , 13., , What are the conditions to achieve the laser action? ( M – 06, J – 07,O-13,J14,J-16 ), i) There must be an inverted population i.e. more atoms in the excited state than, , in the ground state., ii) The excited state must be a metastable state., iii) The emitted photons must stimulate further emission. This is achieved by the, , use of the reflecting mirrors at the ends of the system., , 14., , Write any three applications of laser in industry. ( J – 08 ), i) The laser beam is used to drill extremely fine holes in diamonds, hard sheets, , etc.,, ii) They are also used for cutting thick sheets of hard metals and welding., iii) The laser beam is used to vapourize the unwanted material during the, , www.Padasalai.Net, manufacture of electronic circuit on semiconductor chips., iv) They can be used to test the quality of the materials., , 15., , Write any three medical applications of laser. ( M – 08,O – 09,J – 11,O – 11,O16 ), i) In medicine, micro surgery has become possible due to narrow angular spread, , of the laser beam., ii) It can be used in the treatment of kidney stone, tumour, in cutting and, sealing the small blood vessels in brain surgery and retina detachment., iii) The laser beams are used in endoscopy., iv) It can also be used for the treatment of human and animal cancer., , 16., , What is hologram? ( O – 07 ), , A three dimensional image of an object can be formed by holography., , In ordinary photography, the amplitude of the light wave is recorded on the, photographic film. In holography, both the phase and amplitude of the light, waves are recorded on the film. The resulting photograph is called hologram., , 204, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , OTHER IMPORTANT ‘3’ MARKS:, 17., , Write any three properties of canal rays., i) They travel in straight lines., ii) They affect photographic plates., iii) These rays can produce fluorescence., iv) They ionize the gas through which they pass, , 18., , State the observations and conclusions in Rutherford - particles scattering, experiment, Most of the α particle either passed straight through the gold foil or were scattered, by only small angles of the order of a few degrees. This observation led to the, conclusion that an atom has a lot of empty space, , 19., , What do you meant by the distance of closest approach ?, An particle directed towards the centre of the nucleus will move close upto a, distance ro where its kinetic energy will appear as electrostatic potential energy. After, this, the particle begins to retrace its path. This distance ro is known as the distance of, the closest approach., , www.Padasalai.Net, 20., , What is normal population., , In a system of thermal equilibrium, the number of atoms in the ground state (N1) is, greater than the number of atoms in the excited state (N2). This is called normal, population, , 21., , Define stimulated or induced absorption., sample of free atoms, some of which are in the ground state with energy E1 and, some in the excited energy state with energy E2. If photons of energy hν = E2-E1 are, incident on the sample, the photons can interact with the atoms in the ground state and, are taken to excited state. This is called stimulated or induced absorption, , 22., , What is optical pumping, If the atoms are taken to the higher energy levels with the help of light, it is called, optical pumping., , 23., , What is meant by population inversion?, If the atoms in the ground state are pumped to the excited state by means of external, agency, the number of atoms in the excited state (N2) becomes greater than the number, of atoms in the ground state (N1). This is called population inversion., , 205, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 24., , 2017-2018, , Define spontaneous emission., If the excited energy level is an ordinary level, the excited atoms return to the lower, (or) ground energy state immediately without the help of any external agency. During, this transition a photon of energy E2-E1 = hν is emitted. This is called spontaneous, emission., , 25., , Define stimulated or induced emission., If the excited state is a metastable state, the atoms stay for some time in these, levels. The atoms in such metastable state can be brought to the lower energy levels, with the help of photons of energy hν = E2 – E1. During this process, a photon of, energy E2 – E1 = hν is emitted. This is known as stimulated emission (or) induced, emission, , 26., , What are drawbacks of Thomson atom model?, i) According to electromagnetic theory, the vibrating electron should radiate energy, , and the frequency of the emitted spectral line should be the same as the electron. In, the case of hydrogen atom, Thomson’s model gives only one spectral line of about, 1300 Å. But the experimental observations reveal that hydrogen spectrum consists, of five different series with several lines in each series., ii) It could not account for the scattering of α-particles through large angles., , www.Padasalai.Net, 27., , State the postulates of Bohr atom model?, , i) Bohr’s quantization condition: An electron cannot revolve round the nucleus in all, , possible orbits. The electrons can revolve round the nucleus only in those allowed or, permissible orbits for which the angular momentum of the electron is an integral, multiple of, , •, , (where h is Planck’s constant = 6.626 × 10-34 Js)., These orbits are called stationary orbits or non- radiating orbits and an electron, revolving in these orbits does not radiate any energy., ii) Bohr’s frequency condition: An atom radiates energy, only when an electron, , jumps from a stationary orbit of higher energy to an orbit of lower energy. If the, electron jumps from an orbit of energy E2 to an orbit of energy E1, a photon of, energy hν = E2 – E1 is emitted., 28., , Why ordinary plane grating cannot be used to produce diffraction effects in, X-ray., In order to diffract X – ray, grating with much finer rulings having distance, between rulings comparable to the wavelength of X – rays are required. It is, impossible to construct a grating of such fine dimensions artificially., , 206, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 29., , Define excitation potential energy and excitation potential., Excitation potential energy :, The energy required to raise an atom from its normal state into an excited state is, called excitation potential energy of the atom., Excitation potential :, The energy required to raise an atom from its normal state into an excited state is, called excitation potential energy of the atom., For example, the energy required to transfer the electron in hydrogen atom from the, ground state to the first excited state = (13.6-3.4) = 10.2eV., The energy required to raise it to the second excited state = (13.6 – 1.51) = 12.09 eV., The potentials corresponding to these energies are called as the excitation potentials., , 30., , Define critical potential of an atom, The excitation potential and ionization potential are called as the critical potentials of, the atom. The critical potential of an atom, is defined as the minimum potential required, to excite a free neutral atom from its ground state to higher state., , 31., , How Balmer series is produced?, , www.Padasalai.Net, When the electron jumps from any of the outer orbits to the second orbit, we get a, spectral series called the Balmer series. All the lines of this series in hydrogen have, their wavelength in the visible region., Here n1=2, n2 = 3,4,5 …, , The wave number of the Balmer series is N̅ = R5 8, 32., , m, , 9, , How lyman series formed:, When the electron jumps from any of the outer orbits to the first orbit, the spectral, lines emitted are in the ultraviolet region of the spectrum and they are said to form a, series called Lyman series, Here, n1 = 1, n2 = 2,3,4 …, The wave number of the Lyman series is given by N̅ = R51 8, , 33., , m, , 9, , How can X – rays be detected?, i) blackening of a photographic plate and, ii) the ionization produced by X–rays in a gas or vapour. An ionization chamber,, , which utilizes the property of ionization, is generally used to detect and measure the, intensity of X-rays., , 207, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 34., , 2017-2018, , Write the characteristics of anode in Coolidge tube, i) high atomic weight – to produce hard X-rays, ii) high melting point – so that it is not melted due to the bombardment of fast moving, electrons, which cause lot of heat generation., iii) high thermal conductivity – to carry away the heat generated., , 35., , State Bragg’s law., If this path difference 2d sin θ is equal to integral multiple of wavelength of X-ray, i.e. nλ, then constructive interference will occur between the reflected beams and they, will reinforce with each other. Therefore the intensity of the reflected beam is, maximum., ∴ 2d sin θ = n λ, where, n = 1, 2, 3 … etc., , 36. Write, , any three industrial applications of X-rays., , i) X–rays are used to detect the defects or flaws within a material, ii) X–rays can be used for testing the homogeneity of welded joints, insulating, materials etc., iii) X-rays are used to analyse the structure of alloys and the other composite, bodies., iv) X–rays are also used to study the structure of materials like rubber,, cellulose, plastic fibres etc., , www.Padasalai.Net, 37. Write, , any three scientific research applications of X-rays., , i) X–rays are used for studying the structure of crystalline solids and alloys., ii) X–rays are used for the identification of chemical elements including, determination of their atomic numbers., iii) X–rays can be used for analyzing the structure of complex molecules by, examining their X–ray diffraction pattern., 38., , Distinguish between laser light and ordinary light., Laser light, 1. Coherent with the waves, all exactly, in phase with one another, 2. Monochromatic light, 3. does not diverge at all, 4. extremely intense, , 208, , J.SHANMUGAVELU, , Ordinary light, Incoherent with does not exactly in, phase with one another, May be monochromatic, or, heterochromatic, diverge at all, Does not extremely intense, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PUBLIC ‘3’ MARK PROBLEMS:, 1., , −, , A beam of electrons moving with a uniform speed of 4 × 107 ms 1 is projected, normal to the uniform magnetic field where B = 1 × 10-3 Wb/m2. What is the path, of the beam in magnetic field?[M-12], , Data : v = 4 ×107 ms−-1;, , B = 1 × 10-3 Wb/m2;, , path of the electron, r = ?, , Solution : Since, the electrons are released normally to the magnetic field, the, electrons travel in a circular path., , ∴ Bev =, r=, =, , pË, , pË, S, , T., , S, , «, , ., , r = 0.2275 m, −, , The Rydberg constant for hydrogen is 1.097 × 107m 1. Calculate the short, wavelength limits of Lyman series.[O-06,M-17], , www.Padasalai.Net, 2., , −, , Data : R = 1.097 × 107m 1 ; For short wavelength limit of Lyman Series, n1 = 1, n2 = ∞,, λs= ?, Solution: The wave number for Lyman series is,, N̅ =, , c, , For short wavelength limit,, , ( R 5m 8, N¬Í =, , cW, , =R5, , ∴λs =, =, , 9, , m, , 8, , â, , 9=R, , ì, . TÏ, , «, , λs = 911.6Å, , 209, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PUBLIC ‘5’ MARK PROBLEMS:, 13., , An electron beam passes through a transverse magnetic field of 2 × 10−-3 tesla and, an electric field E of 3.4 × 104 V/m acting simultaneously. If the path of the, electrons remain undeviated, calculate the speed of electrons. If the electric field, is removed, what will be the radius of the electron path?[O-10], , Data: B = 2 × 10−-3 Tesla;, Sol:, , E = 3.4 × 104 V/m;, , V, , =, =, , Ó, , a., , v = 1.7, F=, , r=?, , š, , S, , 107 ms, , -1, , pË, , F = Bev, pË, , www.Padasalai.Net, Bev =, , Be =, , r=, =, , pË, , pË, , T., , S, S, , ., , .Ï, , «, , r = 4.834 10-2 m, , 214, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , FIVE MARKS:(1 – question : Q. No: 57), 1. Write any five properties of cathode rays. (O-08,J-09,O-14,M-17), 2. Write any five Properties of canal rays.(M-14), 3. Derive the expression for radius of the electron in the nth orbit of hydrogen atom. (J-13), 4. Prove that the energy of electron for hydrogen atom in the nth orbit is, #p Ó, (O-07,J-13), En =, m •, 5. Explain the spectral series of hydrogen.(Diagram not necessary), (M-06,M-10,J-10,M-12,M-13,J-14,J-16,O-16), , 6. Mention any five properties of X-rays. (J-06,M-11,O-13, M-15), 7. Describe Laue experiment. What are the facts established by it? (O-06,O-15), 8. State and obtain Bragg’s law. (J-08,O-09,O-11, J-15), 9. Explain the origin of characteristic X-rays. (M-09,J-11,J-12,O-12), , www.Padasalai.Net, TEN MARKS:(1 – question : Q. No: 67), , 1. Describe J.J Thomson method for determine the specific charge of an electron., (O - 09, M - 10, J - 10, O - 10, O – 11, J – 12,O-13,O-14,J-16,O-16), , 2. Describe Millikan’s oil drop experiment to determine the charge of an electron., (J - 08, O - 08), 3. State postulates of Bohr atom model. Obtain an expression for the radius of nth orbit of, hydrogen atom based on the Bohr’s Theory. (M - 06 , M - 08, J - 09, M -12,M-14,O-15), 4. How will you determine the wavelength of x-rays using Bragg’s spectrometer. Write, any five properties of x-rays. (M - 07), 5. Derive Bragg’s law. Explain how a Bragg’s spectrometer can be used to determine the, wavelength of x-rays. (J – 07;J-14), 6. Draw a neat sketch of ruby laser. Explain its working with the help of energy level, diagram. (O - 06, J - 07, M - 09, J -11, O -12, M – 13, M – 15,M-17), 7. Explain the working of He-Ne laser with the help of energy level diagram., (J - 06, M - 11, J – 13,J-15), , 217, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, Only for slow bloomers, ( 5 & 10 Marks answer ), 1. spectral series of hydrogen:, , when the electron jumps from outer orbit (n2) to lower orbit (n1) the following, spectral series are emitted, Series, Lyman, , Lower, orbit n1, 1, , Outer, orbit n2, 2,3,4…., , Region, Ultraviolet, , Balmer, , 2, , 3,4,5…., , Visible, , Paschen, , 3, , 4,5,6…., , Infrared, , Brackett, , 4, , 5,6,7…., , Infrared, , Pfund, , 5, , 6,7,8…., , Infrared, , ¬, Wave number °, ±̅ = R§1 8, , ±̅ = R§ 8, ±̅ = R§ 8, , ±̅ = R§, , T, , E, , m, m, m, , ¨, , ¨, ¨, , 8m ¨, , ±̅ = R§a 8 m ¨, , www.Padasalai.Net, 2. Laue experiment:, , X–rays from the X–ray tube is collimated into a fine beam by two slits S1 and S2, It passes through a ZnS. Crystal, The emergent rays are made to fall on a photographic plate P., The diffraction patten so obtained consists of a central spot at O and a series of spots, arranged in a definite pattern about O. These spots are known as Laue spots., Important facts :, (i), (ii), , X–rays are electromagnetic waves of extremely short wave length., The atoms in a crystal are arranged in a regular three dimensional lattice., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 5. Millikan’s oil drop experiment, , Principle: The study of the motion of uncharged oil drop under free fall due to, gravity and charged oil drop in a uniform electric field, Diagram :, , Construction :, It consists of two horizontal circular, metal plates A and B, about 22 cm in, diameter and separated by a distance of, about 16 mm., The plates are surrounded by a constant, temperature bath D and the chamber C, containing dry air., The plates are connected to a 10000 V, battery. highly viscous liquid is enter the, space between A and B. The terminal, velocity of the droplet is v., , Th, ³ e net downward force µ = weight of the oil drop – upthrust experienced by, acting on the oil drop, , Motion under gravity:, , = Uaa (© 8, a, , a, , Uaa •©, , the oil drop., , www.Padasalai.Net, = Uaa ( 8 • ©, a, , By Stoke’s law,, a, , Uaa ( 8 • © = 6Uaηv, a(5, , T·T, , ð#G ø, , 9, , Motion under electric field:, Eq = Uaa ( 8 • © 7 6Uaηv, a, , Eq -, , a, , Uaa ( 8 • © ( 6Uaηv, , Eq = 6Uy, E=, , aæ, , v+v, , 5, , TT, , ð#G ø, , 9, , f, ³, , e = 1.6, , 10-19 C, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 223 : www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 7., , 2017-2018, , Ruby laser :, , Construction :, It consists of ruby rod of length 10 cm and 0.8 cm in diameter. A ruby is a crystal, of aluminium oxide Al2O3, in which some of aluminium ions ( Al3+ ) are replaced by the, chromium ions (Cr3+). One end is fully silvered and the other is partially silvered. The, ruby rod is surrounded by a helical xenon flash tube., Diagram :, , Working: When the 5500 Å radiation (green colour) photons are absorbed by the, chromium ions which are pumped to the excited state E3., This ions decay without radiation to the meta stable state E2 has a much longer lifetime, (10-3s)., When the excited ion from E2 drops down spontaneously to the ground state E1, it, emits a photon of wavelength 6943 Å., This photon is reflected back and forth by the silvered ends, it stimulates other excited, ion and emit a fresh photon in phase with stimulating photon., Finally, a red light of wave length 6943 Å emerges through the partially silvered end, , www.Padasalai.Net, Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], Contact for subject doubt, Email :
[email protected], Phone No: 9952223467, , JEYAM TUITION CENTRE, [PARAMAKUDI], Contact for tuition, Email :
[email protected], Phone No: 9944888512, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , JEYAM TUITION CENTRE PARAMAKUDI, , www.TrbTnpsc.com, , 2017-2018, , 8. He – Ne laser :, , Construction:, A continuous and intense laser beam can be produced with the help of gas lasers., He – Ne laser consists of a quartz discharge tube containing helium and neon in the, ratio of 1 : 4 at a pressure of about 1 mm of Hg. One end of the tube is fitted with a, perfectly reflecting mirror and the other end with partially reflecting mirror. A powerful, radio frequency generator is used to produce a discharge in the gas, so that the helium, atoms are excited to a higher energy level., Diagram :, , www.Padasalai.Net, Working : When an electric discharge passes through the gas, the electron collide, with the He and Ne atoms and excite them to metastable states of energy 20.61 eV and, 20.66 eV respectively., He atom help in achieving a population inversion in Ne atoms by collision. When an, excited Ne atom drops down spontaneously to 18.70 eV, it emits a 6328 Å photon., This photon is reflected back and forth by the reflector ends, it stimulates an excited, neon atom and emit a fresh 6328 Å photon., The output radiations escape from the partially reflecting mirror., The neon atoms drop down to lower state E, through spontaneous emission emitting, incoherent light., The Ne atoms are brought to the ground state through collision with the walls of the, tube., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 7.DUAL NATURE OF RADIATION, AND MATTER RELATIVITY, PREPARED BY, J.SHANMUGAVELU M.Sc, B.Ed. [P.G. Assist in Physics], , ONE MARK QUESTIONS ( 2- QUESTIONS), BOOK BACK ONE MARKS:, 1. A photon of frequency ν is incident on a metal surface of threshold frequency νο. The, kinetic energy of the emitted photoelectron is, a) h (ν – νο), b) hν, c) hνο, d) h (ν+ νο), 1, , hν = hνο + 2mv2, , Sol:, , ⇒ mv2 = hν - hνο, , K.E = h (ν –νο), 2. The work function of a photoelectric material is 3.3 eV. The threshold frequency will be, equal to, a) 8 × 1014Hz, b) 8 × 1010Hz, c) 5 × 1020Hz, d) 4 × 1014Hz., Sol:, W = hνο, , www.Padasalai.Net, ⇒νο =, , $, , ´, , =, , = 0.8, , a.a, , ., , ., , 1015 = 8, , SÓ, , 1014Hz, , 3. The stopping potential of a metal surface is independent of, a) frequency of incident radiation, b) intensity of incident radiation, c) the nature of the metal surface, d) velocity of the electrons emitted., 4. At the threshold frequency, the velocity of the electrons is, a) zero, b) maximum, c) minimum, , d) infinite, , Hints:At threshold frequency, the photo electric emission is not possible., So velocity of the electron is Zero, , 5. The photoelectric effect can be explained on the basis of, a) corpuscular theory of light, b) wave theory of light, c) electromagnetic theory of light, d) quantum theory of light, 6. The wavelength of the matter wave is independent of, a) mass, b) velocity, c) momentum, Hints: λ, , ", = ü¸, , ", , d) charge, , = ý;, , h – planck constant, v – velocity, m – mass, p – momentum, so independent of charge., , 224, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 7. If the kinetic energy of the moving particle is E, then the de Broglie wavelength is,, , a)λ =, Sol:, , ", , h, =, √2Vem, , √ kö, ´, , b) λ =, , √=ü–, , λ, , kinetic energy, ∴λ=, , mv2 = eV;, , c) λ = h√2mE, , ´, , d) λ = ö√, , k, , given eV = E ;, , ´, , √ kö, , 8. The momentum of the electron having wavelength 2Å is, a) 3.3 × 1024kg m s−1, b) 6.6 × 1024kg m s−1, −24, −1, c) 3.3 × 10 kg m s, d) 6.6 × 10−24kg m s−1, h, , λ = mv =, , Sol:, , ´, , ⇒p= =, ª, , = 3.3, , ´, , ù, SÓ, , ., , 10-24kg m s−1, , 9. According to relativity, the length of a rod in motion, a) is same as its rest length, b) is more than its rest length, c)is less than its rest length, d)may be more or less than or equal to rest length depending on the speed of the rod, Hints: b < b, (b – length of the rod in rest, b - length of the rod in motion), 10. If 1 kg of a substance is fully converted into energy, then the energy produced is, b) 9 × 1024J, c) 1 J, d) 3 × 108J, a) 9 × 1016J, 2, Sol:, E = mc, = 1 ( 3 108) = 9 1016J, , www.Padasalai.Net, PUBLIC ONE MARKS:, 11. The value of stopping potential when the frequency of light is equal to the threshold, frequency is, a) maximum, b) zero, c) minimum, d) infinity, Hints:At a threshold frequency the value of stopping potential is zero., 12. Two photons, each of energy 2.5 eV are simultaneously incident on the metal surface. if, the work function of the metal is 4.5 eV then from the surface of the metal, a) one electron will emitted, b) two electrons will emitted, d) not a single electron will be emitted, c) more than two electrons will be emitted, Hints:The energy of a photon is less than work function. So electrons not emitted, 13. According to special theory of relativity the only constant in all frames is, a) mass, b) length, c) time, d) velocity of light, 14. The work function of a photoelectric material is 6.626 × 10-19J. The threshold frequency is, a) 1 × 1015Hz, b) 10 × 10-19 Hz, c)1 × 10-15Hz, d) 10 × 1019Hz, Sol: W = hνο, , =, , 225, , ., ., , SÓ, , = 1 × 1015Hz, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 15. When a material particle of rest mass ‘mo’ attains the velocity of light its mass becomes, a) 0, b) 2mo, c) 4 mo, d) ∞, k, Sol:, m=, ;, (now v = c);, , m=, m=, , _ #¹, , k, , º, , _ #º, , k, , º, , _, , º, , º, , º, , =, , k, , =∞, , 16. Photon has, a)energy but zero mass, c) zero mass and zero energy, , b) mass but zero energy, d) infinite mass and energy, , 17. An electron of mass m and charge e associated from rest through a potential of V volt,, then its final velocity is, ¯, , a)_ k, Sol:, , b)_, 1, mv2, 2, , = eV ⇒ v = _, , c) _, , =þ», , ¯, , k, , ¯, , d), , ü, , ¯, , k, , k, , 18. Einstein’s photoelectric equation is, a) W + ℎν = ½ mv2max, b) ½ mv2max =W, , c) ℎν+½ mv2max= W, , d)W+ ½ mv2max = ℎν, , 19. Electron microscope works on the principle of, a) photoelectric effect, b) particle nature of electron, c) wave nature of moving electron, d) dual nature of matter, , www.Padasalai.Net, 20. A graph is drawn taking frequency of incident radiation ( ν) along the X – axis and its, stopping potential ( Vo) along the Y – axis . The nature of the graph is, b) a parabola, c) an ellipse, d) a circle, a) a straight line, Hints:If the frequency of the incident radiation is plotted against the corresponding stopping, potential, a straight line is obtained, , 21. photon of energy2E is