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https://t.me/NEET_StudyMaterial, Chapter, , 2, , Units and Measurements, Solutions, SECTION - A, Objective Type Questions, 1., , The base quantity among the following is, (1) Speed, , (2) Weight, , (3) Length, , (4) Area, , Sol. Answer (3), There are seven base quantities,, (i) Mass, , (ii) Length, , (iii) Time, , (iv) Current, , (v) Amount of substance, , (vi) Luminous intensity, , (vii) Temperature, 2., , Which of the following is not a unit of time?, (1) Second, , (2) Minute, , (3) Hour, , (4) Light year, , (3) 3 × 108 m, , (4) 3.08 × 1016 m, , Sol. Answer (4), Light year is the unit of distance, 1 light year = 9.46 × 1015 m, 3., , One astronomical unit is a distance equal to, (1) 9.46 × 1015 m, , (2) 1.496 × 1011 m, , Sol. Answer (2), One astronomical unit is the average distance between earth and sun, 1 astronomical unit (AU)  1.496  1011 m, , 4., , The volume of a cube having sides 1.2 m is appropriately expressed as, (1) 1.728 × 106 cm3, , (2) 1.7 × 106 cm3, , (3) 1.8 × 106 cm3, , (4) 1.73 ×106 cm3, , Sol. Answer (2), The volume of cube is l3, , v  (1.2 cm)3  1.728  106 cm3, v  1.7  106 cm3, Answer should be reported in minimum number of significant figures., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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10, 5., , Units and Measurements, , Solution of Assignment, , Ampere second is a unit of, (1) Current, , (2) Charge, , (3) Energy, , (4) Power, , Sol. Answer (2), Current I , , q, ⇒ q  It, t, , q  Ampere second, , So, ampere second is the unit of charge., 6., , The most precise reading of the mass of an object, among the following is, (1) 20 g, , (2) 20.0 g, , (3) 20.01 g, , (4) 20 × 100 g, , Sol. Answer (3), A measurement having more number of decimal places is the one with the most precision., So, 20.01 g is most precise., 7., , The most accurate reading of the length of a 6.28 cm long fibre is, (1) 6 cm, , (2) 6.5 cm, , (3) 5.99 cm, , (4) 6.0 cm, , (3) nm, , (4) N s, , Sol. Answer (2), Most accurate reading is the one having minimum error., So, 16 – 6.281 = 0.28 cm, 16.5 – 6.281 = 0.22 cm, 15.99 – 6.281 = 0.29 cm, 16.0 – 6.281 = 0.28 cm, So, second reading is most accurate., 8., , Which of the following is a unit that of force?, (1) N m, , (2) mN, , Sol. Answer (2), Nm  Unit of torque, mN  Milli newton  10–3 N, nm  Nano metre, Ns  Unit of momentum, 9., , The value of 60° in radian is, (1), , , 2, , (2), , , 3, , (3), , , 4, , (4), , , 5, , Sol. Answer (2), 180º =  radian, 1º =, , , rad, 180, , 60º =, , 60º , , ,  60 rad, 180, , , rad, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 11, , 10. The total plane angle subtended by a circle at its centre is, (1) rad, , (2) 2 rad, , (3), , 2, rad, 3, , (4), , , rad, 2, , Sol. Answer (2), The total plane angle is 360º or 2 rad., 11. A far off planet is estimated to be at a distance D from the earth. If its diametrically opposite extremes subtend, an angle  at an observatory situated on the earth, the approximate diameter of the planet is, , D, , (1), , (2), , D, , , (3) D, , (4), , 1, D, , Sol. Answer (3), , , Arc length, Radius, , , D, , D, , d, , D, , , d  D, , d, , 12. One unified atomic mass unit represents a mass of magnitude, (1) 10–30 kg, , (2) 1.66 × 1027 kg, , (3) 1.66 × 10–27kg, , (4) 1030 kg, , Sol. Answer (3), 1 amu  1.66  10 27 kg, , 13. If the average life of a person is taken as 100 s, the age of the universe on this scale is of the order, (1) 1010 s, , (2) 108 s, , (3) 1017 s, , (4) 109 s, , Sol. Answer (1), Time span of human life = 109 s, Age of universe = 1017 s, So,, , If,, , , Age of universe 1017, ,  108, Time of human, 109, , Age of universe,  108, 100, Age of universe  1010 s, , 14. Which of the following is the most precise measurement?, (1) 3 × 10–3 m, , (2) 0.0030 m, , (3) 30 × 10–4 m, , (4) 300 × 10–5 m, , Sol. Answer (4), 3 × 10–3, 3.0 × 10–3, 3.0 × 10–3, 3.00 × 10–3, So, fourth measurement is most precise., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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12, , Units and Measurements, , Solution of Assignment, , 15. The number of significant figures in a pure number 410 is, (1) Two, , (2) Three, , (3) One, , (4) Infinite, , Sol. Answer (4), A pure number has infinite number of significant figures., 16. Thickness of a pencil measured by using a screw gauge (least count .001 cm) comes out to be 0.802 cm. The, percentage error in the measurement is, (1) 0.125%, , (2) 2.43%, , (3) 4.12%, , (4) 2.14%, , Sol. Answer (1), The percentage error is, , L, 0.001,  100% =,  100% = 0.1246%  0.125%, L, 0.802, , 17. The percentage error in the measurement of the voltage V is 3% and in the measurement of the current is, 2%. The percentage error in the measurement of the resistance is, (1) 3%, , (2) 2%, , (3) 1%, , (4) 5%, , Sol. Answer (4), V = IR  R , , V, I, , ⎛ R ⎞, ⎛ V I ⎞,  ⎟  100%, ⎟⎠  100%  ⎜⎝,  ⎜⎝, R, V, I ⎠, , , R,  100%  3%  2% = 5%, R, , 18. The relative error in the measurement of the side of a cube is 0.027. The relative error in the measurement of, its volume is, (1) 0.027, , (2) 0.054, , (3) 0.081, , (4) 0.046, , (3) Least count error, , (4) Personal error, , Sol. Answer (3), Volume of cube, V = side3, , V 3 side, , V, side, V, V,  0.081,  3  0.027 , V, V, , 19. Zero error in an instrument introduces, (1) Systematic error, , (2) Random error, , Sol. Answer (1), Zero error is a part of systematic error., 20. A packet contains silver powder of mass 20.23 g ± 0.01 g. Some of the powder of mass 5.75 g ± 0.01 g is, taken out from it. The mass of the powder left back is, (1) 14.48 g ± 0.00 g, , (2) 14.48 ± 0.02 g, , (3) 14.5 g ± 0.1 g, , (4) 14.5 g ± 0.2 g, , Sol. Answer (2), m1 = 20.23 g ± 0.01 g, m2 = (5.75 ± 0.01) g, m1 – m2 = [(20.23 – 5.75) ± 0.02] g, m  (14.48  0.02) g, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 13, , 21. The addition of three masses 1.6 g, 7.32 g and 4.238 g, addressed upto proper decimal places is, (1) 13.158 g, , (2) 13.2 g, , (3) 13.16 g, , (4) 13.15 g, , Sol. Answer (2), m1 = 1.6 g, m2 = 7.32 g, m3 = 4.238 g, m1 + m2 + m3 = 13.158 g, but answer should be reported in one decimal place only., , , m  13.2 g, , 22. The area of a sheet of length 10.2 cm and width 6.8 cm addressed upto proper number of significant figures, is, (1) 69.36 cm2, , (2) 69.4 cm2, , (3) 69 cm2, , (4) 70 cm2, , Sol. Answer (3), l = 10.2 cm, w = 6.8 cm, Area = lw = 10.2 × 6.8 = 69.36,  Area = 69 cm2, 23. The radius of a