incident on a photosensitive surface of photoelectric work, function E. The maximum kinetic energy of photoelectron emitted is, b) 2E, c) 3E, d) 4E, a) E, Sol:, hν =2E and W =E;, Energy of the incident photon =Work function + Kinetic energy of the electron, 1, , hν = W + 2mv2, , ⇒ mv2= hν – W =2E –E =E, 22. The de Broglie wavelength of electron accelerated with a potential V is, , a) λ =, , h, √Vem, , b) λ =, , h, √2Vem, , c) λ =, , ", √=»þü, , d) λ =, , ´, , 2½, , k_ œ, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 23. In the photoelectric phenomenon if the ratio of the frequency of incident radiation, incident on a photosensitive surface is 1: 2 : 3 the ratio of the photoelectric current is, a) 1 : 2 :3, b) √1: √2: √3, c) 1 : 4 : 9, d) 1 : 1: 1, Hints: frequency of incident radiation is independent of photoelectric current, 24. When the momentum of a particle increases its de Broglie wavelength, a) increases, b) decreases, c) does not change, d) infinity, Hints:λ =, , ´, , ´, , = ù; momentum P is inversely proportional to the wave length λ. If momentum, kT, , of a particle increases its de Broglie wavelength is decreases, , 25. The number of de Broglie waves of an electron in the nth orbit of an atom is, a)n, b)n-1, c)n+1, d)2n, Hints:The number of de Broglie waves of an electron in the nth orbit of an atom is integral, multiple of de Broglie wave length, 26. When an electron is accelerated with potential difference V, its de Broglie wavelength is, directly proportional to, a)V, b) V-1, c) V1/2, d) V-1/2, Sol: λ =, , ´, , √ ¯ k, , ⇒ λ∝, , √¯, , = V-1/2, , 27. In photoelectric effect, a graph is drawn taking the frequency of incident radiation along, X - axis and the corresponding stopping potential along the y -axis. The nature of the, graph is:, a) a straight line passing through origin b) a straight line having positive y – intercept, d) a parabola, c) a straight line having negative y – intercept, , www.Padasalai.Net, 28. The length of the rod placed inside a rocket is measured as 1m by an observer inside the, rocket which is at rest. When the rocket moves with a speed of 36 106 km/hr the length, of the rod as measured by the same observer is :, a) 0.997 m, b) 1.003 m, c) 1m, d) 1.006 m, Hints: The rod and an observer placed inside a rocket. When the rocket moves with a speed of, 36 106 km/hr and the rod , observer moves with same speed. So rod does not contract., 29. The threshold frequency of a photosensitive surface is 5 × 1014 Hz. Then which of the, following will produce photoelectric effect from the same surface?, a) Sodium vapour lamp, b) Ruby laser, c) He-Ne laser, d) Both (b) and (c)., Hints : The photo electric effect is possible, if the frequency of the incident photon is greater, than that of threshold frequency, In other words If the wavelength of the incident photon is, smaller than that of threshold wavelength., Wave length of Sodium vapour lamp is 5896Å and 5890Å, ', P, a, Threshold wavelength is Ñ =, =, = 0.6 10# Ñ = 6000 Å, Ó, ν0, , E, , The wavelength of sodium vapour lamp is smaller than threshold wavelength of photosensitive, surface. So sodium vapour lamp will produce photoelectric effect from the same surface, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , Choose the correct answer from the options given, 30. A photon of frequency ν is incident on a metal surface of threshold frequency νₒ The, Kinetic energy of emitted photo electron is __________, (a) h ( ν - νₒ ), (b) hν, (c) hνₒ, (d) h (ν + νₒ), 31. Focal length of the electromagnetic lens used in an electron microscope depends, on_______, (a) The velocity of electrons, (b) The magnitude of the current passing through energizing coils., (c) The medium between the energizing coils, (d) All the above, 32. Photoelectric current depends upon __________, (a) intensity of incident light, (b) frequency of incident light, (c) the potential difference between two plates, (d) all the above, 33. Electron microscope is operated in __________, (a)high pressure, (b) high vacuum, (c) normal pressure, (d) none of the above, , www.Padasalai.Net, 34. Stopping potential of a metal surface is independent of ___________, (a) frequency of incident radiation, (b) intensity of incident radiation, (c) the nature of metal surface, (d) velocity of electron emitted, , 35. If the KE of a moving particle is E, then the de Broglie wavelength is_______, (a) λ = h / √2 t, (b) λ = √2 t / h, , (c) λ = h√2 t, , (d) λ = E√2 ℎ, , 36. In photo cell the light energy is converted into___________, (a) sound energy, (b) magnetic energy, (c) electric energy, (d) heat energy, 37. The mathematical form of Einstein’s photoelectric equation is __________, (a) hν = ½mv² max, (b) hν - hνₒ = ½mv² max, (d) hν = W + hνₒ, (c) hν = W - ½mv² max, 38. The maximum KE of the photoelectrons _____________, (a) Increases with intensity of incident light, (b) decreases with intensity of incident light, (c) varies linearly with the frequency of incident light, (d) varies exponentially with the frequency of the incident light, , 228, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 39. Current produced by a photoelectric cell is __________, (a) proportional to the intensity, (b) inversely proportional to the intensity, (c) independent of intensity, (d) proportional to the frequency, 40. The linear momentum of de Broglie wave is ___________, (a) h / p, (b) λ / p, (c) h / λ, , (d) λ / h², , 41. The resolving power of an electron microscope will be ______________ than that of an, optical microscope, (a) 10,000 times greater, (b) 1,00,000 times greater, (c) 1,00,000 times lesser, (d) 10,000 times lesser, 42. The relation between the mass of a body at rest ( mₒ ) and the mass of the same body, moving with a velocity ν is _____________, pₒ, (a) m = mₒ, (b) m =, _ #P æ, ], pₒ, pₒ, (c) m =, (d) m =, _ #] æ, _] æ #, P, P, , www.Padasalai.Net, 43. When the frequency of incident radiation increases the value of stopping potential _____, (a) will decrease, (b) will increase, (c) remains the same, (d) will not increase, 44. When the intensity of light incident on a photoelectric surface is doubled,, (a) the frequency of emitted photons will be doubled, (b) the number of photoelectrons will be doubled, (c) the number photoelectrons will become 4 times, (d) there is no effect at all, 45. The potential which is just sufficient to bring the photoelectric current to zero is, called_________ potential, (a) photoelectric, (b) threshold, (c) stopping, (d) minimum, 46. The electron microscope is based on the principle of ____________, (a) photoelectric effect, (b) duel nature of electrons, (c) particle nature of electron, (d) wave nature of moving electrons, 47. The rest mass of photon is_______________, (a) hν, (b) hν / C, (c) C/hν, , 229, , J.SHANMUGAVELU, , [P.G. T. in Physics], , (d) zero, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 48. de Broglie wavelength λ of a particle is related to its kinetic energy E by the, relation_____________, (a) λ α E, (b) λ α 1 / E, (c) λ α √t, (d) λ α 1 / √t, 49. The energy required to bring the electrons of maximum velocity to rest is________, ( a) eV, (b) eVo, (c) ½mv², (d) mv², 50. The Phenomenon of photoelectric effect is__________, (a) spontaneous process, (b) instantaneous process, (c) continuous process, (d) stimulated process, 51. The cathode of a photo emissive cell is coated with__________, (a) low work function material, (b) high work function material, (c) light sensitive material, (d) reflecting material, 52. The de Broglie wavelength of an object of mass 0.03 kg moving with a velocity 20 m/s, ____________, -, , -, , (a) 2.1 × 10 34 m, , (b) 1.1 × 10 33 m, , -, , (c) 6.6 × 10 34 m, , -, , (d) 3.3 × 10 33 m, , 53. The frequency of photon having an energy of 413eV is __________, (a) 1018 Hz, (b) 10 17 Hz, (c) 1016 Hz, (d) 1015 Hz, , www.Padasalai.Net, 54. The wave number of light of radiation of wavelength 5000 Å is ___________, -, , -, , -, , (a) 2 × 10 7 m 1, , -, , (b) 2 × 10 6 m 1, , -, , -, , (c) 5 × 10 7 m 1, , -, , (d) 2 × 106 m 1, , 55. Threshold frequency of a metal is 3 × 1013 Hz, then its work function ___________, -, , -, , (a) 4 × 10 19J, , (b) 3 × 10 19J, , -, , -, , (c) 2 × 10 20J, -, , (d) 5 × 10 19J, -, , 56. If the momentum of a radiating photon is 3.3 × 10 29 kg ms 1 then its wavelength, is__________, -, , -, , (a) 6 × 10 3 m, , (b) 3 × 10 3 m, , -, , -, , (c) 2 × 10 3 m, , (d) 2 × 10 5 m, , 57. de Broglie wavelength of a proton moving with 1/15th of velocity of light is ______, -, , -, , (a) 3 × 10 14 m, , (b) 2 × 10 15 m, , -, , (c) 2 × 10 14 m, , -, , (d) 3 × 10 16 m, , 58. The de Broglie wavelength of an electron having KE of 20eV is___________, (a) 0.275 nm, (b) 2.75 Å, (c) 27.5 nm, (d) 0.275 Aº, -, , 59. A particle of mass 10 34 kg is moving with a speed of 1.8 × 108 m/s. The mass of the particle, when it is in motion is_________, -, , (a) 12.5 × 10 24 kg, , 230, , -, , (b) 1.25 × 10 24 Kg, , J.SHANMUGAVELU, , -, , (c) 0.125 × 10 24 kg, , [P.G. T. in Physics], , -, , (d) 12.5 × 10 22 kg, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 60. On a metal surface two photons of energy 1eV and 2.5eV falls consecutively. The work, function of the metal is 0.5eV. What is the ratio of maximum velocity of two photons ?, (a) 1 : 4, (b) 1 : 1, (c) 2:1, (d) 1 : 2, 61. The wavelength of X-rays is about ________ than that of visible light., (a) 2000 to 3000 times greater, (b) 3000 to 4000 times lesser, (c) 2000 to 3000 times lesser, (d) 3000 to 4000 times greater, 62. The wavelength of electrons accelerated by a potential difference of 60,000 V is, about__________, -, , (a) 3 × 10 12 m, , -, , (b) 5 × 10 12 m, , -, , (c) 5 × 10 10 m, , -, , (d) 4 × 10 10 m, , 63. Proton when accelerated through a potential difference of V volt has a wavelength λ, associated with it. An alpha particle in order to have the same wavelength must be, accelerated through the voltage______________, (a) V, (b) V, (c) V/8, (d) 2V, 64. If an electron is accelerated by a potential of 54 kV, then its de Broglie wavelength, is___________, (a) 3.34 Å, (b) 1.67 Å, (c) 16.7 Å, (d) 0.84 Å, 65. If the wavelength of an electron is 7.218 Å, then its velocity is______________, (a) 103 m/s, (b) 106 m/s, (c) 1012 m/s, (d) 109 m/s, , www.Padasalai.Net, -, , 66. If the wavelength of electron is 50 × 10 11 m, then the potential difference applied, is__________, (a) 12,000 V, (b) 60,000 V, (c) 6,000 V, (d) 120,000 V, 67. The momentum of photon of wavelength 6600 Å is ______________, -, , (a) 1027 kgms 1, , -, , -, , (b) 10 27 kgms 1, , -, , (c) 1019 kgms 1, , -, , -, , (d) 10 19 kgms 1, , 68. The wavelength of proton having frequency of 1.5 × 1013 Hz is____________, -, , (a) 2 × 10 10 m, , -, , (b) 2 × 10 5 m, , (c) 0.2 Å, , (d) 20 Å, , 69. The de Broglie wavelength of proton accelerated through a p.d of 823 V is_______, -, , (a) 10 12 m, , -, , (b) 10 10 m, , -, , (c) 10 8 m, , (d) 2.417 Å, , 70. The momentum of a proton and an alpha particle are equal. The mass alpha particle is, four times the mass of proton. The ratio of wavelength associated with them _______, (a) 1:4, (b) 4:1, (c) 1:1, (d) 1:2, 71. An alpha particle and a proton are accelerated through the same potential. The ratio of, their de Broglie wavelengths is ______________, (a) 1:1, (b) 1:2, (c) 1:3, (d) 1 : 2√2, , 231, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 72. The frequency of photon of energy 65 eV is____________, (a) 1.57 × 1016 Hz, (b) 1.57 × 1015 Hz, (c) 1.04 × 1015 Hz, , (d) 1.04 × 1016 Hz, , 73. The energy required for the transition n = 2 to n = ∞, (a) 3.4 eV, (b) 1.7 eV, (c) 6.8 eV, , (d) -13.6 eV, , 74. Threshold frequency of metal is 105 Hz, the frequency of incident light is 2 × 1015 Hz then, the energy of photo electron emitted is ____________, -, , (b) 6.625 × 10 19 J, , (a) 6.6 J, , (c) 12.25 × 1019 J, , -, , (d) 2.25 × 10 19 J, , 75. If threshold wavelength of sodium is 6800 Aº, what is its work function?, (a) 0.91 eV, (b) 13.6 eV, (c) 1.82 eV, (d) 1.72 eV, 76. The energy of incident UV rays on aluminium metal of work function 4.2 eV is, Then the kinetic energy of emitted photoelectron is___________, -, , (a) 3 × 10 17 J, , -, , (b) 3 × 10 19 J, , 6.2 eV., , -, , (c) 3 × 1029 J, , (d) 6 × 10 19 J, , 77. A light of wavelength 4000 Å falls on metal surface of work function 2 eV. Then the, maximum kinetic energy of emitted photon is_________, (a) 2 eV, (b) 1.1 eV, (c) 1.5 eV, (d) 0.5 eV, , www.Padasalai.Net, 78. Newton’s laws are not valid in ____________, (a) Inertial frames, (b) non-inertial frames, (c) all frames, (d) reference frames, 79. In the photo emissive cell, the anode is made up of____________, (a) copper, (b) gold, (c) platinum, , (d) zinc, , 80. Photoelectric cell are used in ______________, (a) reproducing sound in cinematography, (b) controlling the temperature of the furnace, (c) automatic switching on & off of street lights, (d) all the above, 81. In Newton’s mechanics which of the following is treated as absolute?, (a) Mass, (b) time, (c)length & space, (d) all the above, 82. According to the special theory of relativity, the velocity of light in free space, is_____________, (a) dependent on the motion of the source, (b) dependent on the motion of the observer, (c) independent of the motion of the observer, (d) a constant in all frames of reference, , 232, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 83. The clock in moving space will appear to ______________, (a) go slower than the clock on the earth, (b) go faster than the clock on the earth, (c) be the same as on the earth, (d) come to rest compared to the clock on the earth, 84. ______________ demonstrated photoelectric effect experimentally first., (a) J.J Thomson, (b) Hallwachs, (c) Richardson, (d) de Broglie, 85. The relationship between stopping potential Vₒ & KE of electrons is given as _______, (a) eVₒ = mv2max, (b) eVₒ = ½mvmax, (c) eVₒ = ½mv2max, (d) Vₒ = ½mv2max, 86. When a material particle of rest mass mₒ attains a speed C, its mass becomes _______, (a) 0, (b) 2 mₒ, (c) 4 mₒ, (d) ∞, 87. Light of frequency 1.5 times the threshold frequency is incident on a photosensitive, material. If the frequency is halved and intensity is doubled, the photoelectric, becomes__________, (a) quadrupled, (b) doubled, (c) halved, (d) zero, 88. The work function of a photoelectric material is 3.3eV. The threshold frequency will be, equal to __________, (a) 8 × 1014 Hz (b) 8 × 1010 Hz, (c) 5 × 1020 Hz, (d) 4 × 1014 Hz, , www.Padasalai.Net, 89. If the electron is moving with a velocity of 500 km / s then the de Broglie wavelength is, __________, (a) 500 m, (b) 9.11 Å, (c) 14.5Å, (d) 66.2 Å, , 90. An electron of mass ‘m’ and charge ‘e’ accelerated from rest through a potential of V volt,, then its final velocity ____________, , (a) ‘Ve / m, , (b) ‘Ve / 2m, , (c) ‘2Ve / m, , (d) 2Ve / m, , 91. The wavelength of electron having momentum 3.3 × 10-24 m/s Hz is_________, (a) 10 Å, (b) 2 Å, (c) 20 Å, (d) 1 Å, , 233, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , ‘3’ MARKK QUESTIONS: (1 – QUESTON : Q. No: 43), PUBLIC ‘3’ MARKS:, 1. What is photoelectric effect (photo electric emission)? (M- 11), , Photoelectric emission is the phenomena by which a good number of, substances, chiefly metals, emit electrons under the influence of radiation such as, γ- rays, X-rays, ultraviolet and even visible light., 2. What is cut-off or stopping potential? ( O – 09,O-13 ), , The minimum negative (retarding) potential given to the anode for which the, photo electric current becomes zero is called the cut-off or stopping potential., 3. Define : threshold frequency. [ M – 15], , Threshold frequency is defined as the minimum frequency of incident, radiation below which the photoelectric emission is not possible completely,, however high the intensity of incident radiation may be. The threshold frequency, is different for different metals., , www.Padasalai.Net, 4. State any three laws of photo electric emission?( 3 OR 5 MARK QUESTION), , i) For a given photo sensitive material, there is a minimum frequency called the, , threshold frequency, below which emission of photoelectrons stops completely,, however great the intensity may be., ii) For a given photosensitive material, the photo electric current is directly, proportional to the intensity of the incident radiation, provided the frequency is, greater than the threshold frequency., iii) The photoelectric emission is an instantaneous process. i.e. there is no time, lag between the incidence of radiation and the emission of photo electrons., iv) The maximum kinetic energy of the photo electrons is directly proportional to, the frequency of incident radiation, but is independent of its intensity., , 5., , What is a photoelectric cell ? Mention the types of photoelectric cells ?.(O14,J-16), The photoelectric cell is a device which converts light energy into electrical, energy. The photo electric cells are of three types:, i) Photo emissive cell, ii) Photo voltaic cell and, iii) Photo conductive cell, , 234, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 6. Give three applications of photoelectric cells.(J – 06,M–10,O–10,J–12,M-14,J-, , 14), , i) Photoelectric cells are used for reproducing sound in cinematography., ii) They are used for controlling the temperature of furnaces., iii) Photoelectric cells are used for automatic switching on and off the street, , lights., iv) Photoelectric cells are used in the study of temperature and spectra of stars., v) Photoelectric cells are also used in obtaining electrical energy from sunlight, during space travel., 7. What are the uses of an electron microscope? ( M – 07,O-15,O-16 ), , i) It is used in the industry, to study the structure of textile fibres, surface of, , metals, composition of paints etc., ii) In medicine and biology, it is used to study virus, and bacteria., iii) In Physics, it has been used in the investigation of atomic structure and, , structure of crystals in detail., 8. What are the limitations of electron microscope? ( M – 06, M – 09, M – 12 , J-, , 15), , www.Padasalai.Net, An electron microscope is operated only in high vacuum. This prohibits the, use of the microscope to study living organisms which would evaporate and, disintegrate under such conditions., , 9. According to classical mechanics, what is the concept of time ?( J – 10 ), , According to classical mechanics,, i) The time interval between two events has the same value for all observers, irrespective of their motion., ii) If two events are simultaneous for an observer, they are simultaneous for, all observers, irrespective of their position or motion., 10. What are inertial and non-inertial frame of references? (O – 06, M – 08,O – 11, , ), , i) Inertial (or) unaccelerated frames: A frame of reference is said to be a, , inertial, when the bodies in this frame obey Newton’s law of inertia and other, laws of Newtonian mechanics. In this frame, a body remains at rest or in, continuous motion unless acted upon by an external force., ii) Non-inertial (or) accelerated frames: A frame of reference is said to be a, , non-inertial frame, when a body not acted upon by an external force, is, accelerated. In this frame, Newton’s laws are not valid., , 235, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 11. State the fundamental postulates of special theory of relativity? ( O – 07, J –, , 09,M – 11,M-13,M-17 ), , The two fundamental postulates of the special theory of relativity are :, i) The laws of Physics are the same in all inertial frames of reference., ii) The velocity of light in free space is a constant in all the frames of, reference., , OTHER IMPORTANT QUESTIONS:, 12. What is a frame of reference?, , A system of co-ordinate axes which defines the position of a particle in two, or three dimensional space is called a frame of reference., It is classified in to two types (i) Inertial (or) unaccelerated frames, (ii) Non-inertial (or) accelerated frames, 13. What are matter waves?, , Louis de Broglie put forward the bold hypothesis that moving particles should, possess wave like properties under suitable conditions. If radiation shows dual, aspects, so should matter. The waves associated with the moving particles are, called matter waves or de Broglie waves., , www.Padasalai.Net, 14. Define work function., , The work function of a photo metal is defined as the minimum amount of, energy required to liberate an electron from the metal surface., W = h±0, 15. Distinguish between inertial and non – inertial frame of reference., , Inertial frame of reference, 1. Obey Newton’s law of inertia, 2. A body remains at rest or in, continuous motion unless acted upon, by an external force., , 236, , J.SHANMUGAVELU, , Non – inertial frame of reference, Newton’s laws are not valid., A body not acted upon by an, external force, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 16. If a body moves with the velocity of light, what will be its mass? Comment, , on your result., , If a body moves with the velocity of light so that, v = c., The mass of the particle moves with a velocity ‘v’ is given by, p, , m=, , m=, , _ #¾, p, , 5, , _ #5, 5, , =, , p, , =∞, , It is not possible. Hence no particle will move with a velocity equal to or, greater than that of light., 17. Explain the application of photo electric cell in burglar alarm., , In burglar alarm, ultraviolet light is continuously made to fall on the photocell installed at the door-way. A person entering the door interrupts the beam, falling on the photo-cell. The abrupt change in photocurrent is used to start an, electric bell ringing., 18. Explain the application of photo electric cell in fire alarm., , www.Padasalai.Net, In fire alarm, a number of photo-cells are installed at suitable places in a, building. In the event of breaking out of fire, light radiations fall upon the, photo- cell. This completes the electric circuit through an electric bell or a siren, which starts operating as a warning signal., , 19. Write the concept of space in classical mechanics?, , i) Fixed frame of reference by which the position or motion of any object in, the universe could be measured., ii) The geometrical form of an object remains the same irrespective of changes in, position or state of motion of the object or observer., 20. Write the concept of time in classical mechanics?, , i) The time interval between two events has the same value for all observers, irrespective of their motion., ii) If two events are simultaneous for an observer, they are simultaneous for all, observers, irrespective of their position or motion. This means simultaneity is, absolute., 21. Write the concept of time in classical mechanics?, , The mass of the body is absolute and constant and independent of the motion, of the body, , 237, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 8. How fast would a rocket have to go relative to an observer for its length to be, , corrected to 99% of its length at rest? [M-12, O-07,J-11,J-14] [Compulsory], , Data:, , d, , d, , TT, , = 99 % =, , ;, , v=?, , b ( b _1 8, , Sol:, , b(, d, , d, , TT, , (, , TT, , Ë, =, , b0, , TT, , = _1 8, , Ë, =, , V = 0.141 c, , = 0.141 3 108, V = 2.829 108 ms-1, , www.Padasalai.Net, 9. A metallic surface when illuminated by light of wavelength 3333 Å emits, , electrons with energies upto 0.6 eV. Calculate the work function of the metal. [O09,J-12,M-13, O-16] [Compulsory], , Data : λ = 3333 Å, K.E = 0.6 eV ; W = ?, Solution : Work function, W = hν – kinetic energy, or, , W=, , ℎ=, , ª, , – K.E, ., , =§, , SÓ, , aaaa, , a, , ', , ¨ – (0.6, , 1.6, , 10-19), , = (5.96 × 10-19) – (0.96 × 10-19), , W = 5 × 10-19 J, , W=, , E, ., , eV, , W = 3.125 eV., 241, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , −31, , 12. The rest mass of an electron is 9.1 × 10, , kg. What will be its mass if it moves, with 4/5 of the speed of light?