sphere is (2.6 ± 0.1) cm. The percentage error in its volume is, (1), , 0.1,  100 %, 2.6, , (2) 3 ×, , 0.1,  100 %, 2.6, , (3), , 0.1,  100 %, 3  2.6, , (4), , 0.1, %, 2.6, , Sol. Answer (2), r = (2.6 ± 0.1) cm, , V , , 4 3, r, 3, , V, 3 r,  100% ,  100%, V, r, , V, 3  0.1,  100% ,  100%, V, 2.6, 24. The uncertain digit in the measurement of a length reported as 41.68 cm is, (1) 4, , (2) 1, , (3) 6, , (4) 8, , Sol. Answer (4), 41.68 cm, The rightmost digit is most insignificant and leftmost is most significant., So, 8  most insignificant, 4  most significant, 25. We can reduce random errors by, (1) Taking large number of observations, , (2) Corrected zero error, , (3) By following proper technique of experiment, , (4) Both (1) & (3), , Sol. Answer (1), The only method of reducing random errors is by taking more and more number of observations., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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14, , Units and Measurements, , Solution of Assignment, , 26. The number of significant figures in the measured value 0.0204 is, (1) Five, , (2) Three, , (3) Four, , (4) Two, , Sol. Answer (2), The non-zero digits after the decimal places are significant., 27. The number of significant figures in the measured value 26000 is, (1) Five, , (2) Two, , (3) Three, , (4) Infinite, , Sol. Answer (2), The trailing zeros are not significant., So, only two digits are significant., 28. The number of significant zeroes present in the measured value 0.020040, is, (1) Five, , (2) Two, , (3) One, , (4) Three, , Sol. Answer (4), Zeores appearing between and after non-zero numbers are significant., 0.020040, , 29. The number of significant figures in the measured value 4.700 m is the same as that in the value, (1) 4700 m, , (2) 0.047 m, , (3) 4070 m, , (4) 470.0 m, , Sol. Answer (4), 4.700  Four significant figures., Also, 470.0 m  Four significant figures., 30. If a calculated value 2.7465 g contains only three significant figures, the two insignificant digits in it are, (1) 2 and 7, , (2) 7 and 4, , (3) 6 and 5, , (4) 4 and 6, , Sol. Answer (3), 2.7465 g  Last two digits are most insignificant., 31. An object of mass 4.237 g occupies a volume 1.72 cm3. The density of the object to appropriate significant, figures is, (1) 2.46 g cm–3, , (2) 2.463 g cm–3, , (3) 2.5 g cm–3, , (4) 2.50 g cm–3, , (3) 2.80, , (4) 2.83, , Sol. Answer (1), m = 4.237 g, V = 1.72 cm3, Density =, , Mass, 4.237 g, , Volume 1.72 cm3, ,  d  2.46 gcm 3, 32. Round off the value 2.845 to three significant figures., (1) 2.85, , (2) 2.84, , Sol. Answer (2), 2.845  2.84, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 15, , 33. A length 5.997 m rounded off to three significant figures is written as, (1) 6.00 m, , (2) 5.99 m, , (3) 5.95 m, , (4) 5.90 m, , Sol. Answer (1), 5.997 6.00 m, 34. The order of the magnitude of speed of light in SI unit is, (1) 16, , (2) 8, , (3) 4, , (4) 7, , Sol. Answer (2), Speed of light 3 × 108 ms–1, Order of magnitude = 8, 35. The values of a number of quantities are used in a mathematical formula. The quantity that should be most, precise and accurate in measurement is the one, (1) Having smallest magnitude, , (2) Having largest magnitude, , (3) Used in the