[J-14] [Compulsory], th, , Given data:, Sol:, , m=, , m=, , =, , k, , m0 = 9.1, , 10#a kg;, , v = C;, E, , m=?, , _ #¹, º, , S, , T., , Ã, , T., , Ó, § ¿¨, # \, º, S, , _ #Ó, \, , =, , 9.1 10831, , =, , T., , =, , _25816, 25, , 9.1 10831, _9, 25, , S, S, \, , www.Padasalai.Net, =, , S, , T., , a, , m = 15.16, , E, , E.E, , =, , S, , a, , <@#Â< kg, , 13. At what speed is a particle moving if the mass is equal to three times its rest, , mass.(O-14,J-15,J-16), , Data : m = 3 mo ; v = ?, Sol:, , m=, , 3m =, , k, , _ #¹, k, , º, , _ #¹, º, , v = 0.943 c, = 0.943 × 3 × 108, , v = 2.829 × 108 ms−1, 244, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, 5, , −1, , 14. Calculate the de Broglie wave length of an electron, if the speed is 10 ms ., −31, , −34, , (Given m = 9.1 × 10 kg; h = 6.626 × 10 Js) (O-15), Data: m = 9.1 × 10−31 kg; h = 6.626 × 10−34 Js; v = 105 ms−1., wave length Ñ =, , Sol:, , =, , •, , pË, 6.626 10834, , 9.1 10831 105, , Ñ = 72.81, , °, , FIVE MARKS:(2 – question: Q. No: 58, 59), 1. What is photoelectric effect? State the laws of photoelectric emission.(M-11,J-15), 2. What is work function? State the laws of photoelectric emission. (M-07,M-09), 3. Obtain Einstein’s photoelectric equation., , (M-06,J-06,O-09,M-10,J-10,O-11,M-14,J-14), , www.Padasalai.Net, 4. Explain the construction and working of photo – emissive cell with diagram., , (O-08,O-13), , 5. Write any five applications of photoelectric cells., , (O-06,J-07,M-08,J-08,J-09,M-12,J-13,O-16,M-17), 6. Derive an expression for de- Broglie wavelength of matter waves., , (O-06,M-07,M-09,J-10,O-10,J-11,J-12,M-13,O-14,M – 15,M-17), 7. Explain wave mechanical concept of atom. (O-07), 8. Write the uses and limitations of an electron microscope.(M-14,J-16), 9. Explain length contraction (or) Lorentz – Fitzgerald contraction with an example., , (M-06,M-08,M-10,O-10,O-11,J-13, M – 15,O-15), 10. Explain time dilation with an example. (J-06,J-08,J-14), 11. Derive Einstein’s mass energy equivalence. (J-07,O-12), , 245, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , Only for slow bloomers, ( 5 Mark answer), 1. Photo electric effect: Photoelectric emission is the phenomena by which a good, number of substances, chiefly metals, emit electrons under the influence of radiation, such as • rays, X-rays, ultraviolet and even visible light., Laws of photoelectric emission:, i) For a given photo sensitive material, there is a minimum frequency called the, threshold frequency, below which emission of photoelectrons stops completely,, however great the intensity may be., ii) For a given photosensitive material, the photo electric current is directly proportional, to the intensity of the incident radiation, provided the frequency is greater than the, threshold frequency., iii) The photoelectric emission is an instantaneous process. i.e. there is no time lag, between the incidence of radiation and the emission of photo electrons., iv) The maximum kinetic energy of the photo electrons is directly proportional to the, frequency of incident radiation, but is independent of its intensity., 2. Einstein’s photoelectric equation:, Albert Einstein, successfully applied quantum theory of radiation to, photoelectric effect., The emission of photo electron is the result of the interaction between a single photon, of the incident radiation and an electron in the metal. When a photon of energy hν is, incident on a metal surface,, i) A part of the energy of the photon is used in extracting the electron from the surface, of metal. The work function of a photo metal is defined as the minimum amount of, energy required to liberate an electron from the metal surface., ii) The remaining energy of the photon is used to impart kinetic energy to, the liberated electron., , www.Padasalai.Net, hν = W + mv2, If the electron does not lose energy by internal collisions, the entire energy( hν – W), will be exhibited as K.E of electron, , hν = W + m•pµ" (or) h (ν – ν0) = m•pµ", , 3. De Broglie wavelength of matter waves., E = hυ, , Planck’s relation,, , 2, , Einstein’s mass energy relation E = mC, , z(•, , 246, , J.SHANMUGAVELU, , hυ ( mC2, , Ñ=, , Ñ=, , •, , (or) h, , p=, •, , pË, , (or), , P, c, , = mC2, , •, , +, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 4. Applications of photo electric cells:, i) reproducing sound in cinematography., ii) controlling the temperature of furnaces., iii) automatic switching on and off the street lights., iv) study of temperature and spectra of stars., v) These cells are used in opening and closing of door automatically., , 5. Photoelectric cells: It is a device which converts light energy into electrical energy., Construction: It consists of a highly evacuated bulb B made, of glass or quartz ., A semi cylindrical metal plate C, acts as cathode., This plate is coated with a low work function material such, as caesium oxide., A thin platinum wire A is serves as the anode., Working: When a light of suitable wave length falls on the, cathode, photo electrons are emitted., The resulting current is measured by a micro ammeter., The current produced by this type of cell is proportional to, the intensity of the incident light for a given frequency., 6. Uses and limitations of an electron microscope:, Uses:, i), Study the structure of textile fibres, surface of metals, composition of paints etc., ii), Study virus, and bacteria., iii), Investigation of atomic structure and structure of crystals in detail., Limitations:, It is operated only in high vacuum. This prohibits the use of the microscope to study, living organisms which would evaporate and disintegrate under such conditions., , www.Padasalai.Net, 7. Length contraction ( or ) Lorentz – Fitzgerald contraction:, The length of the rod in S′is lo at rest. The, frame of reference S′ moves with a velocity, v. Now, the length of the rod is b, The rod is contracted by a factor _1 8, , T, , ,, , b ( b- _ 1 8, , T, , ,, , i.e.,, , in the direction of motion, , b < bo, , Example : a circular object appears as an ellipse for a fast moving observer, 8. Time dilation:, A clock in the frame S′ at a position gives out signals at an interval to. This interval, is observed by an observer in frame S moving with velocity is t,, t=, , ë, , _<# “=, Æ, , =, , i.e,, , Time lengthened by a factor of, , t > t0, , _<# “=, Æ, , =, , Example : The clock in the moving space ships will appear to go slower than the, clocks on the earth., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 8.NUCLEAR PHYSICS, PREPARED BY, J.SHANMUGAVELU M.Sc, B.Ed. [P.G. Assist in Physics], , ONE MARK QUESTIONS: (4 - Questions), BOOK BACK QUESTIONS:, 1. The nuclear radius of 4Be8 nucleus is, a) 1.3 × 10−15m, b) 2.6 × 10−15m, Sol:, R = r A1/3, , = 1.3, , 10-15, , (8)1/3, , = 1.3, , 10-15, , 2 = 2.6, , c) 1.3 × 10−13m, , 10-15m, , 2. The nuclei 13Al27and 14Si28 are example of, a) isotopes, b) isobars, c) isotones, 27, Hints:, (N = A – Z = 27 -13=14), and, 13Al, 28, 14Si, , d) 2.6 × 10−13m, , d) isomers, , (N = A – Z = 28 -14= 14) having 14 neutrons., , Different element having same number of neutron it is called as isotones, , www.Padasalai.Net, 3. The mass defect of a certain nucleus is found to be 0.03 amu. Its binding energy is, a) 27.93 eV, b) 27.93 KeV, c) 27.93 MeV, d) 27.93 GeV, Sol:When the protons and neutrons combine to form a nucleus, the mass that disappears, (mass defect, Δm) is converted into an equivalent amount of energy (Δmc2). This energy is, called the binding energy of the nucleus. So Δm = Δmc2, BE = Δmc2 =∆m = 0.03 amu, = 0.03 931 MeV, = 27.93 MeV, , 4. Nuclear fission can be explained by, a) shell model, b) liquid drop model, , c) quark model, , 5. The nucleons in a nucleus are attracted by, a) gravitational force b) electrostatic force, , c) nuclear force, , 6. The ionisation power is maximum for, a) neutrons, b) α– particles, Increasing order – γ < ‰ < ™ ;, Decreasing order – α > ‰ > •;, , c) γ– rays, , d) Bohr atom model, d) magnetic force, d) β –particles, , 7. The half life period of a certain radioactive element with disintegration constant 0.0693, per day is, b) 14 days, c) 140 days, d) 1.4 days, a) 10 days, Sol:, , Tæ =, , =, , 248, , . Ta, , ª, . Ta, ., , Ta, , = 10 days, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 8. The radio-isotope used in agriculture is, a) 15P31, b)15P32, , c) 11Na23, , d) 11Na24, , 9. The average energy released per fission is, a) 200 eV, b) 200 MeV, c) 200 meV, 235, Sol:, + 0n1 → 56Ba141 + 36Kr92 + 30n1 + Q;, 92U, , d) 200 GeV, , Mass defect = M.R – M.P, = 236.054398 – 235.829095, = 0.225303 amu, As, 1 amu = 931 MeV,, energy released in a fission =0.225303 × 931, ≃ 200 MeV, , 10. The explosion of atom bomb is based on the principle of, a) uncontrolled fission reaction, b) controlled fission reaction, c) fusion reaction, d) thermonuclear reaction, 11. Anaemia can be diagnosed by, a) 15P31, b)15P32, , c) 26Fe59, , d) 11Na24, , 12. In the nuclear reaction 80Hg198+ X →79Au198+ 1H1 , X-stands for, a) proton, b) electron, c) neutron, Sol: Equate mass no. and atomic no. on both sides, , d) deuteron, , www.Padasalai.Net, 198, 1, 80Hg + 0X, , →79Au198+ 1H1, , Neutron (0n1) has mass no. 1 and atomic no. 0, , 13. In β– decay, a) atomic number decreases by one, c) proton number remains the same, Sol: In β decay; ZXA →Z+1YA + -1e0, , b) mass number decreases by one, d) neutron number decreases by one, , Equate the neutron no. L.H.S = R.H.S, , (A-Z) = (A-xZ 7 1|, (A-Z) = (A-Z-1), , ( So neutron number decreases by one), , = -1, , 14. Isotopes have, a)same mass number but different atomic number, b)same proton number and neutron number, c)same proton number but different neutron number, d)same neutron number but different proton number, For example:, Isotope, 1, 1H, 2, 1H, 3, 1H, , 249, , No. proton, 1, 1, 1, , J.SHANMUGAVELU, , No. neutron, 0, 1, 2, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 15. The time taken by the radioactive element to reduce to 1/e times is, a) half life, b) mean life, c) half life/2, d) twice the mean life, N, N0, , Sol:, , = ……(1);, , but N = N0e#ª±, j, , ⇒, , j, , = e#ª± … … . 2, , Equate 1 & 2 ⇒ e#ª± =, ÍÎ, , =, , eª± ( e ( equate the power of exponential), λt ( 1, ⇒ t=, , =τ, , ª, , (τ is the mean life), , 16. The half life period of N13is 10.1 minute. Its life time is, , a) 5.05 minutes, , b) 20.2 minutes, , c), , ., , . Ta, , minutes, , d) infinity, , Hints: An infinite time is required for the complete disintegration of all the atoms of all, elements, , www.Padasalai.Net, 17. Positive rays of the same element produce two different traces in a Bainbridge mass, spectrometer. The positive ions have, a) same mass with different velocity, b) same mass with same velocity, c) different mass with same velocity, d) different mass with different velocity, Hints: The same element produce two different traces. The velocity selector allows the ions of, a particular velocity to come out of it. So its mass is different and velocity is same, 18. The binding energy of 26Fe56 nucleus is, a) 8.8 MeV, b) 88 MeV, BE, ), A, , Sol:The binding energy per nucleon, , =, , 8.8 MeV=, , c) 493 MeV, , d) 41.3 MeV, , A%²³%²ø ² 6øÐ -Ñ ±´ ² ,°, Ò-±4° ² k, , 6 -Ñ ² ,° 4²., , ., , Binding energy of the nucleus, 56, , ⇒ Binding energy of the nucleus (8.8 MeV × 56, = 493MeV, , 19. The ratio of nuclear density to the density of mercury is about, a) 1.3 × 1010, b) 1.3, c) 1.3 × 1013, d) 1.3 × 104, 17, -3, Sol: Nuclear density = 1.816 10 kg m ;, , 103 kg m-3, , density of mercury = 13.6, j ,° 46 ³ ².%±Ð, , ³ ².%±Ð -Ñ k 6, 6Ð, , =, , ., , «, , a., , =1.3, , 250, , S, , 1013, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PUBLIC QUESTIONS:, , 20. The nuclear force is due to the continuous exchange of particles called, a) Leptons, b) mesons, c) hyperons, d) photons, 21. In the following nuclear reaction 7N14+ 0n1 → X + 1H1, the element X is, a) 6N14, b) 6C14, c) 6O14, d) 7N13, 14, 1, Sol: Equate mass no. and atomic no. on both sides 7N + 0n → X + 1H1, , mass no., , 14 + 1 = A +1, ⇒ A = 14, , atomic no. 7 + 0 = Z + 1, ⇒Z=6, A= 14 & Z = 6, The element is 6C14, 22. Which of the following particles is a lepton?, b) Proton, a) Electron, , c) Neutron, , 23. One amu is equal to, a) 931 eV, c) 1.66 10-27kg, , b) mass of carbon atom, d) mass of oxygen atom, , d) π- meson, , 24. The time taken by the radioactive element to reduce to e-1/2 times is, a) Half life period, b) Half life period /2 c)mean life period, d) mean - life period / 2, , www.Padasalai.Net, N = N0e# æ ……(1);, , Sol:, , but, , N = N0e#ª± … … . 2, , Equate 1 & 2 ⇒N0e# æ = N0e#ª±, , e#, , æ, , = e#ª± equate the power of exponential, = λt, , t=, , =, , ª, , Ó, , =, , k 4² °%Ñ, , (τ (, , ª, , is the mean life), , 25. The value of 1 amu is, a) 931 eV, b) mass of carbon atom c) mass of one proton d) 1.66, 26. The penetrating power is maximum for, a) α- particles, b) β- particles, Hints: Increasing order – α < ‰ < •;, , c) γ- particles, , 10-27kg, , d) protons, , Decreasing order – γ > ‰ > ™;, , 251, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 27. In the following nuclear reaction 13 Al27+ 2 He4 → X+ 0n1,X – stands for, a) 15Si30, b) 15P30, c) 15S30, d) 15Si29, Sol:, mass no. 27 + 4 = A +1, , ⇒ A = 30, atomic no., , 13 + 2 = Z + 0, ⇒ Z = 15, , A= 30 & Z = 15 The element is 15P30, 28. The moderator used in nuclear reactor is, a) Cadmium, b) Boron carbide, , c) Heavy water, , d) Uranium (92U235), , 29. The number of α and β particles emitted when an isotope 92U238undergoes α and β decays, to form 82Pb206are respectively, a) 6, 8, b) 4, 3, c) 8, 6, d) 3, 4, Sol: 92U238→ 82Pb206 + a 2He4 + b -1e0, , Equate mass number, , 238 = 206 + 4a + 0, ⇒ 4a = 32, ⇒ a =8, , Equate atomic number, , 92 = 82 + 2a – b, , ⇒ 2a – b = 10 …….(1), , www.Padasalai.Net, Sub a = 8 in (1) ⇒ (2, , 8) – b = 10, , ⇒ b = 16 – 10, , ⇒b=6, , 30. The particles which exchange between the nucleons and responsible for the origin of the, nuclear force are, a) photons, b) leptons, c) mesons, d) baryons, 31. Which of the following is not a moderator?, a) Liquid sodium, b) Ordinary water, , c) graphite, , d) Heavy water, , 32. An element zXA successively undergoes three α decays and four β – decays and gets, converted to an element Y. the mass number and atomic number of the element Y are, respectively, a) A–12, Z – 2, b) A–12, Z + 2, c) A–12, Z +4, d) A–8, Z + 2, Sol: ZXA→ 32He4 + 4-1e0 + aYb, , Equating atomic number Z = 3, , 2 + (-4) + a, , ⇒Z = 2 +a, ⇒ a = Z -2, Equating mass number A = 3, , 4 + (4, , 0)+b, , ⇒ A= 12 + b, b = A -12;, , 252, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 33. The fuel used in Kamini reactor is, a) 92U235, c) 92U239, , b) 92U233, d) low enriched uranium, , 34. The particle which has zero mass but has energy, is, a) electron, b) photon, c) proton, , d) neutron, , 35. The mean life (τ ) and half life (T 1/2) of a radioactive element are related as, , b) τ =, , a) τ = 2T ½, , !½, , c) τ = 0.6931 T½, , @.j-Â<, , Ò½, , d) τ =, , 36. An element zXA successively undergoes three α particles and four β – particles is, converted to an element Y represented as, a) Z – 6Y A–12, b) Z + 2Y A–12, c) Z – 2Y A–12, d) Z – 10Y A–12, A, 4, 0, b, Sol:, ZX → 32He + 4-1e + aY, , Equating atomic number Z = 3, , 2 + (-4) + a, , ⇒Z=2+a, a = Z -2, Equating mass number, , A=3, , 4 + (4, , 0)+b, , ⇒ A= 12 + b, b = A -12;, The element is Z – 2Y A–12, , www.Padasalai.Net, 37. Which of the following is used to detect the presence of blocks in blood vessels?, a) 15P31, b) 15p32, c) 26Fe59, d) 11Na24, 38. If the nuclear radius is 2.6 × 10–15 m ,then its mass number is, a) 2, b) 4, c) 8, 1/3, -15, Sol:, R=r A, 2.6 10, , 10-15, , = 1.3, , d) 16, , (A)1/3, , 2 = (A)1/3, A = 23 =8, 39. The energy of Slow neutrons, a) 1000 eV – 2000 eV b) 2000 eV-0.5 MeV, , c) 0 eV - 1000 eV, , 40. One amu is equal to, a) 931 eV, b) mass of carbon atom c) 1.66 × 10-24 g, -27, Sol: 1 amu = 1.66 x 10 kg, = 1.66, , 10-27, , = 1.66, , 10-24g, , d) 0MeV-10 MeV, d) mass of electron, , 103g, , 41. Particle that has no charge and no rest mass but travels with velocity of light is, a) baryon, b) meson, c) lepton, d) Photon, , 253, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 42. The half life of a radioactive element is 300 days. The disintegration constant of the, radioactive element is, a) 0.00231 day, b) 0.00231/ day, c) 0.0231/ day, d)0.0231 day, Sol:, , T½ =, , ⇒ λ=, =, , 0.6931, λ, . Ta, Ò½, , . Ta, a, , = 0.00231/day, 43. The mean life of radon is 5.5 days. Its half life is, a) 8 days, b) 2.8 days, c) 0.38 days, Sol:, T½ = 0.6931 τ, , = 0.6931, , d) 3.8 days, , 5.5, , = 3.8 days, 44. The energy equivalent to 1 amu is, a) 931 MeV, b) 931 meV, , c) 931 eV, , d) 913 MeV, , 45. The fuel used in Kamini (Kalpakkam mini reactor) is, a) mixture of carbides of uranium and plutonium, c) mixture of oxides of plutonium and uranium, , b) 92U233, d) 92U235, , www.Padasalai.Net, 46. In the nuclear reaction 4Be9+ X → 6C12+ 0n1 ,X – stands for, a) Proton, b) α – particle(2He4), c) electron, 9, 12, 1, Sol:, 4Be + X → 6C + 0n, , mass no., , d) deuteron, , 9 + A = 12 +1, A=4, , atomic no. 4 + Z = 6 + 0, Z=2, A= 4 & Z = 2 The element is 2He4, 47. Which of the following belongs to Baryon group?, a) photon, b) electron, c) Pion, , d) proton, , 48. According to the law of disintegration N = Nₒ þ#› , the number of radioactive atoms that, have been decayed during a time of t is, a) No, b) N, c) Nₒ-N, d) No/2, Sol: At time = 0; the number of atoms = N0 ;, , Time = t; the number of atoms = N, the number of radioactive atoms that have been decayed during a time of t = N0 -N, 49. The coolant used in fast breeder reactor is, a) ordinary water, b) heavy water, , 254, , J.SHANMUGAVELU, , c) liquid sodium, , [P.G. T. in Physics], , d) boron carbide, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 50. Which of the following are isotones?, a) 92U235and 92U238, b) 8O16 and 7N14, c) 6C14and 7N14, 14, 13, Hints:7N (14 -7 =7), and 6C (13 – 6 = 7) having 7 neutrons., , d) 7N14and 6C13, , Different element having same number of neutron it is called as isotones, 51. Arrange α, β and γ rays in the increasing order of their ionizing power, a) α, β, γ, b) β, α, γ, c) γ, β, α, d) γ, α, β, Hints: Increasing order – γ < ‰ < ™ ;, , Decreasing order – α > ‰ > • ;, , 52. When mass number increases, nuclear density, a) increases, b) decreases, c)remains constant, d) may increase (or) decrease, Hints:The nuclear density is 1.816 × 1017 kg m−3 which is almost a constant for all the nuclei, kÕ, irrespective of its size. ρN= Ó, ., 2 6S, S, This expression is independent on mass number(A). So remains constant, 53. The nuclear force between a proton and another proton inside the nucleus is, a) zero, b) short range, c) repulsive, d) long range, Hints: Nuclear force is charge independent., So it is not repulsive force and Nuclear force is a short range force., It is very strong between two nucleons which are less than 10−15 m apart., , gravitational force is long range force, , www.Padasalai.Net, 54. The cosmic ray intensity is maximum at a latitude of, a) 0˚, b) 45˚, c) 90˚, Hints: The intensity is maximum at the poles (θ = 900),, Minimum at the equator (θ = 00) and, constant between latitudes of 420 and 900., , d) 60˚, , 55. The rays which have the greatest ionising power is, a) neutrons, b) α– particles, c)γ– rays, , d) β– particles, , 56. Hydrogen bomb is based on the principle of, a) nuclear fission, c) nuclear force, , b) nuclear fusion, d) carbon nitrogen cycle, , 57. The unit of disintegration constant is, a) no unit, b) second, , c) second-1, , Sol: τ (, , ⇒λ(, , 1, λ, , Ó, , =, , ., , d) curie, , = s-1, , 58. In proton – proton cycle four protons fuse together to give, a) an ™ particle, two electrons, two neutrinos and energy of 26.7 MeV, b) an particle, two positrons, two neutrinos and energy of 26.7 MeV, c) a helium atom, two positrons, two neutrinos and energy of 26.7 MeV, d) an ™ particle, two positrons, two anti - neutrinos and energy of 26.7 MeV, Hints: 4 1H1 → 2He4 + 2 1e0 + 2ν + energy (26.7 MeV), 4, 2He is a helium nucleus (™ particle ), , 255, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 59. Which of the following is massless and chargeless but carrier of energy and spin ?, a) neutrino, b) Muon, c) Pion, d) Kaon, 60. The radio isotope used in the treatment of skin disease is, a) Na21, b) I131, c) Fe59, , d)P32, , 61. The binding energy per nucleon of 26Fe56 nucleus is, a) 8.8 MeV, b) 88 MeV, c) 493 MeV, 62. 1 curie is, a) activity of one gram uranium, c) 3.7 1010 becquerrel, , d) 41.3 MeV, , b) 1 disintegration/second, d)1.6 1012 disintegration/second, , 63. The half life period of N13is 10.1 minute. Its mean life time is, , a) 5.05 minutes, sol:, , Ö1æ, , b) 20.2 minutes, , c), , <@.<, , @.j-Â<, , minutes, , d) infinity, , 2, % ( 0.6931, , =, , ., , . Ta, , minutes, , 64. The penetrating power is maximum for, a) α- particles, b) β- particles, Hints: Increasing order – α < ‰ < •;, , c) gamma rays, , d) protons, , Decreasing order – γ > ‰ > ™;, , www.Padasalai.Net, 65. The radius of the nucleus which contains 64 nucleons is, a) 2.6 F, b) 5.2 F, c) 10.4 F, , d) 7.8 F, , 66. The decay constant of a free neutron is :, a) 0.013 minute -1 b) 0.053 minute – 1 c) 3 minutes d) 0.069 minute -1, Sol:, , T½ =, , ⇒ λ=, =, , 0.6931, λ, . Ta, Ò½, , . Ta, a, , = 0.053 minute – 1, 67. In a Bainbridge mass spectrometer positive rays of the same element produce different, traces. The traces correspond to :, a) isotopes, b) isobars, c) isotones, d) none of the above, , 256, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, Choose the correct answer from the options given, 68. Mass of proton is __________ times the mass of electron., (a) 2, (b) 1836, (c) 1636, , (d) 1863, , 69. Number of neutrons in 17Cl35 is, (a) 17, (b) 18, , (d) 19, , (c) 35, , 70. Since the atoms of Isotopes have identical electronic structure, they have, (a) equal number of neutrons, (b) identical physical properties, (c) identical chemical properties, (d) dissimilar chemical properties, 71. Emprical relation between radius of nucleus (R) and its mass number (A) is given by, (a) R = rₒA4, (b) R = rₒ3A4, (c) R = rₒ3A1/ 3, (d) R = rₒA1/ 3, 72. Isobars are, (a) different nuclei of same element, (b) having similar physical and chemical properties, (c) identical nuclei of different elements, (d) having different physical and chemical properties, , www.Padasalai.Net, 73. In an empirical relation connecting radius of nucleus ( R ) and its mass number, ( A ) the, value of rₒ is, -, , -, , (a) 1.3 × 10 3 F, , (b) 1.3 × 10 15 F, , -, , (c) 1.3 × 10 15 m, , (d) 13 F, , -, , 74. If the nuclear density of 1H2 nuclei is 1.816 × 1017 kgm 3 then the nuclear density of 2He4, nuclei is, -, , (b) 1.816 / 4 × 1017 kgm 3, , -, , -, , (d) 1.816 × 1017 kgm 3, , (a) 2 × 1.816 × 1017 kgm 3, , -, , (c) 4 × 1.816 × 1017 kgm 3, 75. The charge of 8O16 nuclei is, -, , (a) 1.6 × 10 19 C, , -, , (b) -12.8 × 10 19 C, , 76. Examples of isobars are ___________, (a) 1H1 and 2He4, (b) 1H2 and 2He3, , -, , -, , (c) 1.28 × 10 18 C, , (d) 2.56 × 10 18 C, , (c) 1H3 and 2He3, , (d) 1H1 and 2He4, , 77. The ratio of nuclear density to the density of mercury is about __________, (a) 1.3 × 1010, (b) 1.3, (c) 1.3 × 1013, (d) 1.3 × 104, 78. The electrons in the atom of an element which determine its chemical and electrical, properties are called ___________, (a) active electrons, (b) revolving electrons, (c) excess electrons, (d) valence electrons, , 257, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 79. The ratio of radii of two nuclei is 1 : 2. The ratio of their mass number is __________, (a) 1 : 4, (b) 8 : 1, (c) 1 : 8, (d) 1 : 16, 80. Energy equivalence of 1 amu is, (a) 931 eV, (b) 913 eV, 81. In, , ;.`., , (c) 931 MeV, , (d) 913 MeV, , graph, beyond A = 120, the binding energy per nucleon is, , (a) decreases rapidly, (c) decreases slowly, , (b) increases slowly, (d) a constant, , 82. The energy equivalent of 1 amu is = ____________, (a) 981 MeV, (b) 913 MeV, (c) 931 MeV, , (d) 942 MeV, , 83. The mass defect of certain nucleus is found to be 0.03 amu. Its binding energy is____, (a) 27.93 eV, (b) 27.93 KeV, (c) 27.93 MeV, (d) 27.93 GeV, 84. The mass of proton is 1.007277 amu and that of neutron is 1.008665 amu. If the mass of, 2, 2, 1H = 2.01473 amu. Then the binding energy of 1H is ___________, (a) 1.128 MeV, (b) 0.164 MeV, (c) 1.52 MeV, (d) 2.42 MeV, 85. In Bainbridge mass spectrometer, the velocity of the particle emerging from the velocity, selector is ___________, (a) v = F / q, (b) v = B / q, (c) v = E / B, (d) v = Eq / B, , www.Padasalai.Net, 86. In Bainbridge mass spectrometer, the mass of an ion is _______, ì×, £ì×, £ì×, (a), (b), (c), (d) BB'Rq, £š, š, š, , 87. In Bainbridge mass spectrometer, the distance between the opening of the chamber and, the position of the dark line gives __________, (a) the radius, (b) the diameter, (c) half the radius, (d) twice the diameter, 88. Positive rays of the same element produce two different traces in a Bainbridge mass, spectrometer. The positive ions have ___________, (a) same mass with different velocity, (b) same mass with same velocity, (c) different mass with same velocity, (d) different mass with different velocity, 89. Ratio of strength of nuclear force to that of gravitational force is _________, -, , (a) 10 40, , 258, , -, , (b) 10 20, , J.SHANMUGAVELU, , (c) 1020, , [P.G. T. in Physics], , (d) 1040, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 90. According to present view of the nuclear force, the force that binds the protons and, neutrons is a _________, (a) electrostatic force, (b) magnetic force, (c) secondary force, (d) gravitational force, 91. The nucleons in a nucleus are attracted by ____________, (a) gravitational force, (b) electrostatic force, (c) nuclear force, (d) magnetic force, 92. Nuclear forces were explained by ___________, (a) Chadwick, (b) Bohr, (c) Curie, , (d) Yukawa, , 93. Nuclear force is acting between __________, (a) neutron – neutron only, (b) proton – proton only, (c) neutron – proton only, (d) all the above, 94. Nuclear force is a ____________, (a) long range force, (c) repulsive force, , (b) short range force, (d) charge based force, , 95. During the radioactive disintegration of radium 88Ra226 into Radon 86Rn222, the energy of, gamma ray emitted is about, (b) 0.187 MeV, (c) 1.87 MeV, (d) 187 MeV, (a) 18.7 MeV, , www.Padasalai.Net, 96. According to the concept of Yakawa, the particles which exchange between the nucleon, and responsible for the origin of nuclear force are ________, (a) photons, (b) Leptons, (c) mesons, (d) baryons, 97. Nuclear density, (a) depends on atomic number, (b) is a constant, (c) depends on neutron number, (d) depends on mass number, 98. 1 amu is equal to, -, , -, , (a) 1.494 × 10 10 J, (c) 931 J, , (b) 14.94 × 10 10 J, (d) 931 eV, , 99. Energy released per nuclear fusion, (a) 200 MeV, (b) 26.7 eV, 100., , 259, , Of the following, isotonic nuclei are, (a) 11Na22, 12Mg24, (b) 11Na24, 10Ne23, , J.SHANMUGAVELU, , (c) 2.67 MeV, , (d) 26.7 MeV, , (c) 12Mg24, 11Na24, , (d) 10Ne23, 11Na22, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 101., , Activity of one gram of radium is, (a) 3.7 × 1010 becquerel, (b) 3.7 curie, (c) 3.7 × 1010 curie, (d) 1 becquerel, , 102., , An equation for a free neutron decay is, , (a) 0n1 → 1H1 + -1e0 + ν, , (b) 0n1 → -1e0 + 1H2 + ν, , (c) 0n1 → 1H1 + -1e0 + ν, , (d) 0n1 → 1H1 + 1e0 + ν, , 2017-2018, , 103., , In pressurized heavy water reactors, the fuel used is, (a) uranium, (b) uranium carbide, (c) uranium oxide, (d) Low enriched uranium, , 104., , In the reaction 90Th234 →, (a) α, (b) β, , 105., , The ionization power is minimum for _____________, (a) alpha particles, (b) β – particles, (c) gamma rays, , (d) electrons, , Which one travels with velocity of light?, (a) α – ray, (b) β – ray, (c) γ – ray, , (d) cathode ray, , 106., , 91Pa, , 234, , + X , the particle emitted is, (c) γ, (d) photon, , www.Padasalai.Net, 107., , When a gamma ray is emitted from a radioactive atom, (a) only its mass number changes, (b) only its atomic number changes, (c) both mass number and atomic number changes, (d) neither mass number nor atomic number changes, , 108., , In which of the following decays the element does not change, (a) α – decay, (b) β – decay, (c) γ – decay, , 109., , A alpha particle has _____________, (a) a charge +e, (b) the mass equal to deuteron, (c) a charge -2e, (d) charge to mass ratio equal to that of a deuteron, , 110., , In beta decay _________________, (a) atomic number increases by one, (b) neutron number increases by one, (c) mass number decrease by one, (d) proton number remains the same, , 260, , J.SHANMUGAVELU, , [P.G. T. in Physics], , (d) neutron decay, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 111., , The relation connecting half life and mean life of a radioactive sample is, . Ta, . Ta, (a) = 0.6931T, (b) T =, (c) Ø =, (d) T = 0.6931 Ø, I, , 112., , Activity of one gram of radium is equal to, (a) 1 roentgen, (b) 1 curie, (c) 1 henry, , (d) 1 second, , Isotope used to locate brain tumour is, (a) Na24, (b) I131, , (d) P32, , 113., , 114., , (c) Fe59, , The radio of C14 and C12 atoms in atmosphere is, (a) 106 : 1, (b) 104 : 1, (c) 1 : 106, , (d) 1 : 104, , 115., , The exposure of radiation dosage which couses diseases like leukemia is, (a) 600 R, (b) 100 R, (c) 250 mR, (d) 25 mR, , 116., , The half life period of an isolated neutron is about __________, (a) 31 minutes, (b) 13 minutes, (c) 13 hours, (d) 13 s, , 117., , The natural radioactive gas is, (a) radon, (b) helium, , (c) oxygen, , (d) krypton, , www.Padasalai.Net, 118., , The radio isotope used in agriculture is ____________, (a) 15P31, (b) 15P32, (c) 11Na23, , (d) 11Na24, , 119., , Cobalt 60 is used for the treatment of, (a) Cancer, (b) heart attack, (c) thyroid gland, (d) maintaining blood circulation, , 120., , The half life period of radiocarbon is ____________, (a) 2800 years, (b) 5600 years, (c) 4200 years, , 121., , (d) 5300 years, , A radioactive substance has a half life period of 30 days. The disintegration constant is, (a) 0.023 / day, (b) 0.231 / day, (c) 2.31 / day, (d) 23.1 / day, , 122., The half life of a certain radioactive element with disintegration constant 0.0693 per, day is ___________, (a) 10 days, (b) 14 days, (c) 140 days, (d) 1.4 days, 123., , 261, , The half life period of N13 is 10.1 minute. Its life time is _____________, (a) 5.05 minutes, (b) 20.2 minutes, (c) 10.1 / 0.693 minutes, (d) infinity, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 124. The half life of a radioactive substance is 5 minutes. The amount of substance decayed, in 20 minutes will be ___________, (a) 93.75 %, (b) 75 %, (c) 25 %, (d) 6.25 %, 125. In nuclear fission 0.1 % mass is converted into energy. The energy released by the, fission of 1kg mass will be ___________, (a) 9 × 1016 J, (b) 9 × 1019 J, (c) 9 × 1013 J, (d) 9 × 1017 J, 126. The percentage of initial quantity of a radioactive element remaining undecayed after, six half life period is __________, (a) 3 %, (b) 6.25 %, (c) 1 %, (d) 1.5 %, 127., , Which of the following statement is wrong ? In nuclear reaction, (a) the sum of initial atomic numbers is equal to sum of the final atomic numbers, (b) law of conservation of charge is satisfied, (c) conservation of nucleons is satisfied, (d) the initial rest mass is not equal to the find rest mass, , 128., , An example for electrostatic accelerator is, (a) Cockcroft – Walton accelerator (b) Linear accelerator, (c) cyclotron, (d) Betatron, , www.Padasalai.Net, 129. The class of accelerators which can accelerate particles only upto few million electron, volt are, ( a) Linear accelerator, (b) cyclotron accelerator, (c) spiral type accelerator, (d) electrostatic accelerator, 130., , With spiral type accelerators, the particles are accelerated to an energy in the order of, (a) few million electron volt, (b) 106 eV, , (c) 109 eV, , -, , (d) 10 9 eV, , 131. The first instrument to record the visual observation of the tracks of the charged, particles when they pass through matter is, (a) Geiger – muller counter, (b) wilson’s cloud chamber, (c) Vandergrift generator, (d) Cyclotron, 132., , Average number of neutrons released per fission of uranium is, (a) 2, (b) 3, (c) 2.5, (d) 3.5, , 133., , The instrument used to measure the intensity of radioactive radiation is __________, (a) cyclotron, (b) Bainbridge spectrometer, (c) electron microscope, (d) Geiger – Muller counter, , 134., , The rest mass of mesons vary between, (a) 250 me – 1000 me, (b) 250 mp – 1000 mp, (c) 250 me – 100 me, (d) 250 mp – 100 me, , 262, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 135., , In fast breedor reactors,, (a) heavy water used as moderator, (b) graphite is used as moderator, (c) ordinary water is used as moderator, (d) no moderator is required, , 136., , The average energy released per fission is _________, (a) 200 eV, (b) 200MeV, (c) 200 meV, , (d) 200 GeV, , Which of the following is not a moderator ___________, (a) heavy water, (b) paraffin, (c) graphite, , (d) Ordinary water, , 137., 138., , The principle of on atom bomb is _____________, (a) nuclear fission, (b) nuclear fusion, (c) conservation of momentum, (d) collision of simple particles, , 139., , In the following reaction 4Be9 + 2He4 → bXa + 0n1 , the value of ‘a’ is, (a) 16, (b) 12, (c) 10, (d) 14, , 140., , The fusion reaction in hydrogen bomb is, (a) 1H3 + 1H2 → 2He4 + 0n1 + q, (b) 41H1 → 21He4 + 21e0 + Q, (d) 1H1 + 1H3 → 2He4 + Q, (c) 1H2 + 1H2 → 2He3 + Q, , www.Padasalai.Net, 141., , Total energy radiated by sun is about ____________, -, , (b) 3.8 × 1026 J/S, , -, , (d) 8.3 × 10 26 J/S, , (a) 3.8 × 10 26 J/S, , (c) 8.3 × 10 26 J/S, , -, , 142., , The neutrons with energy range 0.5 MeV to 10 MeV are called _____________, (a) slow neutrons, (b) fast neutrons, (c) thermal neutrons, (d) none of the above, , 143., , Slow neutrons are neutrons having energies between __________, (a) 1000 eV and 2000 eV, (b) 2000 eV and 0.5 MeV, (c) zero to 1000 eV, (d) 0.5 MeV and 10 MeV, , 144., , Between latitudes of 42º and 90º, the cosmic ray intensity is, (a) minimum, (b) maximum, (c) a constant, (d) none of the above, , 145., , ______________ is the reason for production of carbon-14 in the atmosphere, (a) X – rays, (b) UV – rays, (c) Cosmic rays, (d) gamma rays, , 146., , The cosmic ray intensity is maximum at a place of latitude, (a) 10 km, (b) 20 km, (c) 40 km, , 147., , 263, , (d) 60 km, , Particles having mass equal to or less than about 207 times the mass of an electron are, (a) mesons, (b) leptons, (c) baryons, (d) Hyperons, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 148., , Particles possessing rest mass intermediate between 250me to 1000me are known as, (a) mesons, (b) leptons, (c) baryons, (d) Hyperons, , 149., , Mass of hyperons vary from, (a) 1000 me and 270 me, (b) 207 times of mass of electron to zero, (c) 2180 me and 3275 me, (d) me and mp, , 150., , Natural uranium consists of, (a) 99.28% of U235 and 0.72%, (b) 99.28% of U235 and 0.72%, (c) 99.28% of U238 and 0.72%, (d) 99.28% of U238 and 0.72%, , 151., , 152., , of, of, of, of, , U238, Pu239, U235, Pu239, , Energy of the primary cosmic rays is in order of, (a) 108 MeV, (b) 108 eV, (c) 1018 MeV, , (d) 1018 eV, , Intensity of cosmic ray is maximum at a height of about, (a) 20 m, (b) 20 km, (c) 200m, , (d) 45 km, , 153. In a nuclear reaction, the mass of product nuclei is 0.03 amu less than the mass of, reactant nuclei, then the energy released in the nuclear reaction is, (a) 2783 eV, (b) 2.793 MeV, (c) 27.93 MeV, (d) 0.2793 MeV, , www.Padasalai.Net, 154. A radioactive material of mass 40 milligram becomes 5 milligram in 6 hours then the, half life period of the element is, (a) 1 hour, (b) 90 min, (c) 2 hours, (d) 3 hours, , 155., A radioactive substance is allowed to decay for a time equal to its mean life. Then the, fraction of the element that has decayed is, #, (a), (b), (c) e, (d) e2 – 1, 156., , The safe limit of a person to receive radioactive radiation per week is, (a) 250 R, (b) 25 mR, (c) 250 mR, (d) 250 μR, , 157., , Which of the following is the strongest force in nature ________, (a) electrostatic force, (b) gravitational force, (c) nuclear force, (d) magnetic force, , 158., , Muons belong to _____________, (a) photons, (b) leptons, , 264, , J.SHANMUGAVELU, , (c) baryons, , [P.G. T. in Physics], , (d) mesons, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, ‘3’ MARK QUESTIONS:(2 – Questions : Q. No: 44, 45), PUBLIC ‘3’ MARKS:, , 1. Select the pairs of isotopes, isobars and isotones from the following nuclei:, 11, , Na 22, 12 Mg 24, 11 Na 24, 10 Ne 23( M – 12 ), Isotopes are 11 Na 22,11 Na 24, Isobars are12 Mg 24, 11 Na 24, Isotones are11 Na 24, 10 Ne 23, , 2. Define: mass defect. ( O -10, O-16 ), , The difference in the total mass of the nucleons and the actual mass of the, nucleus is known as the mass defect., Zmp+ NmN – m = ∆m, Here, Δm is the mass defect., 3. Define: binding energy. ( O – 09 ), , When the protons and neutrons combine to form a nucleus, the mass that, disappears (mass defect, Δm) is converted into an equivalent amount of energy, (Δmc2). This energy is called the binding energy of the nucleus., ∴ Binding energy = [ZmP+ Nmn– m] c2 = Δmc2., , www.Padasalai.Net, 4. Write any three findings of binding energy curve. ( O – 06 ), , i) The binding energy per nucleon reaches a maximum of 8.8 MeV at A = 56,, corresponding to the iron nucleus (26Fe56). Hence, iron nucleus is the most, stable., ii) The average binding energy per nucleon is about 8.5 MeV for nuclei having, mass number ranging between 40 and 120. These elements are comparatively, more stable and non radioactive., š, , iii) For higher mass numbers the curve drops slowly and the, is about, ¡, 7.6 MeV for uranium. Hence, they are unstable and radioactive., 5. State any three properties of the nuclear forces. ( J – 08 ), , i) Nuclear force is charge independent. It is not electrostatics in nature., ii) Nuclear force is the strongest known force in nature., iii) Nuclear force is not a gravitational force. Nuclear force is about 1040 times, stronger than the gravitational force., iv) Nuclear force is a short range force. It is very strong between two nucleons, which are less than 10−15 m., , 265, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 6. Write any three properties of Ù 8 ¿¤ ( gamma rays ). ( O-16), , i) The β–particles emitted from a source have velocities over the range of 0.3, c to 0.99 c, where c is the velocity of light., ii) They are deflected by electric and magnetic fields., iii) The ionisation power is comparatively low, iv) They affect photographic plates., v) They penetrate through thin metal foils and their penetrating power is, greater than that of α−rays, vi) They produce fluorescence when they fall on substances like barium, platinocyanide., 7. Write any three properties of Ú 8 ¿¤ ( J-16), , i) They, ii) They, iii) They, iv) They, v) They, vi) They, vii) They, , are electromagnetic waves of very short wavelength., are not deflected by electric and magnetic fields., travel with the velocity of light., produce very less ionisation., affect photographic plates., have a very high penetrating power, greater than that of β-rays., produce fluorescence., , 8. State radioactive law of disintegration?(O-13, J-16), , Rutherford and Soddy found that the rate of disintegration is independent of, physical and chemical conditions. The rate of disintegration at any instant is directly, proportional to the number of atoms of the element present at that instant. This is, known as radioactive law of disintegration., , www.Padasalai.Net, −, , äo, äë, , ∝N, , 9. What is α - decay? Give an example. ( M – 06,M-13 ), , When a radioactive nucleus disintegrates by emitting an α – particle , the, atomic number decreases by two and mass number decreases by four., Example : Radium (88Ra226) is converted to radon (86Rn222) due to α – decay, 226, → 86Rn222 + 2He4, 88Ra, 10. Define: activity and Curie ( O – 06, M – 08, M-10, O – 10,O-13 ), , Activity: The activity of a radioactive substance is defined as the rate at which, the atoms decay., If N is the number of atoms present at a certain time t, the activity R is given, ³j, by R = –, ³±, The unit of activity is Becquerel, Curie : Curie is defined as the quantity of a radioactive substance which gives, 3.7 × 1010 disintegrations per second or 3.7 × 1010 becquerel. This is equal to the, activity of one gram of radium, , 266, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 11. Define: Curie [ M- 15,M-17], , Curie is defined as the quantity of a radioactive substance which gives, 3.7 × 1010 disintegrations per second or 3.7 × 1010 becquerel. This is equal to the, activity of one gram of radium, 12. What are the precautions to be taken by the people working in radiation, , laboratories. [ M – 15], , i) Radioactive materials are kept in thick – walled lead container., ii) Lead aprons and lead gloves are used while working in hazardous area., iii) All radioactive samples are handled by a remote control process., iv) A small micro – film badge is always worn by the person and it is checked, periodically for the safety limit of radiation., 13. Write any three properties of the neutrons. ( J – 06, M – 08, M – 09, J – 11,M-, , 14,M-17 ), i) Neutrons are the constituent particles of all nuclei, except hydrogen., ii) As they are neutral particles, they are not deflected by electric and magnetic, fields., iii) As neutrons are neutral, they can easily penetrate any nucleus., iv) Neutrons are stable inside the nucleus. But outside the nucleus they are, unstable., , www.Padasalai.Net, 14. How do you classify the neutrons in terms of its kinetic energy? ( J – 09 ), , Neutrons are classified according to their kinetic energy as, i) Slow neutrons:, Neutrons with energies from 0 to 1000 eV are called slow neutrons., Thermal neutrons:, The neutrons with an average energy of about 0.025 eV in thermal, equilibrium are called thermal neutrons., ii) Fast neutrons:, Neutrons with energies in the range between 0.5 MeV and 10 MeV are, called fast neutrons. In nuclear reactors, fast neutrons are converted into slow, neutrons using moderators., 15. What is artificial radioactivity? ( J – 12 ), , The phenomenon by which even light elements are made radioactive by, artificial or induced methods is called artificial radioactivity., , 267, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 16. Write the methods of production of artificial radio isotopes. (J-14), , i) Artificial radio-isotopes are produced by placing the target element in the, nuclear reactor, where plenty of neutrons are available., 31, 15P, , + 0n1 → 15P32* + γ, , ii) Another method of production of artificial radio-isotope is to bombard the, target element with particles from particle accelerators like cyclotron., 23, 11Na, , + 1H2 → 11Na24* + 1H1, , 17. Define: roentgen. ( J – 07, O – 08 ), , One roentgen ( 1R ) is defined as the quantity of radiation which produces, 1.6 × 1012 pairs of ions in 1 gram of air., 18. Define: critical mass and critical size. ( O – 08,O-15 ), , The minimum size in which atleast one neutron is available for further, fission reaction. The mass of the fissile material at the critical size is called, critical mass. The chain reaction is not possible if the size is less than the, critical size., , www.Padasalai.Net, 19. What is the use of control rods ? Mention any two control rods. ( O – 07 ), , The control rods are used to control the chain reaction. They are very good, absorbers of neutrons. The commonly used control rods are made up of elements, like boron or cadmium. In our country, boron carbide (B4C) is used as control, rod., 20. What is a breeder reactor ? ( M – 09,J-15 ), 238, 92U, , and 90Th232 are not fissile materials but are abundant in nature. In the, reactor, these can be converted into a fissile material 94Pu239 and 92U233, respectively by absorption of neutrons. The process of producing more fissile, material in a reactor in this manner than consumed during the operation of the, reactor is called breeding. A fast reactor can be designed to serve as a good, breeder reactor., 21. Write the uses of nuclear reactor. (M-13), , i) Nuclear reactors are mostly aimed at power production, because of the large, , amount of energy evolved with fission., ii) Nuclear reactors are useful to produce radio-isotopes., iii) Nuclear reactor acts as a source of neutrons, hence used in the scientific, research., , 268, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 22. Write short notes on proton – proton fusion in sun. ( M -11 ), 1H, , 1, , + 1H1 → 1H2 + 1e0 + ν (emission of positron and neutrino), , 1H, , 1, , + 1H2 → 2He3 + γ (emission of gamma rays), , 2 2He3 → 2He4 + 2 1H1, The reaction cycle is written as, , 4 1H1 → 2He4 + 2 1e0 + 2ν + energy (26.7 MeV), Thus four protons fuse together to form an alpha particle and two Positrons with, release of large amount of energy., , 23. What are cosmic rays? ( J – 08, J -10,O-14 ), , The ionising radiation many times stronger than γ-rays entering the earth, from all the directions from cosmic or interstellar space is known as cosmic rays., The name, cosmic rays was given by Millikan. The cosmic rays can be broadly, classified into primary and secondary cosmic rays., , www.Padasalai.Net, 24. What is pair production and pair annihilation ? ( M – 07, M – 06, J – 06,M-14 ), , i) Pair production: The conversion of a photon into an electron –positron pair, on its interaction with the strong electric field surrounding a nucleus is called, pair production., ii) Pair annihilation : The converse of pair production in which an electron and, positron combine to produce a photon is known as annihilation of matter., 25. Write a note on leptons? ( J – 07, M – 12,O-15 ), , Leptons are lighter particles having mass equal to or less than about 207, times the mass of an electron except neutrino and antineutrino., This group contains particles such as electron, positron, neutrino, antineutrino,, positive and negative muons. The electron and positron are the antiparticles., Neutrino and antineutrino are also associated with β-ray emission. The neutrinos, and antineutrinos are massless and chargeless particles, but carrier of energy and, spin. Muons were discovered in cosmic ray studies., , 269, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , OTHER IMPORTANT ‘3’ MARKS:, 26. What are isotopes ? Give an example., , Isotopes are atoms of the same element having the same atomic number Z but, different mass number A. The nuclei 1H1, 1H2 and 1H3 are the isotopes of hydrogen., 27. What are isotones ? Give an example., , Isotones are atoms of different elements having the same number of, neutrons. 