numerator, , (4) Used in the denominator, , Sol. Answer (1), The quantity having smallest magnitude should be measured very precisely as it is likely to contribute the, maximum relative error., 36. What are the dimensions of the change in velocity?, (1) [M0L0T0], , (2) [LT–1], , (3) [MLT–1], , (4) [LT–2], , Sol. Answer (2), The dimensions of change in velocity is same as that of velocity [M0LT–1]., 37. The dimensional formula for energy is, (1) [MLT–2], , (2) [ML2T–2], , (3) [M–1L2T], , (4) [M L2 T], , Sol. Answer (2), The dimensional formula is [ML2T–2], 38. The pair of the quantities having same dimensions is, (1) Displacement, velocity, , (2) Time, frequency, , (3) Wavelength, focal length, , (4) Force, acceleration, , Sol. Answer (3), Wavelength and focal length both are have units of length., 39. The dimensional formula for relative refractive index is, (1) [M1L1 T1], , (2) [M0L0T0], , (3) [M1L0 T 0], , (4) [MLT–1], , Sol. Answer (2), Refractive index is a pure number, hence dimensionless., 40. The dimensional formula [ML–1T–2] is for the quantity, (1) Force, , (2) Acceleration, , (3) Pressure, , (4) Work, , Sol. Answer (3), The dimensional formula for pressure, P, , Force MLT 2, , ⇒ [ML1T 2 ], 2, Area, L, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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16, , Units and Measurements, , Solution of Assignment, , 41. If the buoyant force F acting on an object depends on its volume V immersed in a liquid, the density  of the, liquid and the acceleration due to gravity g. The correct expression for F can be, (1) V g, , (2), , g, V, , (3)  gV2, , (4), , gV, , Sol. Answer (1), , F  V a b g c, F  [L3 ]a [ML3 ]b [LT 2 ]c, [MLT 2 ]  F  [MbL3a 3b  c T 2c ], On comparing,, , b 1 ,, , –2c = – 2, , , c 1, , 3a – 3b + c = 1,  3a – 3 + 1 = 1,  3a – 2 = 1,  3a = 3  a  1, So, on putting all these values,, F  V g, , 42. The dimensionally correct expression for the resistance R among the following is, [P = electric power, I = electric current, t = time, V = voltage and E = electric energy], (1) R =, , (2) R , , PI, , E, I 2t, , (3) R = V2P, , (4) R = VI, , Sol. Answer (2), Dimensional formula of power =, Current  [A], V , , W ML2 T 2, , = [ML2T–3], t, T, , W ML2 T 2, , = [ML2T–3A–1], q, AT, , E  [ML2 T 2 ], So, R , , E, I 2t, , , , ML2 T 2, A 2T, ,  [ML2T–3A–2], , ML2 T 3 A 1,  [ML2T–3A–2], A, So, (2) is the correct formula., and V = IR  R , , 43. Which of the following does not have dimensions of force?, (1) Weight, , (2) Rate of change of momentum, , (3) Work per unit length, , (4) Work done per unit charge, , Sol. Answer (4), , W,  [ML2 A 1T 3 ], q, which is different from dimension of force [MLT–2], , Dimension of, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 44. The potential energy u of a particle varies with distance x from a fixed origin as u , , 17, , A x, , where A and B, xB, , are constants. The dimensions of A and B are respectively, (1) [ML5/2T–2], [L], , (2) [MLT–2], [L2], , (3) [L], [ML3/2 T–2], , (4) [L2], [MLT–2], , Sol. Answer (1), , u, , A x, xB, , By the principle of homogeneity, x = B (dimensionally), , , B  [L], , and [ML2 T 2 ] , , AL1/2, L, , [ML2 T 2 ] =AL1/2, A  [ML3/2 T 2 ], 2, 45. A physical quantity P is given by