6C14 and 8O16 are some examples of isotones., 28. What are isobars? Give an example., , Isobars are atoms of different elements having the same mass number A, but, different atomic number Z. The nuclei 8O16 and 7N16 represent two isobars. Since, isobars are atoms of different elements, they have different physical and chemical, properties, 29. Define half life period., , The half life period of a radioactive element is defined as the time taken for, one half of the radioactive element to undergo disintegration., , www.Padasalai.Net, Ö, , æ, , =, , . Ta, c, , 30. Define radioactivity., , The phenomenon of spontaneous emission of highly penetrating radiations, such as α, β and γ rays by heavy elements having atomic number greater than 82, is called radioactivity and the substances which emit these radiations are called, radioactive elements, 31. Define mean life., , The mean life of a radioactive substance is defined as the ratio of total life, time of all the radioactive atoms to the total number of atoms in it., , ∴ Mean life =, , . k -Ñ °%Ñ ±%k -Ñ 4°° ±´ 4±-k., ±-±4° ² k, , 6 -Ñ 4±-k., , 32. Define atomic mass unit. Write its energy equivalent., , One atomic mass unit is considered as one twelfth of the mass of carbon, atom 6C12. Carbon of atomic number 6 and mass number 12 has mass equal to 12, amu., 1 amu = 1.66 × 10−27 kg, The energy equivalent of 1 amu = 931 MeV, , 270, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 33. What are thermo nuclear reactions?, , The fusion process can be carried out only at a extremely high temperature, of the order of 107 K because, only at these very high temperatures the nuclei, are able to overcome their mutual repulsion. Therefore before fusion, the lighter, nuclei must have their temperature raised by several million degrees. The nuclear, fusion reactions are known as thermo-nuclear reactions., 34. What are nuclear force?, , There is some other force in the nucleus which overcomes the electrostatic, repulsion between positively charged protons and binds the protons and neutrons, inside the nucleus. This force is called nuclear force, 35. What is, , Ù – decay ? Give an example., , When a radioactive nucleus disintegrates by emitting a β – particle , the atomic, number increases by one and the mass number remains the same., β – decay can be expressed as, A, zX, , →, , A, Z+1Y, , +, , −1e, , 0, , Example : Thorium (90Th234) is converted to protoactinium (91Pa234) due to β – decay, , www.Padasalai.Net, 90Th, , 36. What is, , 234, , → 91Pa234 + −1e0, , Ú – decay ? Give an example., , When a radioactive nucleus emits γ – rays , only the energy level of the, nucleus changes and the atomic number and mass number remain the same., During α or β – decay , the daughter nucleus is mostly in the excited state., It comes to ground state with the emission of γ – rays ., Example : During the radioactive disintegration of radium (88Ra226) into radon, (86Rn222), gamma ray of energy 0.187 MeV is emitted, when radon returns from, the excited state to the ground state, 37. Distinguish between natural and artificial radioactivity., , Natural radioactivity, 1. The radioactive elements are found, to exist in nature, 2. Heavy elements having atomic, number greater than 82 like Uranium,, Thorium exhibit natural radioactivity, 3. Positron are not emitted, , 271, , J.SHANMUGAVELU, , Artificial radioactivity, They are not usually found in, nature, Both light and heavy elements, exhibit artificial radioactivity, Positron are emitted, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 38. What are the four groups of elementary particles., , Elementary particles are classified into four major groups as photons, leptons, mesons and baryons., 39. What are the biological effects of nuclear radiation?, , i) Short term recoverable effects, ii) Long term irrecoverable effects and, iii) Genetic effect, 40. What is meant by artificial transmutation?, , Artificial transmutation is the conversion of one element into another by, artificial methods. The first successful artificial transmutation was carried out by, Rutherford. When nitrogen was bombarded with α-particles of sufficient energy, a, rare isotope of oxygen (8O17) and a proton were formed., 14, 7N, , + 2He4 → 8O17 + 1H1, , This process is called nuclear reaction., , www.Padasalai.Net, 41. What are used as a)moderator b)coolants?, , a) moderator :, Ordinary water and heavy water are the commonly used moderators., Graphite is also used as a moderator in some countries., , b) coolants :, Ordinary water, heavy water and liquid sodium are the commonly used, coolants, 42. What are primary and secondary cosmic rays?, , The primary cosmic rays are those coming from outer space and enter the, outer boundary of the earth’s atmosphere. The primary cosmic rays consist of, 90% protons, 9% helium nuclei and remaining heavy nuclei. The energy of the, primary cosmic rays is of the order 108 MeV., The secondary cosmic rays are produced when primary cosmic rays interact, with gases in the upper layers of the atmosphere. They are made up of particles, like α - particles , protons, electrons, positrons, mesons, photons, etc. in different, proportions., , 272, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 43. State the concept of Yukawa about nuclear force., , The nuclear force existing between any two nucleons may be due to the, continuous exchange of particles called mesons, just as photons, the exchange, particle in electromagnetic interactions., 44. What are particle accelerators?, , A particle accelerator is a device used to accelerate the charged particles, which, are required in the study of artificial transmutation of elements. Hence the accelerator is, the basic device in high energy particle physics., Accelerators can be divided broadly into two types., Electrostatic accelerators:, The Cockcroft – Walton and Van de Graaff generators belong to this class., Cyclic or synchronous accelerator:, Linear accelerator, cyclotron, betatron, synchrocyclotron and synchrotron., 45. Define nuclear fission., , The process of breaking up of the nucleus of a heavier atom into two, fragments with the release of large amount of energy is called nuclear fission., Example: Nuclear reactor, atom bomb, , www.Padasalai.Net, 46. Nuclear fusion?, , Nuclear fusion is a process in which two or more lighter nuclei combine to, form a heavier nucleus., Example: Hydrogen bomb, , 47. What are mesons?, , Mesons are fundamental particles carrying a single unit of charge and, possessing mass intermediate between electron and proton (me and mp)., The name meson was given by Yukawa in 1935. The three types of mesons are, 1) π- meson (pion) 2) K- meson (kaon) and 3) y - meson. The mesons are the, interaction agents between nucleons. The rest mass of mesons varies between 250, me and 1000 me., 48. What are baryons?, , Baryons form the heavier particle group. Protons and neutrons are called, nucleons and the rest of the heavier particles other than nucleons are known as, hyperons. There are four types of hyperons which are lambda, sigma, xi and, omega hyperons. Protons and neutrons are around 1836 times the mass of the, electron , whereas the mass of the hyperons vary from, 2180 me and 3275 me., , 273, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PUBLIC ‘3’ MARK PROBLEMS:, , 1. The half life of radon is 3.8 days. Calculate its mean life[M-07, O-09, J – 09, J-10], , Data: T½, , Sol:, , = 3.8 days;, , τ=?, , Half Life Periods T½ = 0.6931 / λ, , = 0.6931 τ, Mean life τ = T½ / 0.6931, = 3.8 / 0.6931, , τ = 5.482 days., 2., , The half-life of 84Po218 is 3 minute. What percentage of the sample has decayed in, 15 minutes?[O-07,J-13], , Data: T ½ = 3 minutes;, , Number of T ½ in 15 minutes = 5, , www.Padasalai.Net, Sol:, , [, , m, , E, , The remaining part of radioactive substance [ = § ¨ = § ¨, 0, , The percentage of remaining part of radioactive substance = ( ½ ) 5, = 0.03125, , 100%, , 100% = 3.125, , Decayed percentage = 100 – 3.125, , Decayed percentage = 96.875 %, 214, , undergoes a successive disintegration of two α–decays, and two β-decays. Find the atomic number and mass number of the resulting isotope.[J09,O-14,J-15], , 3. The radioactive isotope, , Sol:, , 84Po, , ,, , ,, , —, , —, , For two α decays: 84Pa214 ÛÜÜÝ 82 A 210 ÛÜÜÜÝ 80 B 206, , For two β decays : 80 B 206 ÛÜÜÜÝ 81 C 206 ÛÜÜÝ 82 D 206, , The resulting isotope is lead (Pb)with mass number 206 and atomic number 82 ., , 274, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 4., , Tritium has a half life of 12.5 years. What fraction of the sample of will be left, over after 25 years?[M-10], , Data: T ½ = 12.5 years;, Sol:, , Number of T ½ in 25 years = 25 / 12.5 = 2, remaining part of radioactive substance, , o, , o, , m, , =§ ¨, < =, , <, , =>, , Fraction of the sample left over after 25 years =§ ¨, =, , 5., , Tritium has a half life of 12.5 years. What fraction of the sample of will be left, over after 50 years?[J-12,O-12], , Data: T ½ = 12.5 years;, Sol:, , Number of T ½ in 50 years = 50 / 12.5 = 4, remaining part of radioactive substance, , o, , o, , m, , =§ ¨, , www.Padasalai.Net, < >, , Fraction of the sample left over after 50 years = § ¨, =, , 6., , <, , = <j, , What percentage of given radioactive substance will be left after 5 half life, periods?[M-11], , Sol:, , remaining part of radioactive substance, , m, , o, , 1 E, , = § ¨ = §2¨, o, , 1 E, , percentage of remaining part of radioactive substance = §2¨, , 100%, , percentage of remaining part of radioactive substance = 3.125 %, 7., , Calculate the number of atoms in one gram of 3Li6 ?, ( Avagadro number = 6.023 1023)[O-11], , Sol: According to Avagadro’s principle,, 6 g of 3Li6 = 6.023 × 1023 atoms, 6, , number of atoms in one gram of 3Li =, , ., , a, , = 1.003, 275, , J.SHANMUGAVELU, , [P.G. T. in Physics], , S, , 10 a atoms, Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, PUBLIC ‘5’ MARK PROBLEMS:, 8., , Calculate the binding energy and binding energy per nucleon of20Ca40 nucleus., Given, mass of 1 proton = 1.007825 amu ;, mass of 1 neutron = 1.008665 amu ;, mass of 20Ca40 nucleus = 39.96259 amu ( J-16 ), , Data : Mass of 20Ca40 nucleus = 39.96259 amu, Mass of 1 proton = 1.007825 amu, Mass of 1 neutron = 1.008665 amu, Number of protons A = 20, Number of neutrons N = 20, Binding energy and binding energy per nucleon = ?, Solution :, Mass of 20 Protons, Mass of 20 neutrons, Total mass of the nucleons, , = 20 × 1.007825, , = 20.15650 amu, , = 20 × 1.008665, , = 20.17330 amu, , = 20.15650 + 20.17330 = 40.32980 amu, , www.Padasalai.Net, Actual mass of the 20Ca40 Nucleus = 39.96259 amu, , ∴ Mass defect ∆, , = total mass of the nucleons - actual mass of the nucleus, , = 40.32980 – 39.96259 = 0.36721 amu, , But 1 amu = 931 MeV, Binding energy = 0.36721 × 931 = 341.87251 MeV, š, , Binding energy per nucleon § ¡ ¨ =, , 276, , J.SHANMUGAVELU, , a, , . Ï E, ;`, , = 8.547 MeV, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 15., , 2017-2018, , Calculate the time required for 60% of a sample of radon to undergo decay., Given T½ of radon = 3.8 days[M-08,M-13] [Compulsory], Data : Half life of radon (T½) = 3.8 days, Amount of sample disintegrated = 60%, Time required = ?, , Solution :, , λ =, , . Ta, a., , per day, , Amount of sample disintegrated = 60%, Amount of sample present = 40%, Let No be the original amount of the sample present., From law of disintegration,, N = Noe–λt, Substituting for, , N = 40% of No ,, No = Noe–λt, , www.Padasalai.Net, e λt =, , loge 2.5 = λ × t, t=, , a., , log 10 2.5× 2.3026, , . Ta, , t = 5.022 days, 16., , A piece of bone from an archaeological site is found to give a count rate of 15, counts per minute. A similar sample of fresh bone gives a count rate of 19 counts, per minute. Calculate the age of the specimen., Given : T½ = 5570 years [M-06,O-11,M-14,J-16] [Compulsory], , Data : Count rate of fresh sample, No = 19 counts per minute, Count rate of bone N = 15 counts per minute, T½ = 5570 years ; Age of the sample, t = ?, Solution :, , N = N0e−λt, λ =, , 280, , J.SHANMUGAVELU, , . Ta, , I æ, , (, , . Ta, , EEÏ, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 18., , A reactor is developing energy at the rate of 32 MW. Calculate the required, number of fissions per second of 92U235. Assume that energy per fission is 200, MeV.[J-06,J-08,J-11,O-09,M-14,M-17] [Compulsory], , Given data:, , Energy per fission = 200 MeV;, Developing energy = 32 MW, , Sol: Let N be the number of fissions per second, producing 32 MW., , Energ, y per Þission, Number of, ³, µ = Total energy released per second., Þission per second N, Number of fission second =, , ±-±4°, , ² 6øÐ ù 6 Þ%..%-², R, , a, , N=, , =, , ² 6øÐ 6 ° 4. ³ ù 6 . ,-²³, , R, , a, , ., , R, , a., , www.Padasalai.Net, N = 1 <@<› fission/ second, , 19., , Calculate the energy released when 1 kg of 92U235 undergoes nuclear fission., Assume, energy per fission is 200 MeV. Avagadro number = 6.023 × 1023., Express your answer in kilowatt hour also.[M-06,J-08,O-13] [Compulsory], , Data : Energy produced per fission of 92U235 = 200 MeV, Amount of 92U235 = 1 kg, Avagadro number = 6.023 x 1023, Energy released = ?, Solution : Energy released in one fission = 200 MeV, According to Avagadro’s hypothesis,, Number of atoms in 235 g of uranium = 6.023 × 1023, , ∴ Number of atoms in 1 kg (1000 gm) of Uranium =, , 282, , J.SHANMUGAVELU, , [P.G. T. in Physics], , ., , a, , R, , aE, , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, Energy produced by 1 kg of, , ., , uranium during fission, E =, , R, , a, , aE, , 2017-2018, , 200 MeV, , = 5.126 × 1026 MeV, = 5.126 × 1026 × 106 eV, = 5.126 × 1032 × 1.6 × 10−19 J, , [∵1 eV=1.6 × 10−19 J], , 1 kilowatt hour = 3.6 × 106 joule, E=, , S, , E., , ., , R, , a., , kWh, , E = 2.2782 × 107 kWh, 20., , A carbon specimen found in a cave contained a fraction of 1/8 of C14 to that, present in a living system. Calculate the approximate age of the specimen. Given, T1/2 for 6C12 = 5560 years[O-13] [Compulsory], , www.Padasalai.Net, Given data: N = N0 ⇒, , Sol:, , o, , o, , m, , o, , =§ ¨, , o, , = (§ ¨, , o, , o, , Total time = n, , = ;, , T1/2 = 5560 years, , here n =, , Ò-±4° ±%k, ß4°Ñ °%Ñ, , a, , Half life, , = 3 5560, , t = 16680 years, , 283, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , FIVE MARKS:(1 – question : Q. No: 60), 1. Explanation the binding energy curve( Graph is not necessary) (J-10,O-16), 2. Write the properties, , α - rays. (O-10,J-15), , 3. Write any five properties, , • - rays.(O-15), , 4. Explain the Soddy- Fajan’s radioactive displacement laws. (M-11), 5. State the properties of neutrons. (J-13), 6. Write a note on biological hazards of nuclear radiations. (O-08), 7. Explain the principle and working of an atom bomb.(J-14,), 8. Explain how carbon – nitrogen cycle can account for the production of stellar, , energy.(O-14), 9. Explain the latitude effect of cosmic rays. (O-07,J-09), 10. Explain how cosmic ray shower is formed. (M-07,J-12), , www.Padasalai.Net, TEN MARKS:(1 – question : Q. No: 68), , 1. Describe the principle and action of a Bainbridge Mass spectrometer in determining the, isotopic masses. (J - 06, J - 07, J - 08, O - 06, M - 09, J - 10, O -10, J -11, O -12,O13,M-14), , 2. Obtain an expression for the amount of the radioactive substance present at any moment., Obtain the relation between half life period and decay constant. (O - 08, O – 09, M –, 12,), [OR], State the radioactive disintegration. Obtain the relation N = N0 { #cë . Derive the, relation between half – life period and decay constant. (O-11,J-12,J-15), 3. Describe the discovery of neutrons. Mention the properties of neutrons.(O-14), 4. Explain the principle construction and working of a Geiger-Muller [G.M] counter. (M 07, O - 07, J - 09, M -11, M – 13,J-14,O-15,M-17), 5. What is a nuclear reactor? Explain the function of (i) moderator (ii) control rod and (iii), neutron reflector. Mention the uses of nuclear reactor. (diagram not necessary) (M –, 06,O-16), 6. What are cosmic rays? Explain, (i) latitude effect, (ii) altitude effect of cosmic rays. (M - 08,M - 10, J – 13, M-15), , 285, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, Only for slow bloomers, ( 5& 10 Mark answer ), 1. Radioactive displacement law, , à-decay :, , The atomic number decreases by two and mass number decreases by four, Example: 88Ra226 → 86Rn222 + 2He4, , á−decay:, , The atomic number increases by one and the mass number remains the same., Example: 90Th234 → 91Pa234 + −1e0, , â−decay:, , Only the energy level of the nucleus changes and the atomic number and, mass number remain the same., Example: During the radioactive disintegration of radium into radon , gamma ray of, energy 0.187 MeV is emitted, when radon returns from the excited state to the ground, state, 2. Atom bomb:, , Principle: uncontrolled fission chain reaction., Construction: An atom bomb consists of two, hemispheres of U235 (or 94Pu239), each smaller, than the critical size and are kept apart by a, separator aperture, , www.Padasalai.Net, Operation : When the bomb has to be exploded, a third well fitting cylinder of U235, (or 94Pu239) whose mass is also less than the critical mass, is propelled so that it fuses, together with the other two pieces., Now the total quantity is greater than the critical mass and an uncontrolled chain, reaction takes place resulting in a terrific explosion., 3. Properties of Cathode ray, Canal ray, X – ray, , Ù, Ú – rays:, Cathode ray, deflected by, electric and, magnetic fields, affect, photographic, plates, ionize the gas, through which, they pass, produce, fluorescence, travel in straight, lines, , Canal ray, deflected by, electric and, magnetic fields, affect, photographic, plates, ionize the gas, through which, they pass, produce, fluorescence, travel in straight, lines, , X - ray, not deflected by, electric and, magnetic fields, affect, photographic, plates, ionize the gas, through which, they pass, liberate photo, electrons, travel in straight, lines, , - ray, deflected by, electric and, magnetic fields, affect, photographic, plates, ionize the gas, through which, they pass, produce, fluorescence, travel in straight, lines, , Ù - ray, deflected by, electric and, magnetic fields, affect, photographic, plates, ionize the gas, through which, they pass, produce, fluorescence, velocities over, the range of 0.3 c, to 0.99 c, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html, , Ú - ray, not deflected by, electric and, magnetic fields, affect, photographic, plates, ionize the gas, through which, they pass, produce, fluorescence, travel with the, velocity of light
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 4. Bainbridge mass spectrometer:, Uses : accurate determination of atomic masses., Diagram :, , Construction: A beam of positive ions produced in, a discharge tube is collimated in to a fine beam by, two narrow slits S1 and S2., This fine beam enters into a velocity selector. The, velocity selector allows the ions of a particular velocity, to come out of it. It consists of two electric field E and, an electromagnet, to produce uniform magnetic field B, two fields and direction of the beam are at right angles, to each other., , Working: The electric field and magnetic field are adjusted the ions do not suffer any, deflection. The force exerted by the electric field is equal to qE and the force exerted by, the magnetic field is equal to Bqv., , Bqv = qE, ö, , A, , V=, , ions having this velocity v, through the slit S3, to enter the evacuated chamber D. These, positive ions are subjected to strong uniform magnetic field of induction B′ at right, angles to the plane of the paper acting inwards. These ions are deflected along circular, path of radius R and strike the photographic plate. The centripetal force is B’qv, , www.Padasalai.Net, B’qv =, m=, Substituting V =, , ö, , A, , m=, , kT, ô, , A’ ô, ¯, , AA’ ô, ö, , Ions with different masses trace semi-circular paths of different radii and produce dark, lines on the plate. The distance between the opening and the dark line gives the diameter, 2R from which radius R can be calculated., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 5. Radioactive law of disintegration and Half life period:, The rate of disintegration at any instant is directly proportional to the number of atoms, of the element present at that instant., Let [ be the number of radioactive atoms present initially and N, the number of atoms, at a given instant t. Let dN be the number of atoms undergoing disintegration in a small, interval of time dt, -, , äo, äë, , äo, , ∝[, , äë, , ( 8 Ñ[ (Ñ – decay constant ), , q[, ( 8 Ñ qï, [, , Integrating, loge N = – λt + C, At t = 0, N = [, , b}© [ = C, , b}© N = −λt + b}© [, , b}© § ¨= 8 Ñï, o, o, , o, , [ = [ { # cë, , = { # cë ., , www.Padasalai.Net, o, , The number of atoms of a radioactive, substance decreases exponentially with, increase in time. Initially the disintegration, takes place at a faster rate., As time increases, N gradually decreases, exponentially. An infinite time is required, for the complete disintegration of all the, atoms., , Half life period:, The half life period of a radioactive element is defined as the time taken for one half of, the radioactive element to undergo disintegration., [ = [ { # cë, , ï( Ö, , æ, , ,, , N=, , o, , o, , =[ {, , b}© 2 = ÑÖ, Ö, , æ, , (, , # cI æ, , æ, , dxHã, c, , =, , dxH, , c, , .a, , =, , . Ta, c, , The half life period is inversely proportional to its decay constant., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 6. Neutron – Discovery, In 1930 Bothe and Becker found that when beryllium was bombarded with ™ particles, a highly penetrating radiation was emitted., This radiation was capable of traversing through a thick layer of lead and was, unaffected by electric and magnetic fields., In 1932 Irene Curie and F. Joliot found that those radiations were able to knock out, protons from paraffin and similar substances that are rich in hydrogen., Chadwick in the same year discovered that the emitted radiation consists of particles, of mass nearly equal to proton and no charge. He called them as neutrons., 9, 4Be, , + 2He4 → 6C12 + 0n1, , Properties of neutrons:, i) Neutrons are the constituent particles of all nuclei, except hydrogen., ii) They are not deflected by electric and magnetic fields., iii) Neutrons are stable inside the nucleus. But outside the nucleus they are unstable., iv) As neutrons are neutral, they can easily penetrate any nucleus., v) Neutrons are classified according to their kinetic energy as, a) slow neutrons - 0 to 1000 eV ( thermal neutrons - 0.025 eV), b) fast neutrons - 0.5 MeV and 10 MeV, 7. Geiger – Muller counter:, , www.Padasalai.Net, Uses : Measure the intensity of the radioactive radiation., , Principle : When nuclear radiations pass through gas, ionisation is produced., Diagram :, , Construction: It consists of a metal tube (C), , acting as the cathode and a fine tungsten wire (W), acts as anode. The tube is well insulated from the, anode wire. The tube is filled with an inert gas like, argon at a low pressure. One end is fitted with a, thin mica sheet and this end acts as a window, through which radiations enter the tube., A high potential difference of about 1000 V is applied between the electrodes through a, high resistance R of about 100 mega ohm., Operation: When an ionising radiation enters the counter, primary ionisation takes place ., These ions are accelerated due to the high potential difference and these ions are multiplied, by further collisions., Thus an avalanche of electrons is produced in a short interval of time and reaching the, anode generates a current pulse., This is amplified by electronic circuits and is used to operate an electronic counter., The counts in the counter is directly proportional to the intensity of the ionising radiation., The ionisation of the gas is independent of the type of the incident radiation., G.M. counter does not distinguish the type of radiation ., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 8. Nuclear reactor., A nuclear reactor is a device in which the nuclear fission reaction takes place in a, self sustained and controlled manner., Fissile material or fuel, The fissile material or nuclear fuel generally used is 92U235. Pu239 and U233 are, used as fissile material., PHWR - natural uranium oxide, PWR - low enriched uranium, PFBR - mixture of oxides of plutonium and uranium, FBTR - mixture of the carbides of uranium and plutonium, Kamini reactor - U233 is used., Moderator, It is to slow down fast neutrons of energy about 2 MeV to thermal neutrons of, energy about 0.025 eV, E.g. Ordinary water and heavy water. Graphite, FBR – No moderator, Neutron source, A source of neutron is required to initiate the fission chain reaction for the first, time., E.g. A mixture of beryllium with plutonium or radium or polonium, , www.Padasalai.Net, Control rods, It is used to control the chain reaction and very good absorbers of neutrons, E.g. Boron, Cadmium, Boron Carbide., The cooling system, It removes the heat generated in the reactor core., A good coolant: large specific heat capacity and high boiling point., E.g. Heavy water, ordinary water, liquid sodium., , Neutron reflectors, It prevent the leakage of neutrons to a large extent, by reflecting them back., E.g. Depleted Uranium, Thorium., Shielding, Reactor is surrounded by a concrete wall of thickness about 2 to 2.5m., Uses of reactors :Power production, To produce radio – isotopes, Sources of neutrons - scientific research., , 290, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 9. Cosmic Rays: The ionising radiation many times stronger than •-rays entering the, earth from all the directions from cosmic or interstellar space is known as cosmic, rays., Latitude effect: The variation of cosmic ray intensity with geomagnetic latitude is, known as latitude effect., The intensity is maximum at the poles (θ=900), minimum at the equator (θ = 0) and, constant between latitudes of 420 and 900., Diagram :, , The decrease in cosmic ray intensity at the earth’s equator is explained to be due to the, earth’s magnetic field. The charged particles approaching the earth near the poles along, the direction of the magnetic lines of force., They experience no force and hence maximum intensity at poles., But the charged particles at the equator have to travel in a perpendicular direction to, the field and are deflected away and hence minimum intensity at the equator., , www.Padasalai.Net, Altitude effect:, , The study of variation of cosmic, ray intensity (I) with altitude (h) is, known as altitude., The intensity increases with altitude, and reaches a maximum at a height of, about 20 km. Above this height there, is a fall in intensity., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 9.SEMICONDUCTOR DEVICES AND THEIR, APPLICATIONS, PREPARED BY, J.SHANMUGAVELU M.Sc, B.Ed. [P.G. Assist in Physics], , ONE MARK QUESTION: (3 – Questions), BOOK BACK QUESTIONS:, 1. The electrons in the atom of an element which determine its chemical and electrical, properties are called, b) revolving electrons, a) valence electrons, c) excess electrons, d) active electrons, 2. In an N−type semiconductor, there are, a) immobile negative ions, c) immobile positive ions, , b) no minority carriers, d) holes as majority carriers, , www.Padasalai.Net, 3. The reverse saturation current in a PN junction diode is only due to, a) majority carriers, b) minority carriers c) acceptor ions, d) donor ions, 4. In the forward bias characteristic curve, a diode appears as, a) a high resistance, b) a capacitor, c) an OFF switch, , d) an ON switch, , 5. Avalanche breakdown is primarily dependent on the phenomenon of, a) collision, b) ionization, c) doping, d) recombination, 6. The colour of light emitted by a LED depends on, a) its reverse bias, b) the amount of forward current, c) its forward bias, d) type of semiconductor material, 7. The emitter base junction of a given transistor is forward biased and its collector - base, junction is reverse biased. If the base current is increased, then its, a) VCE will increase, b) IC will decrease, c) IC will increase, d) VCC will increase., 8. Improper biasing of a transistor circuit produces, a) heavy loading of emitter current, b) distortion in the output signal, c) excessive heat at collector terminal, d) faulty location of load line, 9. An oscillator is, a) an amplifier with feedback, c) nothing but an amplifier, , 292, , J.SHANMUGAVELU, , b) a convertor of ac to dc energy, d) an amplifier without feedback, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 10. In a Colpitt’s oscillator circuit, a) capacitive feedback is used, c) no tuned LC circuit is used, , b) tapped coil is used, d) no capacitor is used, , 11. Since the input impedance of an ideal operational amplifier is infinite,, a) its input current is zero, b) its output resistance is high, c) its output voltage becomes independent of load resistance, d) it becomes a current controlled device, 12. The following arrangement performs the logic function of ______ gate, , a) AND, b) OR, c) NAND, Sol: Input of the first NAND gates is A , B and output is ¢¢¢¢¢, A. B, , d) EXOR, , äääää, Input of the second NAND gates is ¢¢¢¢¢, A. B and output is A., B = A.B i.e, , AND gate, , 13. If the output (Y) of the following circuit is 1, the inputs A B C must be, , a) 0 1 0, b) 1 0 0, Sol: The output Y = (A + B) . C;, , c) 1 0 1, , d) 1 1 0, , www.Padasalai.Net, If C=0,, , Y = (A + B) . 0 =0, , If C =1,, , Y = (1 + 0) . 1=1 (OR) Y = (0 + 1) . 1=1, , The input is 101 (OR) 011, , 14. According to the laws of Boolean algebra, the expression (A + AB) is equal to, ¬, b) AB, c) B, d) A, a) A, Sol: A + AB = A (1 + B), , = A (1) = A;, , ( since 1 + B = 1), , 15. The Boolean expression ¢¢¢¢¢¢, å¦Ç can be simplified as, ¬. B., ¬ C¢, a) AB + C¢, b) A, c) AB + BC + CA, Sol: ABC is the product of A and BC,, , ¬ +¦, ¬ + Ç¢, d) å, , Using De Morgan’s theorem, ABC = A. BC = A + BC, =A+B+C, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , PUBLIC QUESTIONS, 16. The forbidden energy gap for germanium is of the order of, a) 1.1 eV, b) 0.7 eV, c) 0.3 eV, , d) 10 eV, , 17. Condition for oscillator is, a) A β = 0, b) A =1/β, , d) A + β= 0, , Sol:, , c) Aβ = ∞, , Aβ=1, ⇒ A =1/ β, , 18. The potential barrier of silicon PN junction diode is approximately, a) 0.3 V, b) 0.7 V, c) 1.1 V, d) 10 V, 19. The Boolean expression to represent NAND operation is, ¬, ¢¢¢¢¢¢¢, ¢¢¢¢¢¢, a) Y=A, 7B, b) Y= å, .¦, c) Y=A, , 20. The forbidden energy gap for silicon is the order of, a) 0.7 eV, b) 0.4 eV, c) 1.1 eV, , d) Y= A.B, d) 10 eV, , 21. In CE single stage amplifier, the voltage gain at mid – frequency is 10. The voltage gain at, upper cut off frequency is, a) 10, b) 14.14, c) 7.07, d) 20, , www.Padasalai.Net, Sol:, , f U = AM, = 10, , √, , 0.707 = 7.07, , ( since, , √, , = 0.707), , 22. Barkhausen condition for maintenance of oscillation is, b) A β = ∞, c)A= β, a) β =1/A, Sol:, , d) A β =, , √, , Aβ=1, ⇒ β =1/ A, , 23. Find the voltage across the resistor as shown in the figure ( silicon diode is used ), , a) 2.4 V, , b) 2.0 V, , c) 1.8 V, , d) 0.7V, , Sol: Barrier potential for silicon = 0.7 V ;, , Voltage across the resistor = 2.7 – 0.7 = 2.0 V, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 24. The output (Y) of the logic circuit given below is, , ¬+B, ¢¢¢¢¢¢¢, ¬, b) A.B, c)A, 7B, d)A, a) A+B, Sol: Output of the first NOT gate = A; Output of the second NOT gate = B;, ¬.B, ¬, Input of the AND gate = A and B; Output of AND gate = A, Output of the final NOT gate ( Y ) = A . B, , æ7B, æ=A+B, Using De Morgan’s theorem A . B = A, 25. The output of the given operational amplifier is, , a) – 2 sin ωt, Sol: Vout =, , =, , #ôç, ô•š, , b) 2 sin ωt, , c) – 2 sin ( ωt+10°), , d) 2 sin ( ωt+10°), , Vin, , #, , 0.2 sin ωt = – 2 sin ωt, , www.Padasalai.Net, 26. The symbol to represent LED is, , a), , b), , c), , d), , 27. The logic gate for which there is ‘low’ output only when both the inputs are ‘High’ is, a) AND, b) NAND, c)NOR, d) EXOR, Hints:, Inputs, A, 0, 0, 1, 1, , B, 0, 1, 0, 1, , Output, Y = .;, 1, 1, 1, 0, , 28. The output voltage of the operational amplifier( Op-Amp ) given below is, , a) – 1 V, b) + 1 V, c) +5V, Sol:This circuit diagram is summing amplifier; Vout = - (V1 + V2), , d) -5V, , = - (2 + x83|) = -2 +3 = +1 V, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 29. In a junction transistor the emitter region is heavily doped since emitter has to supply to, the base, a) minority carriers, b) majority carriers c)acceptor ions, d)donor ions, 30. A logic gate which has an output ‘1’ only when the inputs are complement to each other is, a) AND, b) NAND, c) NOR, d) EXOR, Hints: Complement of 0 is 1 and Complement of 1 is 0, Output, Y = A⨁B, 0, 1, 1, 0, , Inputs, A, B, 0, 0, 0, 1, 1, 0, 1, 1, , 31. Of the following, the donor atoms are, a) silicon and germanium, c)bismuth and arsenic, , b)aluminium and gallium, d)boron and iridium, , 32. In CE amplifiers, the phase difference between input and the output voltage is, a) 0°, b) 90°, c) 270°, d) 180°, 33. In common emitter transistor circuit, the base current ( IB) of the transistor is 50μA and, the collector current (IC) is 25 mA. Then the current gain is, a) 50, b) 500, c) 20, d)200, , www.Padasalai.Net, Sol:, , β ( IC, I, , B, , S, , E, , =E, =, , R, , =, , 103, , = 500, , 34. The following arrangement performs the logic function of, , b) EXOR, a) AND, Sol: Output of the first NOT gate = A;, , c)OR, , d)NAND, , Output of the second NOT gate = B;, Input of the OR gate = A and B; and Output ( Y ) = A 7 B, æ. B, æ=A.B, Using De Morgan’s theorem A 7 B = A, (A.B) is the AND gate formula, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 35. The following arrangement performs the logic function of, , a) AND gate, b) NAND gate, Sol: Output of the first NOT gate = A;, , d) NOR gate, , c) OR gate, , Output of the second NOT gate = B;, Input of the first NAND gate = A and B; and Output = A . B, Using De Morgan’s theorem,, , æ7B, æ=A+B, A . B=A, , Input of the second NAND gate = A + B ;, , and Output = A 7 B, , A 7 B is the NOR gate formula, 36. An example for non – sinusoidal oscillator is, a) multivibrator, b) RC oscillator, c)colpitts oscillator, , d)crystal oscillator, , 37. The Boolean expression of represent AND operation is, ¬, a) Y =A + B, b) Y = A . B, c)Y=A, , ¢¢¢¢, d)Y=AB, , 38. In the pin configuration of IC 741, pin 3 represents, a) inverting input, b) non – inverting input, c) -Vcc, Hints:, , www.Padasalai.Net, Pin number, 1,5, 2, 3, 4, 6, 7, 8, , Function, Offset null, Inverting input, Non – inverting input, Power supply (-Vcc), Output, Power supply (+Vcc), No connection, , 39. The forbidden energy gap for conductors is, a)0.7eV, b) 1.1eV, c) zero, <, , 40. In a transistor the value of 5 8, , a) α, Sol:, , 1, 1, 8 β, α, , I, , b) β, , ( IE =, , d)output, , C, , à, , <, , á, , d)3 eV, , 9 is equal to, , ë, ¿, , é, , c)ê, , d) 1, , ì# ë, ¿, , = ¿= 1;, ¿, , (Iö 8 IA ( Ii ), , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , JEYAM TUITION CENTRE PARAMAKUDI, , www.TrbTnpsc.com, , 2017-2018, , 41. A logic gate for which there is an output only when both the Inputs are zero is, a) NAND, b) NOR, c)EXOR, d) AND, Hints:, Inputs, Output, A, B, Y= 7;, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 42. The phase reversal between the input and output voltages in single stage CE amplifier is, a) π/2, b) 2 π, c) π, d)3 π/2, 43. The following arrangement performs the logic function of, , a) AND, Sol:, , b) OR, c) NAND, Output of the first NANT gate = A;, , d) EXOR, , Output of the second NANT gate = B;, Output of the third NAND gate = A . B, , æ7B, æ=A+B, Using De Morgan’s theorem A . B = A, , www.Padasalai.Net, The output Y is A + B ( OR gate), , 44. In a PN junction diode on the side of N but very close to the junction there are, a) donor atoms, b) acceptor atoms, c) immovable positive ions, d) immovable negative ions, 45. In an N type semiconductor donor level lies, a) just below the conduction band, b) just above the conduction band, c) just below the valence band, d) just above the valence band, 46. For a transistor connected in common emitter mode (CE) the slope of the input, characteristics curve gives, a)input impedance, b)current gain, c) reciprocal of input impedance, d)voltage gain, 47. Which of the following devices has a source of emf inside it ?, a)voltmeter, b)ammeter, c)ohm-meter, , d)rectifier, , 48. Improper biasing of a transistor circuit produces, a)heavy loading of emitter current, b)distortion in the output signal, c)excessive heat at collector terminal, d)faulty location of load line, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 49. The following arrangement performs the logic function of, , a)NOT, b)EXOR, Sol: Output of the first OR gate = A 7 B;, , c)OR, , d)AND, , Output of the second OR gate = A 7 B = A + B, (A + B) is the OR gate formula;, 50. In CE single stage amplifier if the voltage gain at mid frequency AM , then the voltage gain, at lower cut off frequency is, , a), , Üí, , b) √2 AM, , c), , √, Üí, , d), , åï, √=, , 51. The forbidden energy gap for semiconductor Ge and Si are respectively, a) 1.1 eV and 0.7 eV b) 0.7 eV and 1.1 eV, c) 4 eV and 0.7eV, d) 1.1 eV and 7 eV, 52. The potential barrier of germanium PN junction diode is approximately, a) 0.3 V, b) 0.7 V, c) 1.1 V, d) 10 V, 53. A(å + B) = ?, a) A, b) B, Sol:, A(A + B) = A A + AB, , = AB, , c) AB, , d) A + B, , [A A = 0], , www.Padasalai.Net, 54. The voltage at B in the figure is : (Germanium diode ), , a) 5.3 V, b) 5.7 V, Sol: Barrier potential for Germanium = 0.3 V, , c) 6.3 V, , d) 6 V, , Voltage at B = Total potential – Barrier potential, = 6 – 0.3, V= 5.7 V, , 55. In a multimeter, when the current scale shows full scale deflection, the ohmmeter scale, reads, a) Maximum but not infinity, b) infinity, c) zero, d) Minimum but not zero, 56. The Logic gate for which the output is '1', only when both the inputs are '0' is :, a) OR, b) NAND, c) EXOR, d) NOR, Hints:, Inputs, Output, A, B, Y= 7;, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, Choose the correct answer from the options given, , 57. A solid state electronic device mainly consists of a ___________ material, (a) conducting, (b) insulating, (c) super conducting, (d) semiconducting, 58. Resistivity of a semiconductor is ____________, (a) greater than that of conducting material, (b) smaller than that of insulators, (c) greater than that of insulator but lesser than conductors, (d) equal to that of conductors, 59. Of the following, choose the wrong statement., (a) Resistivity of semiconductor is approximately 10-2 – 104 ohm m, (b) Resistance of semiconductor decreases with increase of temperature, (c) Resistance of conductor increases with increase of temperature, (d) Resistivity of a conductor is 10-2 – 104 ohm-m, 60. _____________ is an example for semiconductor, (a) Fe, (b) C, (c) Ag, , (d) Ge, , 61. Electrons in the atom of an element which determine its chemical and electrical properties, are called, (a) valence electrons, (b) revolving electrons, (c) excess electrons, (d) active electrons, , www.Padasalai.Net, 62. For an insulator, the forbidden energy gap is, (a) less than 3 eV, (b) less than 0.7 eV (c) more than 3 eV, , (d) 0.7 eV, , 63. In glass, the energy gap between valence band and conduction band is of the order of, (a)3 eV, (b) 6 eV, (c) 10 eV, (d) 0.7 eV, 64. In case of good conductors, forbidden energy gap is, (a) more than 3 eV (b) equal to 3 eV, (c) Zero, , (d) 0.7 eV, , 65. Electrons in an intrinsic semiconductor, which move into the conduction band at high, temperatures are called, (a) valence electrons, (b) hole, (c) intrinsic carrier, (d) donor, 66. In intrinsic semiconductors,, (a) number of free electrons is equal to number of holes, (b) number of free electrons is greater than number of holes, (c) number of free electrons is lesser than number of holes, (d) number of free electrons is zero, , 300, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 67. Amount of impurity to be added to a intrinsic semiconductor is of the order of, (a) 50 ppm, (b) 100 ppm, (c) 500 ppm, (d) 1000 ppm, 68. Donor atom is a, (a) tetravalent, , (b) trivalent, , (c) pentavalent, , (d) divalent, , 69. Of the following, the donor atoms are, (a) Si and Ge, (b) aluminium and gallium, (c) bismuth and arsenic, (d) boron and indium, 70. Acceptor atom is a, (a) tetravalent, , (b)trivalent, , (c) pentavalent, , 71. _____________ is an example for acceptor atom, (a) Al, (b) Bi, (c) P, , (d) divalent, , (d) As, , 72. In an N type semi conductor, there are, (a) immobile negative ions, (b) no minority carriers, (c) immobile positive ions, (d) holes as majority carriers, 73. In an N type semiconductors,, (a) holes are the majority carriers, (b) both holes and electrons are the majority carriers, (c) electrons are the majority carriers, (d) electrons are the minority carriers, , www.Padasalai.Net, 74. The difference of potential from one side of the barrier to the other side of a PN junction, is known as, (a) depletion region, (b) potential gradient, (c) potential barrier, (d) contact potential, 75. Width of the potential barrier in a PN junction diode depend on, (a) number of electrons, (b) potential difference, (c) number of holes, (d) nature of the material, 76. In forward biased PN junction diode, (a) Potential barrier is reduced, (b) Potential barrier is increased, (c) Width of the potential barrier increases, (d) Potential barrier remains same, 77. In a forward biased junction diode, the voltage at which current starts to increase rapidly, is known as, (a) leakage voltage, (b) reverse saturation voltage, (c) knee – voltage, (d) cutoff voltage, , 301, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 78. Rectifier efficiency of half wave rectifier is approximately, (a) 60.4%, (b) 81.2%, (c) 40.6%, , (d)30.2%, , 79. The output of an half-wave rectifier is, (a) unidirectional and constant, (b) alternating and pulsating, (c) alternating and constant, (d) unidirectional and pulsating, 80. Variations of DC output voltage as a function of DC load current is called, (a) rectification, (b) filter, (c) regulation, (d) full wave rectification, 81. Percentage of regulation is given by –, flÐ ðЕ{ # fðЕ{, flÐ ðЕ{ # fðЕ{, (a), × 100, (b), flÐ ðЕ{, flÐ ðЕ{ # fðЕ{, fðЕ{ # flÐ ðЕ{, (c), × 100, (d), × 100, fðЕ{, fðЕ{, 82. Zener diode are used as ________, (a) rectifiers, (b) filters, 83. Zener current is __________, (a) dependent of applied voltage, (c) dependent on knee voltage, , (c) regulators, , (d) amplifiers, , (b) independent of applied voltage, (d) dependent on material of the diode, , www.Padasalai.Net, 84. ___________ is a reverse biased, heavily doped semiconductor PN junction diode, (a) LED, (b) LCD, (c) Zener diode, (d) transistor, , 85. A Zener diode working in the ____________ region can act as voltage regulators, (a) normal, (b) saturated, (c) breakdown, (d) constant volt, 86. The reverse saturation current in a PN junction diode is only due to, (a) majority carriers, (b) minority carriers, (c) acceptor ions, (d) donor ions, 87. A forward biased diode will act as _____________, (a) A high resistance device, (b) a capacitor, (c) an OFF switch, (d) an ON switch, 88. Avalanche breakdown primarily depends on, (a) collision, (b) ionization, (c) doping, , (d) recombination, , 89. Colour of light emitted by a LED depends on, (a) Its reverse bias voltage, (b) The amount of forward current, (c) Its forward bias voltage, (d) Type of semi conducting materials, , 302, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 90. When an electron in the conduction band recombines with a hole in the valence band then, energy is, (a) stopped, (b) absorbed, (c) released, (d) needed, 91. Emitter’s main function is to supply __________, (a) electrons, (b) majority charge carriers, (c) minority charge carriers, (d) holes, 92. The thickness of the base of a transistor is of the order of, (a) 100 μm, (b) 50 μm, (c) 25 μm, , (d) 200 μm, , 93. In all transistors, _____________ region is made physically larger than any other region, because it has to dissipate more power, (a) base, (b) collector, (c) emitter, (d) either (b) or (c), 94. In any transistor, which of the following is true?, (a) IB = IE / IC, (b) IC = IE + IB, (c) IE = IB + IC, , (d) IC = IE / IB, , 95. Ratio of the collector current to the emitter current is denoted by, , (a), , α, , (b) β, , (c) A, , 96. Value of β interms of α is given by, ,, ,, (a) β =, (b) β =, F,, #,, , (d) β A, F,, , #,, , www.Padasalai.Net, (c), , β=, , ,, , 97. The value of α in any transistor lies between, (a) 50 – 300, (b) upto 1000, (c) 0 – 9, , 5, , ∆¢;`, , ,, , (d) 0.95 – 0.99, , (b) output admittance, (d) output impedance, , 99. Common emitter configuration has, (a) high input impedance, (c) higher current gain, 100., , =, , 9 is called, , ∆Ã;, (a) input impedance, (c) input admittance, , 98. The value of, , (d) β, , (b) low output admittance, (d) all the above, , In CE amplifiers, the phase reversal between input and output voltages is, (a) 0º, (b) 90º, (c) 180º, (d) 270º, , 101. ____________ is the most widely used method of providing bias and stabilization to a, transistor, (a) base bias, (b) base bias with emitter feedback, (c) base bias with collector feedback (d) voltage divider bias, , 303, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 102. Voltage gain of a RC amplifier depends on the _________ over which the amplifier, operates, (a) current, (b) Voltage, (c) frequency, (d) phase, 103., , The gain of the amplifier is constant at _____________ range, (a) low frequency, (b) mid frequency, (c) high frequency, (d) both low and high frequency, , 104. At lower and upper cut-off frequencies, the gain of the amplifier is _________ times of, mid frequency gain, , (a) 2, , 105., , (b), , ½, , (c) √2, , (d), , √, , Improper biasing of a transistor circuit produces, (a) heavy loading of emitter current, (b) distortion in output signal, (c) excessive heat at collector terminal, (d) faulty location of load line, , 106. When number of amplifiers are connected in cascade, the overall voltage gain is equal, to _______________ of individual stages, (a) sum of voltage gain, (b) difference of voltage gain, (c) product of voltage gain, (d) mean voltage gain, , www.Padasalai.Net, 107. When a fraction β of output voltage Vₒ is fed into input voltage Vi , the new input, voltage will be, (a) V'i = Vi + βVₒ, (b) V'i = Vi - βVₒ, (d) V'i = Vi • β√eₒ, (c) V'i = Vi ± βVₒ, 108., , Voltage gain of the amplifier with positive feedback is, ¡, ¡, # —¡, (a), (b), (c), F —¡, # —¡, ¡, , (d), , 109., , Essential conditions for the maintenance of oscillation is, (a) β = 1/A with positive feedback, (b) β = 1/A with negative feedback, (c) β = 1 with positive feedback, (d) β = A with negative feedback, , 110., , Frequency of oscillations in Colpitts oscillators is, , (a) f =, (c) f =, , 304, , ‘, , _P, , F, , P P, , F ñ P, , FP, , J.SHANMUGAVELU, , (b) f =, (d) f =, , F —¡, ¡, , ‘ P P, , _, , P FP, P P, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, 111., , The Boolean expression to represent EXOR operation is, (a) Y = A + B, (b) Y = ¢¢¢¢¢¢¢¢¢¢¢¢, 6 7 ¢¢¢¢, 6, (c) Y = A6¢ + ̅ 6, , 2017-2018, (d) Y = A•B, , 112., , OP – AMP consists of, (a) 20 transistors, 11 resistors, 1 capacitor, (b) 1 transistor, 11 resistors, 20 capacitors, (c) 20 transistors, 1 resistor, 11 capacitors, (d) 11 transistors, 20 resistors, 1 capacitor, , 113., , Since the input impedance of an ideal operational amplifier is infinity,, (a) its input current is zero, (b) its output resistance is high, (c) its output voltage become independent of load, (d) it become a current controlled device, , 114. In a common base connection ∝ = 0.95, IE = 1mA, then the value of collector current is, ____________, (a) 0.05 mA, (b) 0.95 mA, (c) 1.05 mA, (d) 1 mA, 115., , In a transistor, the value of ∝ is 0.99, then the value of β is, (a) 49, (b) 90, (c) 99, , (d) 9.9, , www.Padasalai.Net, 116. In the transistor with β = 40 , the base current is 25 μA. Then the collector current IC, is ____________, (a) 100μA, (b) 1000 mA, (c) 1 mA, (d) 0.1 mA, , 117. In a common base configuration of a transistor, IC = 12.5 mA and IE = 13 mA, then the, base current of the transistor is, (a) 25.5 mA, (b) 0.5 mA, (c) 50 mA, (d) 50 μA, 118. The input impedance of a transistor is 1000 ohm and β = 100 then, the base of emitter, voltage required for collector current of 1 mA is, (a) 1 V, (b) 100 mV, (c) 10 mV, (d) 1 mV, 119. In CE configuration, the IC changes from 2 mA to 4 mA. If VCE is increased from 5V, to 10V, output admittance must be, (a) 8 × 10-4 mho, (b) 4 × 10-3 mho, (c) 2.5 × 103 mho, (d) 1.25 × 103 mho, 120. Three amplifier have gains 10, 50 and 80 respectively, when they are connected in, cascade the overall gain is, (a) 4000, (b) 400, (c) 40000, (d) 140, 121., , If an inductor of inductance, , 305, , J.SHANMUGAVELU, , <, , H and a capacitance 4pF are connected in parallel to, >?