the relation. P  P0e – t  If t denotes the time, the dimensions of constant,  are, , (2) [T2], , (1) [T], , (3) [T–1], , (4) [T–2], , Sol. Answer (4), , P  P0 e t, , 2, , The power of exponent is dimensionless,, , t 2  [M0L0 T0 ],   [T 2 ], , 46. The dimensions of potential energy of an object in mass, length and time are respectively, (1) 2, 2, 1, , (2) 1, 2, – 2, , (3) –2, 1, 2, , (4) 1, – 1, 2, , Sol. Answer (2), The dimensional formula of energy, , E  [ML2 T 2 ], So, dimensions of i) Mass  1, , ii) Length  2, , iii) Time –2, , 47. Which of the following is a dimensional constant?, (1) Magnification, , (2) Relative density, , (3) Gravitational constant, , (4) Relative error, , Sol. Answer (3), Gravitational constant is a dimensional constant., [G] = [M–1L3T–2], Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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18, , Units and Measurements, , Solution of Assignment, , 48. The dimensions of solar constant (energy falling on earth per second per unit area) are, (1) [M0L0 T0], , (2) [MLT–2], , (3) [ML2T–2], , (4) [M T–3], , Sol. Answer (4), Solar constant [S] =, , Energy, ML2 T 2, =,  [MT–3], Area × Time, L2 T, , 49. The amount of heat energy Q, used to heat up a substance depends on its mass m, its specific heat capacity, (s) and the change in temperature T of the substance. Using dimensional method, find the expression for s is, ( Given that [s] = [L2T–2K–1] ) is, (1) QmT, , (2), , Q, mT, , (3), , Qm, T, , (4), , m, Q T, , Sol. Answer (2), Q = ma sb c, [ML2T–2] = [Ma][L2bT–2bK–b][Kc], , , a 1,, , 2b = 2  b  1, –b+c=0,  b = c  c 1, , Q = msT, , , s, , Q, m T, , 50. The focal power of a lens has the dimensions, (1) [L], , (2) [ML2T–3], , (3) [L–1], , (4) [MLT–3], , Sol. Answer (3), Focal length  f = [L], , SECTION - B, Objective Type Questions, 1., , The exchange particles responsible for weak interactions are, (1) Gluons, , (2) -mesons, , (3) Photons, , (4) W and Z bosons, , Sol. Answer (4), Weak interaction takes place through the exchange of BOSONS  W and Z bosons, 2., , Maxwell unified, (1) Electricity with gravitation, , (2) Electricity with magnetism, , (3) Electromagnetism with optics, , (4) Electromagnetism with weak interaction, , Sol. Answer (3), Maxwell unified electromagnetism with optics., 3., , Which of the following is not a derived force?, (1) Tension in a string, , (2) van der Waal forces, , (3) Nuclear force between proton-proton, , (4) Electrostatic force between proton-proton, , Sol. Answer (4), Electrostatic force between proton-proton is a fundamental force., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 4., , Units and Measurements, , 19, , Which one of the following does not experience strong nuclear force?, (1) Leptons, , (2) Baryons, , (3) Hadrons, , (4) Proton, , Sol. Answer (1), Leptons does not experience strong nuclear force., 5., , Which of the following practical units of length is not correct?, (1) 1 fermi = 10–15 m, , (2) 1 astronomical unit = 1.496 × 1011 m, , (3) 1 parsec = 3.26 light year, , (4) 1 light year = 9.46 × 1012 m, , Sol. Answer (4), 1 light year = 9.46 × 1015 m, , 6., , If y represents pressure and x represents velocity gradient, then the dimensions of, (1) [ML–1T–2], , (2) [M2L–2T–2], , (3) [ML–1T0], , d 