=, form LC tank circuit, then the frequency of oscillation is, (a) 5 MHz, (b) 0.5 MHz, (c) 50 MHz, (d) 500 MHz, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , PUBLIC ‘3’ MARKS:( 4 – Questions : Q. No: 46, 47, 48, 49), 1. What is an intrinsic semiconductor ? Give any two examples. ( M – 06 ,O-, , 15,M-17), , A semiconductor which is pure and contains no impurity is known as an, intrinsic semiconductor., ii) In an intrinsic semiconductor, the number of free electrons and holes are, equal. Common examples of intrinsic semiconductors are pure germanium and, silicon., i), , 2. What is meant by doping? ( O – 08 ), , The process of addition of very small amount of impurity into an intrinsic, semiconductor is called doping. The impurity atoms are called dopants., 3. Write the different methods of doping in a semiconductor. ( O – 09, M-15 ), , There are three different methods of doping a semiconductor., i) The impurity atoms are added to the semiconductor in its molten state., ii) The pure semiconductor is bombarded by ions of impurity atoms., iii) When the semiconductor crystal containing the impurity atoms is heated, the, , www.Padasalai.Net, impurity atoms diffuse into the hot crystal., , 4. What is an extrinsic semiconductor? ( J – 06, J – 08, O-10, M-11 ,O-13,J-14), , An extrinsic semiconductor is one in which an impurity with a valency, higher or lower than the valency of the pure semiconductor is added, so as to, increase the electrical conductivity of the semiconductor., Extrinsic semiconductor can be classified as N-type and P-type., 5. What is rectification? ( M – 07, M – 09,O-14 ), , The process in which alternating voltage or alternating current is converted, into direct voltage or direct current is known as rectification. The device used for, this process is called as rectifier., 6. What is Zener breakdown? ( O – 06, J – 07, M – 08,J-16 ), , When both sides of the PN junction are heavily doped, consequently the, depletion layer is narrow. Zener breakdown takes place in such a thin narrow, junction. When a small reverse bias is applied, a very strong electric field is, produced across the thin depletion layer. This field breaks the covalent bonds,, extremely large number of electrons and holes are produced, which give rise to, the reverse saturation current (Zener current). Zener current is independent of, applied voltage., , 306, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 7. What is a zener diode ? Draw its symbol. ( O – 09, O-10, O – 11), , Zener diode is a reverse biased heavily doped semiconductor (silicon or, germanium) PN junction diode, which is operated exclusively in the breakdown, region., , 8. What is light emitting diode (LED) ? Draw its symbol ? (O-15,O-16 ) Give any, , one of its uses. (M – 07,J-15 ), , A light emitting diode (LED) is a forward biased PN junction diode, which, emits visible light when energized., Uses: LEDs are used for instrument displays, calculators and digital watches., Its symbol :, , 9. Draw the circuit diagram for NPN transistor in common collector ( CC ) mode., , ( M- 08), , www.Padasalai.Net, 10. Draw the circuit diagram for NPN transistor in common emitter ( CE ) mode. (M –, , 06, O – 06 ), , 11. Define input impedance of a transistor in CE mode. ( J – 06, J – 11 ), , The input impedance of the transistor is defined as the ratio of small change, in base – emitter voltage to the corresponding change in base current at a given, VCE, ∆fDò, , ∴Input impedance, ri=5, , 9, , ∆õD fPš, , The input impedance of the transistor in CE mode is very high, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 12. Define output impedance of a transistor. ( O – 08, O – 09 ), , The output impedance is defined as the ratio of variation in the collector, emitter voltage to the corresponding variation in the collector current at a, constant base current in the active region of the transistor characteristic curves., ∆f, , ∴output impedance, ro= 5 ∆õYò 9, Y, , õ, , The output impedance of a transistor in CE mode is low., 13. Define band width an amplifier? [M – 07, O – 11, O – 12,O-16,M-17], , Bandwidth is defined as the frequency interval between lower cut off and, upper cut off frequencies., ∴BW = fU – fL, 14. What are advantages of negative feedback? ( J – 07, J – 08, O – 07, M-11 ,M-, , 13,J-15,J-16), i), ii), iii), iv), v), , Highly stabilised gain., Reduction in the noise level., Increased bandwidth, Less distortion., Increased input impedance and decreased output impedance., , www.Padasalai.Net, 15. What are the Barkhausen conditions for oscillations? (O – 07, O – 08, J – 09, M-, , 10,O-15,J-16 ,O-16), , The gain of the amplifier with positive feedback is given by Af =, , ¡, , #¡—, , This condition means that, i) The loop gain Aβ = 1 and, o, ii) The net phase shift round the loop is 0 or integral multiples of 2π., 16. What are the essential components of an oscillator ? draw its block, , diagram.(M-14), The essential components of an oscillator are;, i) Tank circuit,, ii) Amplifier and, iii) Feedback circuit., Diagram:, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 17. What is an integrated circuit? ( J – 08, J – 09,O-14 ), , An integrated circuit (IC) consists of a single – crystal chip of silicon,, containing both active (diodes and transistors) and passive (resistors, capacitors), elements and their interconnections., 18. Mention any three advantages of integrated circuit. (M – 06, O – 06, J -10, J –, , 11,M-14 ), i), ii), iii), iv), v), vi), , Extremely small in size, Low power consumption, Reliability, Reduced cost, Very small weight, Easy replacement, , 19. Distinguish between analog signal and digital signal(J-14), , Analog signal: The signal current or voltage is in the form of continuous, time, varying voltage or current (sinusoidal). Such signals are called continuous or, analog signals., , www.Padasalai.Net, Digital signal: It has two discrete levels, ‘High’ and ‘Low’. In most cases, the, more positive of the two levels is called HIGH and is also referred to as, logic 1. The other level becomes low and also called logic 0. This method of, using more positive voltage level as logic 1 is called a positive logic system., A voltage 5V refers to logic 1 and 0 V refers to logic 0. On the other hand, in, a negative logic system, the more negative of the two discrete levels is taken as, logic 1 and the other level as logic 0. Both positive and negative logic are used, in digital systems., , 20. Draw circuit diagram for OR gate using diodes. ( J – 06 ), , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 21. Draw circuit diagram for AND gate using diodes and resistors. ( J – 07, M-15), , 22. Give the circuit diagram of NOT gate using a transistor. ( J -10, O – 11,O-, , 13,J-15), , www.Padasalai.Net, 23. State de Morgan’s theorem. ( M – 08, M – 09, J-10,M-13, M-15 )., , First theorem:, “The complement of a sum is equal to the product of the complements.”, If A and B are the inputs, then ¢¢¢¢¢¢¢¢, 7 6 = ̅ . 6¢, Second theorem:, “The complement of a product is equal to the sum of the complements.”, . 6 = ̅ + 6¢, If A and B are the inputs, then ¢¢¢¢¢, , 24. What are universal gates ? Why are they called so ? (O-13), , NAND and NOR gates are called Universal gates because they can perform, all the three basic logic functions (NOT, OR and AND)., 25. Give the important parameters (characteristics) of an operational, , amplifier.(O– 07,M-14 ), , The most important characteristics of OP-AMP are :, i) Very high input impedance or even infinity which produces negligible current, at the inputs,, ii) Very high gain,, iii) Very low output impedance or even zero, so as not to affect the output of, the amplifier by loading., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , JEYAM TUITION CENTRE PARAMAKUDI, , www.TrbTnpsc.com, , 2017-2018, , 26. Draw the circuit diagram of inverting amplifier using operational amplifier, , (M-11), , 27. Draw the circuit diagram of a summing amplifier using an operational, , amplifier. ( M – 09,M-13,M-17 ), , www.Padasalai.Net, 28. Draw the circuit diagram of difference amplifier using operational, , amplifier.(J-11), , 29. Mention any three uses of cathode ray oscilloscope ( J- 09,O-14), , i) It, ii) It, iii) It, iv) It, , is, is, is, is, , used, used, used, used, , to, to, to, to, , measure a.c and d.c voltage., study the waveforms of a.c voltages., find the frequency of a.c voltage., study the beating of heart in cardiology., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , OTHER IMPORTANT ‘3’ MARKS:, 30. Define forbidden energy gap?, , The separation between valence band and conduction band is known as, forbidden energy gap. If an electron is to be transferred from valence band to, conduction band, external energy is required, which is equal to the forbidden, energy gap., 31. Distinguish between N- type and P – type semiconductors., , N – type semiconductor, , P – type semiconductor, When a small amount of trivalent, pentavalent impurity such as, impurity (such as indium, boron or, arsenic is added to a pure, gallium) is added to a pure, germanium semiconductor crystal,, semiconductor crystal, the resulting, the resulting crystal is called Nsemiconductor crystal is called, type semiconductor., P-type semiconductor, The impurity atoms added provide The impurity atoms added create, extra electrons are called donor, valencies of electron (holes) are called, atoms, acceptor atoms, The majority carriers are electrons The majority carriers are holes and, and minority carriers are holes, minority carriers are electrons, The donor energy level is close to The donor energy level is close to the, the conduction band, valence band, The donor energy level is far, The acceptor energy level is far away, away from valence band, from conduction band, , 1. When a small amount of, , 2., , 3., 4., , www.Padasalai.Net, 5., , 32. Define rectifier efficiency., , The ratio of d.c. power output to the a.c. power input is known as rectifier, efficiency., 33. Draw the circuit diagram for NPN transistor in common base ( CB ) mode., , 34. Define current gain., , The current gain is defined as the ratio of a small change in the collector, current to the corresponding change in the base current at a constant VCE, , ∴ current gain, β = §, , ∆õY, , ¨, , ∆õD f, Yò, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 35. Define operating point or quiescent point for the amplifier., , The point of intersection Q of this line in the active region of the output, characteristics with a suitable value of the base current IB, such that the output, voltage is symmetrical is called operating point or quiescent point for the, amplifier., 36. Define lower and upper cut off frequency., , Lower cut off frequency (fL ) is defined as the frequency in the low, frequency range at which the gain of the amplifier is, , √, , times the mid frequency, , gain (AM). Upper cut off frequency (fU) is defined as the frequency in the high, frequency range at which the gain of the amplifier is, times the mid frequency, √, , gain (AM)., , 37. What is meant by feedback ? Name and define different types of it., , Feedback is said to exist in an amplifier circuit, when a fraction of the, output signal is returned or fed back to the input and combined with the input, signal., Two types of feedback 1. negative or degenerative 2. Positive or regenerative, , www.Padasalai.Net, Negative or degenerative: If the magnitude of the input signal is reduced by the, feedback, the feedback is called negative or degenerative., Positive or regenerative: If the magnitude of the input signal is increased by the feedback,, such feedback is called positive or regenerative., , 38. State basic laws of Boolean algebra., , Commutative laws, A+B=B+A, AB = BA, Associative Laws, A + (B + C) = (A + B) + C, A (BC) = (AB) C, Distributive law, A (B+C) = AB + AC, 39. What is an oscillator?, , An oscillator may be defined as an electronic circuit which converts energy, from a d.c. source into a periodically varying output. Oscillators are classified, according to the output voltage, into two types viz. sinusoidal and non-sinusoidal, oscillators., , 313, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , JEYAM TUITION CENTRE PARAMAKUDI, , www.TrbTnpsc.com, , 2017-2018, , 40. What are sinusoidal and non sinusoidal oscillators?, , If the output voltage is a sine wave function of time, the oscillator is said to, be sinusoidal oscillator., If the oscillator generates non-sinusoidal waveform, such as square, rectangular, waves, then it is called as non-sinusoidal oscillator (multi vibrator), 41. What is transistor bias ? And write most commonly used methods of, , transistor biasing., , The proper selection of operating point of a transistor and maintenance of, proper emitter voltage during the passage of the signal is known as transistor, biasing., The most commonly used methods of obtaining transistor biasing are, i) Base bias,, ii) Base bias with emitter feedback,, iii) Base bias with collector feedback and, iv) Voltage divider bias., 42. What the three coupling devices used in multi stage amplifiers., , i) Resistance - capacitance (RC) coupling., ii) Transformer coupling, iii) Direct coupling, , www.Padasalai.Net, 43. Draw the circuit diagram of non inverting amplifier using operational, , amplifier.( J – 11 ), , 44. Determine the output wave form for the circuit given below, if the input, , waveforms are as indicated by A and B., , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 8. Find the output F of the logic circuit given below. [M-09, M-14], , Output of 1st OR gate is A+B, Output of 2nd OR gate is A 7 C, Output of AND gate is (A+B).(A+B), =AA+BA+AC+BC, (∴ AA ( A, , =A(1+B+C) +BC, =A(1) + BC, F = A+BC, , ∴17B7C, , 9. What is the Boolean expression for the logic diagram shown in figure. Evaluate, , its output if A = 1, B = 1 and C = 1.[M-10], , www.Padasalai.Net, Output of AND gate is A.B, Output of NOT gate is C¢, , Output of OR gate is (A.B)+ C¢, , Y = (1 . 1)+ 1¢ = (1 . 1)+0, , Output Y = 1, , 10. Give the Boolean equation for the given logic diagram.[J-11], , Output of NAND gate is ¢¢¢¢¢, A. B, Output of AND gate is C. D, , ¢¢¢¢¢, Output of NOR gate R =¢¢¢¢¢¢¢¢¢¢¢¢¢, A. B 7 C. D, , Applying De Morgan’s first theorem, ¬), =äääää, A. B.¢¢¢¢¢, C. D = A.B .(C¢ +D, ¬), R=AB(Ç¢+÷, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 11. The gain of the amplifier is 100. If 5% of the output voltage is fed back into the input, , through a negative feedback network. Find out the voltage gain after feedback. [M07,O-11], Given data:, , Sol:, , β=5%=, , A = 100;, , Af =, =, , ¡, , E, , F¡—, \, , F, , =, , Af = 16.66, 12., , The gain of an amplifier without feedback is 100 and with positive feedback is, 200, calculate the feedback fraction[M-06], Given data: A = 100;, Af = 200; β = ?, , Sol:, , Af =, , ¡, , #¡—, , www.Padasalai.Net, 200 =, , #, , 200 – 20000 β = 100, , ê, , 20000 β = 100, β=, , β = 0.005, 13., , When the negative feedback is applied to an amplifier of gain 50,the gain after, feedback falls to 25. Calculate the feedback ratio.]J-06,O-09,10,M-10, J-14], , Data : A = 50 ; Af = 25, Solution : Voltage gain after feedback,, Af =, 25 =, , ¡, , F¡—, E, , FE ê, , Hence, the feedback ratio β = 0.02, , 320, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, 20., , Prove the Boolean equation ( A + B ) ( A + C ) = A + BC, , ( M- 11,O-14 ), , Applying the law of distribution on LHS of the equation, we get, (A+B) (A+C) = AA + AC + BA + BC, = A+ AC + AB + BC, = A(1+C+B) + BC [ ∵ 1 + C + B = 1], = A + BC, , 21., , ¬ +B) (A+B) = B (O-13), Prove the following logic expression (å, Using the distributive law twice, and get, , ¬ +B) (A+B) =A, ¬A + A, ¬ B + BA + BB, (A, ¬ B + BA + B, =A, ¬ +A+1), =B(A, , ¬A =0 and BB =B), (sinceA, , ¬ +A+1=1), (since A, , =B, 22. What is the Boolean expression for the logic diagram shown in figure. Evaluate, its output if, A = 1, B = 1 and C = 1, , www.Padasalai.Net, Sol :, , Output of AND gate AB, Output of NOT gate C¢, Output of OR gate is Y = AB+C¢, If A = 1, B = 1 and C = 1, Then Y = 1.1 + 1¢ = 1+0, Y=1, 23. A, , galvanometer of resistance 100 Ω which can measure a maximum current of 1, mA is converted into an ohmmeter by connecting a battery of emf 1 V and a fixed, resistance of 900 Ω in series. When an external resistance is measured the current, reading is 0.1 mA. Calculate the value of the resistance., Data :, Ig = 0.1 mA ; V = 1V ; G = 100 Ω ; R1 = 900 Ω ; G+R1 = 1000 Ω, , Sol :, , R=, , ¯, , Q, , =, , ., , S, , =, , Ó, , ( 104, , R = 10000, , External resistance = R – (G+R1 ) =10000 – 1000 = 9000Ω, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, FIVE MARKS:(1 – question: : Q. No: 61), , 1. Explain the working of a half wave diode rectifier.(M-09,J-11,M-12,M-13,O-13), 2. Explain the working of bridge rectifier. (J-06,M-10), 3. Explain the action of Zener diode as a voltage regulator with a necessary circuit., , (O-12,J-15), 4. Obtain the relation between the current amplification factors, , (M-14, J-07), 5. Define current amplification factors, , α and β of a transistor., , α and β of a transistor. Obtain the relation, , between them.(J-16), 6. Explain the function of a transistor as a switch. (O-07,O-14, M-15), 7. With a neat circuit diagram, explain voltage divider biasing of a transistor. (M-07,J-09), 8. Draw the frequency response curve of a single stage CE amplifier and discuss the, , results. (M-08), 9. What is AND gate? Explain the function of AND gate using electrical circuit and, , www.Padasalai.Net, diodes. (J-08), , 10. State and prove De Morgan’s theorems. (M-06,O-06,O-10,J-13,J-14,M-17), , 11. Explain the pin-out configuration of an operational amplifier with the diagram., , (O-08,O-16), 12. Explain the function of an operational amplifier as a summing amplifier.(J-12), 13. Explain how multimeter is used as ohm meter. (M-11,O-15), , 324, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 326 : www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , TEN MARKS:(1 – question: : Q. No: 69), 1. What is rectification ? with a neat diagram, explain the working of a Bridge rectifier., Draw the input and output signals. (M-06,J - 07 ,M -10, J -10, O - 11, J – 12,M-14,J-16), 2. Explain with neat circuit diagram, the working of single stage CE amplifier. Draw the, frequency response curve and discuss the result. (J - 08, O - 08, M- 11, M – 13,O-16), 3. Explain the output characteristics of NPN transistor connected in common emitter, configuration with the help of a neat circuit diagram. (O -10,O 12), 4. Explain the input characteristics of an NPN transistor connected in common emitter, configuration with the help of a neat circuit diagram.(O-14), 5. Sketch the circuit of a Colpitt’s oscillator and explain its working., (J - 06, O - 06, M - 08, J - 09, J - 11, M – 12,J-14,O-15), 6. What is an operational amplifier? With a circuit diagram, explain the working of an, operational amplifier as a summing amplifier. (M – 07, J-15 ), 7. Explain the action of an operational amplifier as difference amplifier. (O -07), , www.Padasalai.Net, 8. What is meant by feedback? Derive an expression for voltage gain of an amplifier with, negative feedback. (M -09,J-13, M-15,M-17), , 9. What is operational amplifier? Explain its action as (i) inverting amplifier (ii) noninverting amplifier. (O – 09,O-13), , Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], Contact for subject doubt, Email :
[email protected], Phone No: 9952223467, , JEYAM TUITION CENTRE, [PARAMAKUDI], Contact for tuition, Email :