2y, dx 2, , are, , (4) [M2L–2T–4], , Sol. Answer (3), d 2y, dx, , will have dimensions of, , 2, , y, x2, , y  pressure, x  velocity gradient, x, y, x2, , 7., , , , ML1T 2, T 2, , V, LT 1, ,  T–1, L, L, ,  [ML–1], , The unit of length, velocity and force are doubled. Which of the following is the correct change in the other, units?, (1) Unit of time is doubled, , (2) Unit of mass is doubled, , (3) Unit of momentum is doubled, , (4) Unit of energy is doubled, , Sol. Answer (3), p=F×t, p = 2F × t, p   2p, , 8., , The dimensions of, , ,  t2, in the equation F , , where F is the force, v is velocity and t is time, is, , v 2, , (1) [MLT–1], , (2) [ML–1T–2], , (3) [ML3T–4], , (4) [ML2T–4], , Sol. Answer (3), , F, ,   t2, v 2, , Dimensionally,  = [T2], [MLT–2] =, , [T 2 ],  [L2 T 2 ], , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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20, , Units and Measurements, , =, , Solution of Assignment, , T2, [MLT 2  L2 T 2 ], ,   = [M–1L–3T6], Dimensions of, 9., , , T2, = 1 3 6 = [ML3T–4], , M L T, , Even if a physical quantity depends upon three quantities, out of which two are dimensionally same, then the, formula cannot be derived by the method of dimensions. This statement, (1) May be true, , (2) May be false, , (3) Must be true, , (4) Must be false, , Sol. Answer (3), This statement is completely correct. If a quantity depends upon two other quantities which are dimensionally, same then formula's validity can be checked but it can't be derived by the method of dimensions., 10. The unit of “impulse per unit area” is same as that of, (1) Viscosity, , (2) Surface tension, , (3) Bulk modulus, , (4) Force, , Sol. Answer (1), Impulse, MLT 1, =,  [ML–1T–1], Area, L2, , Coefficient of viscosity    [ML1T 1], So,, , Impulse,  coefficient of viscosity, Area, , 11. In a practical unit if the unit of mass becomes double and that of unit of time becomes half, then 8 joule will, be equal to ............. unit of work., (1) 6, , (2) 4, , (3) 1, , (4) 10, , Sol. Answer (3), Work  [ML2T–2], n1v1 = n2v2, (8)M1 L21 T12, M2 L22 T22, ,  n2, 2, , ⎡M ⎤ ⎡L ⎤ ⎡ T ⎤,  8⎢ 1 ⎥⎢ 1 ⎥ ⎢ 1 ⎥, ⎣ M2 ⎦ ⎣ L2 ⎦ ⎣ T2 ⎦, 2, , 2, , ⎡ M ⎤ ⎡ L ⎤ ⎡ 2T ⎤,  8⎢ 1 ⎥ ⎢ 1⎥ ⎢ 1⎥, ⎣ 2M1 ⎦ ⎣ L1 ⎦ ⎣ T1 ⎦,  8, , ,  n2, 2, ,  n2, , 1 1,   n2, 2 4, , n2  1, , So, unit of 8 joule = 1  new units, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 21, , 12. In a new system of units energy (E), density (d) and power (P) are taken as fundamental units, then the, dimensional formula of universal gravitational constant G will be, (1) [E–1d–2P2], , (2) [E–2d–1P2], , (3) [E2d–1P–1], , (4) [E1d–2P–2], , Sol. Answer (2), G = [Ea db Pc], E = [ML2T–2], d = [ML–3], P = [ML2T–3], G = [M–1L3T–2], [M–1L3T–2] = [ML2T–2]a [ML–3]b [ML2T–3]c, a + b + c = –1, 2a – 3b + 2c = 3, –2a – 3c = –2  2a + 3c = 2, On solving,, a = –2, b = –1, c=2, So, G  [E 2d 1P 2, 13. In equation y  x 2 cos2 2, are, (1) m2, ms–2, , , , the units of x, ,  are m, s–1 and (ms–1)–1 respectively. The units of y and , , (2) m, ms–1, , (3) m2, m, , (4) m, ms–2, , Sol. Answer (1), , ⎛  ⎞, y  x 2 cos2 2 ⎜ ⎟, ⎝ ⎠, The argument of a trigonometric ratio is always dimensionless., T 1, ,  [M0L0 T0 ] or    ⇒   1  [LT–2], , L T, , and y = x2  [L2],  = s–1  [T–1],  = [LT–1]–1  [L–1T],   ms 2, , y  m2, , 14. A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity ‘’ flowing per second, through, a tube of radius r and length l and having a pressure difference P across its ends, is, (1) V , , Pr 4, 8l, , (2) V , , , 8Pr, , 4, , (3) V , , 8P , r, , 4, , (4) V , , P , 8r 4, , Sol. Answer (1), On checking the dimensionality the correct relation is, V , , Pr 4, 8l, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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22, , Units and Measurements, , Solution of Assignment, , 15. E, m, J and G denote energy, mass, angular momentum and gravitational constant respectively. The dimensions, of, , EJ 2, , are same as of, , m5G 2, , (1) Angle, , (2) Length, , (3) Mass, , (4) Time, , Sol. Answer (1), , EJ 2, 5, , mG, , 2, , ⇒, , ML2 T 2  (ML2 T 1 )2, 5, , 1 3, , 2 2, , M  (M L T ), , =, , ML2 T 2M2L4 T 2, M5M2L6 T 4, ,  [M0L0T0] = Angle (Dimensionless), 16. Let P represent radiation pressure, c represent speed of light and I represent radiation energy striking a unit, area per second, then P x I y c z will be dimensionless for, (1) x = 0, y = z, , (2) x = y = z, , (3) x = z = –y, , (4) x = y = –z, , Sol. Answer (3), P xI yc z, P  Pressure  [ML–1T–2], I  Intensity , , ML2 T 2, E, ,  [MT–3], AT, L2 T, , c  Speed of light = [LT–1], [M0L0T0] = [ML–1T–2]x [MT–3]y [LT–1]z, x   y  x + y = 0, – x + z = 0  x  z, , x  z  y, , 17. The number of particles crossing per unit area perpendicular to Z axis per unit time is given by, (N2  N1 ), (Z2  Z1 ) , where N2 and N1 are the number of particles per unit volume at Z2 and Z1 respectively. What, , N  D, , is the dimensional formula for D?, (1) [M0L–1T2], , (2) [M0L–1T–1], , (3) [M0L2T–1], , (4) [M0L2T2], , Sol. Answer (3), N  D, , (N2  N1 ), (Z2  Z1 ), , Dimensionally,, , D, , N (Z2  Z1 ), (N2  N1 ), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 23, , Given,, N2, N1  Number of particles per unit volume., N2, N1 , , N, ⇒ [L3 ], V, , Z2 – Z1  [L], N, , Number of particles, Area  (T), , N  [L–2T–1], So, D , , L2 T 1  L, L3, ,  [L2T–1], , 18. The frequency of vibrations f of a mass m suspended from a spring of spring constant K is given by a relation, of type f = cmxKy, where c is a dimensionless constant. The values of x and y are, (1) x , , 1, 1, , y, 2, 2, , (2), , x, , 1, 1, , y, 2, 2, , (3) x , , 1, 1, , y, 2, 2, , (4), , x, , 1, 1, , y, 2, 2, , Sol. Answer (4), f  Frequency  [T–1], m  Mass  [M], c  Constant, , K, , f MLT 2, ,  [MT–2], x, L, , [M0L0T–1] = c[Mx My T–2y], x + y = 0,, , , x, , 1, 2, , –2y = – 1, , , y, , 1, 2, , ⎛ 2ct ⎞, ⎛ 2x ⎞, 19. The equation of a stationary wave is y  2 A sin ⎜, ⎟ cos ⎜  ⎟ . Which of the following statements is, , ⎝, ⎠, ⎝, ⎠, incorrect?, (1) The unit of ct is same as that of , (3) The unit of, , 2c, 2x, is same as that of, , t, , (2) The unit of x is same as that of , (4) The unit of, , c, x, is same as that of, , , , Sol. Answer (4), , ⎛ 2ct ⎞, ⎛ 2x ⎞, y  2 A sin ⎜, ⎟ cos ⎜  ⎟, , ⎝, ⎠, ⎝, ⎠, ct,  dimensionless ⇒ ct  , , x,  dimensionless ⇒ x  , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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24, , Units and Measurements, , Solution of Assignment, , 