[email protected], Phone No: 9944888512, 325, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 10.COMMUNICATION SYSTEMS, PREPARED BY, J.SHANMUGAVELU M.Sc, B.Ed. [P.G. Assist in Physics], , ONE MARK QUESTIONS: (2 – Questions:), BOOK BACK QUESTIONS:, 1. High frequency waves follow, a) the ground wave propagation, c) ionospheric propagation, , b) the line of sight direction, d) the curvature of the earth, , 2. The main purpose of modulation is to, a) combine two waves of different frequencies, b) acquire wave shaping of the carrier wave, c) transmit low frequency information over long distances efficiently, d) produce side bands, 3. In amplitude modulation, a) the amplitude of the carrier wave varies in accordance with the amplitude of the, modulating signal., b) the amplitude of the carrier wave remains constant, c) the amplitude of the carrier varies in accordance with the frequency of the modulating, signal, d) modulating frequency lies in the audio range, , www.Padasalai.Net, 4. In amplitude modulation, the band width is, a) equal to the signal frequency, b) twice the signal frequency, c) thrice the signal frequency, d) four times the signal frequency, Sol: Band width =(ωc + ωs) - (ωc - ωs), = ωc + ωs - ωc + ωs = 2 ωs, (ωs,fs are signal frequency. So ωs= fs), =2 fs;, 5. In phase modulation, a) only the phase of the carrier wave varies, b) only the frequency of the carrier wave varies., c) both the phase and the frequency of the carrier wave varies., d) there is no change in the frequency and phase of the carrier wave, 6. The RF channel in a radio transmitter produces, a) audio signals, c)both audio signal and high frequency carrier waves, , b) high frequency carrier waves, d)low frequency carrier waves., , 7. The purpose of dividing each frame into two fields so as to transmit 50 views of the, picture per second is, a) to avoid flicker in the picture, b) the fact that handling of higher frequencies is easier, c) that 50 Hz is the power line frequency in India, d) to avoid unwanted noises in the signals, , 326, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 8. Printed documents to be transmitted by fax are converted into electrical signals by the, process of, a) reflection, b) scanning, c) modulation, d) light variation, , PUBLIC QUESTIONS:, 9. In television, blanking pulse is applied to, a) horizontal plates, b) vertical plates, , c)control grid, , d)filament, , 10. In a AM superheterodyne receiver, the local oscillator frequency is 1.245MHz. The tuned, station frequency is, a) 455 kHz, b) 790 kHz, c)690kHz, d)990kHz, Sol:Tuned station frequency = local oscillator frequency – intermediate frequency, , = 1.245 MHz – 455 kHz, =1245 kHz – 455 kHz =790 kHz, 11. The radio waves after refraction from different parts of ionosphere on reaching the earth, are called as, a) ground waves, b) sky waves, c) space waves, d)microwaves, 12. The principle used for transmission of light signals through optical fibre is, a) refraction, b) diffraction, c) polarization, d)total internal reflection, , www.Padasalai.Net, 13. Digital signals are converted into analog signals using, a) FAX, b) modem, c)cable, , d)coaxial cable, , 14. In interlaced scanning time taken to scan one line is, a) 20 ms, b) 64μs, c) 50 ms, , d) 100 μs, , 15. The first man – made satellite is, a) Aryabhatta, b) Sputnik, , c) Venera, , d) Rohini, , 16. The audio frequency range is, a) 20 Hz – 200000 Hz, c) 20 Hz to 2000000 Hz, , b) 20 Hz – 2000 Hz, d) 20 Hz to 20000 Hz, , 17. Skip distance is the shortest distance between, a) The point of transmission and the point of reception, b) The uplink station and the downlink station, c) The transmitter and the target, d) The receiver and the target, 18. An FM signal has a resting frequency of 105 MHz and highest frequency of 105.03 MHz ,, when modulated by a signal. Then the carrier swing is, a) 0.03 MHz, b) 0.06 MHz, c) 0.03 kHz, d) 60 MHz, Sol: frequency deviation ∆f = 105 MHz – 105.03MHz, , ∆f = 0.03MHz., , carrier swing = 2, , =2, , 327, , J.SHANMUGAVELU, , ∆f, , 0.03 = 0.06 MHz, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , 2017-2018, , JEYAM TUITION CENTRE PARAMAKUDI, , 19. In the AM superheterodyne receiver system the value of the intermediate frequency is, equal to, a) 445 kHz, b) 455 kHz, c) 485 kHz, d) 465 kHz, 20. In an AM receiver, the local oscillator frequency is 2750kHz. The tuned station frequency, is, a) 2905 kHz, b) 2295 kHz, c) 3055 kHz, d) 2250 kHz, Sol:Tuned station frequency = local oscillator frequency – intermediate frequency, , = 2750kHz – 455 kHz, = 2295 kHz, 21. For FM receivers, the intermediate frequency is, a) 455 kHz, b) 455 MHz, c) 10.7 kHz, , d) 10.7 MHz, , 22. intermediate frequency in FM receivers, a) 455 kHz, b) 10.7 MHz, , d) 22 MHz, , c) 40 MHz, , 23. In AM receiver, if 900 kHz station is tuned then the local oscillator will have to produce a, frequency of, a) 600 kHz, b) 455 kHz, c) 10.7MHz, d) 1355 kHz, sol: local oscillator frequency = Tuned station frequency + intermediate frequency, , = 900 kHz + 455 kHz, , www.Padasalai.Net, = 1355 kHz, , 24. The maximum carrier swing allowed in frequency modulation is :, a) 455 kHz, b) 10. 7 MHz, c) 75 kHz, d) 150 kHz, Sol: frequency deviation ∆f = 75 kHz, , carrier swing = 2, , =2, , ∆f, , 75 = 150 kHz, , 25. The resting frequency of FM transmitter is 98.5 MHz. The allowed minimum and, maximum frequency on either side of the centre frequency are respectively:, (a) 98.400 MHz and 98.600 MHz, (b) 98.450 MHz and 98.550 MHz, (c) 98.425 MHz and 98.575 MHz, (d) 98 MHz and 99 MHz, Sol: Resting frequency of FM transmitter = 98.5 MHz, , frequency deviation ∆f = 75 kHz = 0.075 MHz, , minimum and maximum frequency on either side of the centre frequency, = (98.5 MHz - 0.075 MHz) And (98.5 MHz + 0.075 MHz), = 98.425 MHz and 98.575 MHz, , 328, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , Choose the correct answer from the option given, 1) High frequency waves follow __________, (a) the ground wave propagation, (b) the line of sight direction, (d) the curvature of the earth, (b) ionosphere propagation, 2) The main purpose of modulation is to ___________, (a) combine two waves of different frequencies, (b) acquire wave shaping of the carrier wave, (c) transmit low frequency information over long distances efficiently, (d) produce side bands, 3) In amplitude modulation, the bandwidth is ___________, (a) equal to the signal frequency, (b) twice the signal frequency, (c) thrice the signal frequency, (d) four times the signal frequency, 4) In phase modulation ___________, (a) only the phase of the carrier wave varies, (b) only the frequency of the carrier wave varies, (c) both phase and frequency of the carrier wave varies, (d) there is no change in frequency and phase of the carrier wave, , www.Padasalai.Net, 5) Propagation of electromagnetic wave depends on, (a) nature of wave, (b) environment, (c) medium, (d) both (a) & (b), , 6) Communication refers to ______________, (a) Sending information, (b) receiving the information, (c) processing the information, (d) sending, receiving and processing of information electronically, 7) Ground wave propagation takes place ______________, (a) when the transmitting antenna is close to the ground, (b) when the receiving antenna is close to the ground, (c) when the transmitting and receiving antennas are far off from the ground, (d) when the transmitting and receiving antennas close to the ground, 8) Ground wave propagation is of prime importance for _____________, (a) short wave signals only, (b) long wave signals only, (c) medium wave signals only, (d) medium and long wave signals, , 329, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 9) Space wave propagation is particularly suitable for the waves having frequency, (a) above 40 MHz, (b) below 30 MHz, (c) below 20 MHz, (d) above 30 MHz, 10) The mechanism involved in sky wave propagation is ___________, (a) reflection, (b) refraction, (c) interference, (d) polarization, 11) Long distance radio communication is possible through the ______________, (a) ground wave propagation, (b) surface wave propagation, (c) the sky wave propagation, (d) all the above, 12) The refractive indices of the various layers in the ionosphere varies with respect, to____________, (a) electron density only, (b) frequency of the incident wave only, (c) intensity of the incident wave only, (d) electron density and the frequency of the incident wave, 13) As the ionization density increases for a wave approaching the given layer at an angle,, the refractive index of the layer is ____________, (a) increased, (b) reduced, (c) increased or reduced, (d) constant, , www.Padasalai.Net, 14) The music, speech etc., are converted into audio signals using a ___________, (a) loud speaker, (b) photocell, (c) diode, (d) microphone, 15) The audio frequency range is ___________, (a) 20 Hz to 200000 Hz, (b) 20 Hz to 2000 Hz, (c) 20 Hz to 2000000 Hz, (d) 20 Hz to 20000 Hz, 16) The radiation of electrical energy is practicable only at _____________, (a) low frequencies, (b) very low frequencies, (c) moderate frequencies, (d) high frequencies, 17) Which signals can be sent through thousands of kilometers with comparatively small, power ?, (a) audio signals, (b) video signals, (c) high frequency signals, (d) low frequency signals, 18) In amplitude modulation, which is changed in accordance with the intensity of the, signal?, (a) frequency of the carrier wave, (b) phase of the carrier wave, (c) amplitude of the carrier wave, (d) both frequency and phase of the carrier wave, , 330, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 19) Factor that determine the strength and quality of the transmitted signal is, (a) Q – factor, (b) frequency of the carrier wave, (c) frequency of the modulating wave, (d) modulation factor, 20) For effective modulation, the degree of modulation should ____________, (a) exceed 100 %, (b) exceed 200 %, (c) never exceed 50 %, (d) never exceed 100 %, 21) A carrier wave of amplitude 10 mV is modulated by a sinusoidal audio signal wave of, amplitude 6 mV, the modulation factor is _____________, (a) 0.6, (b) 6, (c) 60, (d) 0.06, 22) A 5MHz sinusoidal carrier wave of amplitude 10 mV is modulated by a 5 kHz, sinusoidal audio signal wave of amplitude 6 mV. Find the lower and upper side band, frequencies ____________, (a) 4.995 MHz, 5.005 MHz, (b) 9.995 MHz, 10.005 MHz, (c) 4.5 MHz, 5.5 MHz, (d) 10 MHz, 15MHz, 23) The magnitude of both the upper and lower side bands is ___________, (a) 2 times the carrier amplitude Ec, (b) 1/2 times the carrier amplitude Ec, (c) ’m’ times the carrier amplitude Ec, (d) m/2 times the carrier amplitude Ec, , www.Padasalai.Net, 24) If the modulation factor ‘m’ is equal to unity, then each side band has amplitude equal, to ____________, (a) 2 times the carrier amplitude, (b) √2 times the carrier amplitude, (c) 1/√2 times the carrier amplitude, (d) half of the carrier amplitude, 25) The human voice or music contains waves with a frequency range of __________, (a) 3 – 30 Hz, (b) 30 – 300 Hz, (c) 3000 – 30000 Hz, (d) 300 – 3000 Hz, 26) Which modulation facilitates highest transmission speeds on a given bandwidth?, (a) amplitude modulation, (b) frequency modulation, (c) Phase modulation, (d) all the above, 27) For the purpose of coupling the transmitter and the receiver to the space link, We, use____________, (a) Amplifier, (b) oscillator, (c) Antenna, (d) FAX, , 331, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 28) Transmitting antenna converts the ___________, (a) electrical signal into electrical energy, (b) electrical signal into magnetic energy, (c) electrical signal into electromagnetic energy, (d) electromagnetic energy into electrical signal, 29) Receiving antenna converts the ___________, (a) electrical signal into electromagnetic energy, (b) electrical signal into electrical energy, (c) electromagnetic energy into electrical signal, (d) electromagnetic energy into magnetic signal, 30) In FM broadcast, the frequency deviation of sound signal is ___________, (a) 25 kHz, (b) 50 kHz, (c) 75 kHz, (d) 100 kHz, 31) In TV transmission sound signals are ____________, (a) amplitude modulated, (b) frequency modulated, (c) phase modulated, (d) none of the above, 32) Vidicon camera tube works on the principle of __________, (a) photoconductivity, (b) thermoelectric effect, (c) thermionic emission, (d) Seebeck effect, , www.Padasalai.Net, 33) When exposed to light, the resistance of the photo conductive material, (a) decreases, (b) increases, (c) increases or decreases, (d) is not altered, , 34) In Vidicon camera tube the front face of the target plate is coated with ________, (a) antimony tri sulphide, (b) aluminium oxide, (c) zinc sulfide, (d) tin oxide, 35) The frequency of scanning is _____________, (a) 20 per second, (b) 50 per second, (c) 100 per second, , (d) 25 per second, , 36) How many synchronizing pulses are used for scanning?, (a) one, (b) two, (c) three, , (d) four, , 37) Blanking pulse used for TV scanning is, (a) high frequency saw tooth potential, (b) low frequency saw tooth potential, (c) high positive potential, (d) high negative potential, 38) In TV scanning, blanking pulse is applied to, (a) horizontal deflector plates, (b) vertical deflector plates, (c) control grid, (d) filament, , 332, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 39) In interlaced scanning, the vertical scanning frequency is _________, (a) 10 fields per second, (b) 25 fields per second, (c) 50 fields per second, (d) 100 fields per second, 40) In radar receiver, the returning echo pulse appears slightly displaced from the, transmitted pulse which measures _____________ of the target, (a) nature, (b) shape, (c) power, (d) range, 41) A signal which can take any value with in the given range, (a) Digital signal, (b) analog signal, (c) AM signal, , (d) FM signal, , 42) In TV transmission, the picture should not be scanned during the return journey of, the scanning. This is done by ______________, (a) vertical scanning pulse, (b) horizontal scanning pulse, (c) blanking pulse, (d) triggering pulse, 43) The greatest technical problem is analog communication, (a) noise, (b) nature of signal, (c) wider band width, (d) power of system, 44) Any form of information, that has been put into digital form is called, (a) signal, (b) amplitude, (c) power, (d) data, , www.Padasalai.Net, 45) In twisted pair cable, wire is twisted to, (a) decreasing external noise, (b) speedy data transfer, (c) increasing external noise, (d) both (a) & (b), , 46) Fax machine cannot be used for transmitting, (a) sound messages, (b) live scenes and motion, (c) either (a) & (b), (d) both (a) & (b), 47) A modem is used for ______________, (a) modulation only, (b) demodulation only, (c) modulation and demodulation, (d) printing the information, 48) Optical fiber works on the principle of ____________, (a) total internal reflection, (b) refraction, (c) reflection, (d) polarization, 49) Satellite orbiting the earth will be geo-stationary when it is at a height of, (a) 36,000 km from the earth, (b) 3600 km from the earth, (c) 360 km from the earth, (d) 36 km from the earth, , 333, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 50) For uplink transmission commercial communication satellites use, (a) 5 MHz bandwidth near 6 GHz, (b) 50 MHz bandwidth near 6 GHz, (c) 500 MHz bandwidth near 5 GHz, (d) 500 MHz bandwidth near 6 GHz, 51) For downlink transmission commercial communication satellites use, (a) 500 MHz bandwidth near 4 GHz, (b) 50 MHz bandwidth near 4 GHz, (c) 500 MHz bandwidth near 5 GHz, (d) 5 MHz bandwidth near 4 GHz, 52) In actual practice the band width used for uplink transmission by the satellite is, (a) 5.725 – 7.075 GHz, (b) 3.4 – 4.8 GHz, (c) 6.725 – 7.075 GHz, (d) 5.725 – 6.075 GHz, , ‘3’ MARK QUESTIONS: (1 – Question: Q. N.o: 50), , www.Padasalai.Net, PUBLIC ‘3’ MARKS:, 1., , What are the different types of radiowave propagation (O – 11,M-17), i) Ground (surface) wave propagation, ii) Space wave propagation, iii) Sky wave (or) ionospheric propagation, , 2. What is meant by skip distance? (J–0,M–08,J – 09,O – 10 ,M – 11,O-13,M-14,O-, , 15), , In the sky wave propagation, for a fixed frequency, the shortest distance, between the point of transmission and the point of reception along the surface is, known as the skip distance., 3. What is meant by skip zone ( J-14 ), , The region between the point where there is no reception of ground waves, and the point where the sky wave is received first is known as skip zone. In the, skip zone, there is no reception at all., , 334, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 4. What is the necessity of modulation? ( O – 07 ), , In radio broadcasting, it is necessary to send audio frequency signal, (eg. music, speech etc.) from a broadcasting station over great distances to a, receiver. The music, speech etc., are converted into audio signals using a, microphone. The energy of a wave increases with frequency. So, the audio, frequency (20 – 20000 Hz) is not having large amount of energy and cannot be, sent over long distances. The radiation of electrical energy is practicable only at, high frequencies e.g. above 20 kHz. The high frequency signals can be sent, through thousands of kilometres with comparatively small power., 5. Define: modulation factor in amplitude modulation. (M– 06, J – 08, O-09, J– 11,, , M– 10,M-15 ), Modulation factor defined as the ratio of the change of amplitude in carrier, wave after modulation to the amplitude of the unmodulated carrier wave., modulation factor,, Ükù°%± ³ ,´4²ø -Ñ ,466% 6 d4T 4ѱ 6 k-³ °4±%-², m=, Ükù°%± ³ -Ñ ,466% 6 d4T, Ñ-6 k-³ °4±%-², , 6. What are the advantages and disadvantages of frequency modulation? ( M –, , 09 ), , www.Padasalai.Net, Advantage:, i) It gives noiseless reception. Noise is a form of amplitude variation and a FM, receiver will reject such noise signals., ii) The operating range is quite large., iii) The efficiency of transmission is very high., Disadvantages:, i) A much wider channel is required by FM., ii) FM transmitting and receiving equipments tends to be more complex., 7. What are the advantages and disadvantages of digital communication ? (M–, , 06,07,J–06,08,09,10,M-13 ) [3 OR 5 MARK QUESTION], Advantages:, i) The transmission quality is high and almost independent of the distance, between the terminals., ii) The capacity of the transmission system can be increased., iii) The newer types of transmission media such as light beams in optical fibers, and wave guides operating in the microwave frequency extensively use digital, communication., Disadvantages, i) A digital system requires larger bandwidth., ii) It is very difficult to gradually change over from analog to digital, transmission., , 335, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 8. What is FAX ? Mention its uses. ( O – 06,O-14 ), , Fax (or) Facsimile, is an electronic system for transmitting graphical, information by wire or radio. It is used to send printed material by scanning and, converting it into electronic signals. These signals modulate a carrier to be, transmitted over the telephone lines. Since modulation is involved, fax, transmission can also take place by radio., 9. Give any three advantages of fiber optical communication. ( O – 08,J-15 ), , i) Transmission loss is low., ii) Fiber is lighter and less bulky than equivalent copper cable., iii) More information can be carried by each fiber than by equivalent copper, cables., iv) There is no interference in the transmission of light from electrical, disturbances or electrical noise., 10. Write any three applications of RADAR. ( J – 07 ), , i) Air and sea navigation is made entirely safe, with radar installations., ii) Radar systems are used for the safe landing of air crafts., iii) The pulses can be used for discovering the position of buried metals, oils and, ores., , www.Padasalai.Net, 11. Define channel width (band width), , The channel width is given by the difference between extreme frequencies i.e., between maximum frequency of USB and minimum frequency of LSB., ∴ Channel width = 2 × maximum frequency of the modulating signal, = 2 × (fs)max, , 12. Define amplitude modulation (AM).[ M – 12], , When the amplitude of high frequency carrier wave is changed in accordance, with the intensity of the signal, the process is called amplitude modulation., 13. What are the advantage and limitations of amplitude modulations, , Advantages:, i) Easy transmission and reception, ii) Lesser band width requirements, iii) Low cost, Limitations:, i) Noisy reception, ii) Low efficiency, iii) Small operating range, , 336, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 14. Define Frequency modulation (FM)., , When the frequency of carrier wave is, intensity of the signal, the process is called, In frequency modulation, the amplitude and, constant. Only, the frequency of the carrier, the signal., , changed in accordance with the, frequency modulation., phase of the carrier wave remains, wave is changed in accordance with, , 15. Define carrier swing, , The total variation in frequency from the lowest to the highest is called, carrier swing (CS). Hence, Carrier swing = 2 × frequency deviation, , = 2 × Δf, 16. Define Phase modulation (PM)., , In phase modulation, the phase of the carrier wave is varied in accordance, with the amplitude of the modulating signal and the rate of variation is, proportional to the signal frequency, 17. What is an optical fiber ? What are the applications of fiber optical, , www.Padasalai.Net, communication?, , An optical fiber is a thin transparent rod, usually made of glass or plastic,, through which light can propagate. The light signals travel through the rod from, the transmitter to the receiver and can be easily detected at the receiving end of, the optical fiber. The principle of total internal reflection is used for the, transmission of light signals through the optical fiber., Applications:, The various applications of fiber in communication area are, voice, telephones, video phones, message services, data network etc., , 18. Write any three merits of Satellite communication. (J-16, O-16), , i) Mobile communication can be easily established by satellite communication., ii) Satellite communication is economical compared with terrestrial communication, particularly where long distances are involved., iii) Compared to the optical fiber communication, satellite communication has the, advantages that, quality of transmitted signal and location of sending and, receiving stations are independent of distance., iv) For thin traffic remote areas like north east regions in India, Ladakh etc.,, satellite communication is most economical., v) For search, rescue and navigation, satellite communication is far superior and, economical compared to other systems., , 337, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , OTHER IMPORTANT ‘3’ MARKS:, , 19. What is scanning?, , Scanning is the process by which an electron beam spot is made to move, across a rectangular area, so as to cover it completely. This rectangular area may, be the target surface in a television camera or the screen of a picture tube in a, television receiver., 20. What are the uses of modem?, , A modem is used to convert digital signals into analog signals capable of, being transmitted over telephone lines. At the receiving end of the system,, modem is used to demodulate the analog signals and reconstruct the equivalent, digital output., 21. State the demerits of Satellite communication., , i) Between talks there is a time gap which becomes quite annoying. This time, delay also reduces the efficiency of satellite in data transmission., , www.Padasalai.Net, ii) An imperfect impedance match may cause echo, received back after a delay., Echo suppressor has to be used., iii) Repair of satellite is almost impossible, once it has been launched., , 338, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
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www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , PUBLIC ‘5’ MARK PROBLEMS:, 1. A 10 MHz sinusoidal carrier wave of amplitude 10 mV is modulated by a 5 kHz, , sinusoidal audio signal wave of amplitude 6 mV. Find the frequency components of, the resultant modulated wave and their amplitude.[M-11,M-17]., , www.Padasalai.Net, FIVE MARKS:(1-question: Q. No: 62), 1. What are the advantages and disadvantages of digital communication., (M-06,J-06,J-08,J-09,J-12), 2. Explain the space wave propagation of radio waves. (O-06), 3. Mention the principle of radar and write its applications., (M-07,M-08,M-10,M-12,M-13), 4. Write a short notes on optical fiber communication and give its uses.(J-07,O-11), 5. Draw the functional block diagram of amplitude modulated radio transmitter., (O-07,O-10,M-14,O-15), 6. Explain the function of a frequency modulated radio transmitter with a block diagram., (O-08,O-09,J-10,J-11,O-12,J-14,J-16), 7. What are the advantages and disadvantages of frequency modulation? ( O-16), 8. With the help of a block diagram, explain the operation of FM superheterodyne, receiver. (M-09, M-15), , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
Page 341 : www.Padasalai.Net, , www.TrbTnpsc.com, , JEYAM TUITION CENTRE PARAMAKUDI, , 2017-2018, , 9. What are the merits and demerits of satellite communication?(J-13,O-13,J-15), 10. Draw the block diagram of a simple RADAR system.(O-14), , TEN MARKS:(1-question: Q. No: 70), 1. Explain the analysis of amplitude modulated wave. Plot the frequency spectrum and, band width. (J - 06, O - 06, M - 08, M - 09, J - 10, O - 10,J – 12,M-13,J-14,J-16), 2. With the help of a functional block diagram. Explain the operation of a super heterodyne, AM receiver. (J - 07, M - 11, M -12,O-13,O-16), 3. With the help of a functional block diagram, explain the function of a monochrome TV, receiver. (M - 06, M - 07, M -10, J - 11, J – 13,O-14,O-15), 4. With the help of block diagram, explain the function of various units in the monochrome, television transmitter. (J - 08, O - 08, O - 09), 5. Explain the construction and working of a vidicon camera tube with neat diagram., (J – 09,M-14), 6. With the help of block diagram, explain the function of RADAR system., , (OR), , Explain transmission and reception of RADAR with a block diagram., (O - 07, O – 12,M -15 ), , www.Padasalai.Net, 7. Explain the principle and working of RADAR with neat block diagram. (O -11, M-17), , E ß Ú õÀ • i ² ®, E ß Ú õÀ © m k ÷ © • i ² ®, E ß Ú õÀ • i ¯ õu x J ß Ö ª À ø », , Prepared by, J. SHANMUGAVELU [P.G.Assist. in Physics], Contact for subject doubt, Email :
[email protected], Phone No: 9952223467, , JEYAM TUITION CENTRE, [PARAMAKUDI], Contact for tuition, Email :
[email protected], Phone No: 9944888512, 340, , J.SHANMUGAVELU, , [P.G. T. in Physics], , Ph. No:9952223467, , http://www.trbtnpsc.com/2018/02/latest-12th-study-materials-2018.html