20. If energy E, velocity V and time T are taken as fundamental units, the dimensional formula for surface tension, is, (1) [EV–2T–2], , (2) [E–2VT–2], , (3) [E–2V–2T], , (4) [E–2V–2T–2], , Sol. Answer (1), , Force, MLT 2, =,  [MT–2], Length, L, Surface tension = [MT–2], , Surface tension =, , E  [ML2T–2], V  [LT–1], T  [T], Surface tension = [Ea Vb Tc], [MT–2] = [ML2T–2]a [LT–1]b [T]c, On comparing,, , a 1,, , 2a + b = 0, , , 2+b=0, , , , b  2, , –2a – b + c = –2,  –2 + 2 + c = –2, , , c  2, , Surface tension = [EV–2T–2], 21. If force F, area A and density D are taken as the fundamental units, the representation of Young’s modulus, ‘Y’ will be, (1) [F–1A–1D–1], , (2) [FA–2D2], , (3) [FA–1D], , (4) [FA–1D0], , Sol. Answer (4), Young's modulus =, , Stress, = [ML–1T–2], Strain, , F  [MLT–2], A  [L2], D  [ML–3], [ML–1T–2] = [MLT–2]a [L2]b [ML–3]c, a + c = 1,, , , a  1 c, , a + 2b – 3c = –1,  –2 = –2a – 3c,  2 = 2a + 3c,  2 = 2 – 2c + 3c,  0 = +c  c  0, , , , a 1, , 1 + 2b = – 1, 2b = –2, , , b  1, , Young's modulus = [FA–1D0], Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 25, , 22. If velocity, time and force (V, T & F) are considered as fundamental quantities, the dimensional formula for, mass will be, (2) [F–1TV ], , (1) [FTV ], , (3) [FTV–1 ], , (4) [FT–1V ], , Sol. Answer (3), M  Mass  [M], V  Velocity  [LT–1], T  Time  [T], F  Force  [MLT–2], [M] = [MLT–2]a [LT–1]b [T]c, , a 1,, , a + b = 0,, , –2a – b + c = 0, , , ,  –2 + 1 + c = 0, , b  1, , , , c 1, , 1, , M  [FV T], , 23. If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of, the sphere will be, (1) 2%, , (2) 4%, , (3) 6%, , (4) 8%, , Sol. Answer (3), Volume of sphere =, , 4, R 3, 3, , , , V, R,  100% = 3 ,  100% = 3 × 2%, V, R, , , , V,  100%  6%, V, , 24. A set of defective observation of weights is used by a student to find the mass of an object using a physical, balance. A large number of readings will reduce, (1) Random error, , (2) Systematic error, , (3) Random as well as systematic error, , (4) Neither random nor systematic error, , Sol. Answer (1), Random errors can be reduced by taking a large number of observations., 25. A force F is applied on a square area of side L. If the percentage error in the measurement of L is 2% and, that in F is 4%, what is the maximum percentage error in pressure?, (1) 2%, , (2) 4%, , (3) 6%, , (4) 8%, , Sol. Answer (4), Pressure =, , Force, Area, , P, F, 2L,  100% =,  100% ,  100% = 4% + 2 × 2%, P, F, L, P,  100%  8%, P, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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26, , Units and Measurements, , Solution of Assignment, , 26. The radius of a sphere is (5.3 ± 0.1) cm. The percentage error in its volume is, (1), , 0 .1,  100, 5.3, , (2) 3 , , 0 .1,  100, 5 .3, , (3), , 3 0 .1, ,  100, 2 5.3, , (4) 6 , , 0 .1,  100, 0 .3, , Sol. Answer (2), r  (5.3  0.1) cm, , V , , 4 3, r, 3, , V, 3 r,  100% ,  100%, V, r, , V, 3  0.1,  100% ,  100, V, 5.3, 27. If the percentage error in the measurement of momentum and mass of an object are 2% and 3% respectively,, then maximum percentage error in the calculated value of its kinetic energy is, (1) 2%, , (2) 1%, , (3) 5%, , (4) 7%, , Sol. Answer (4), KE =, , Momentum, p2, =, Mass, 2m, , ⎛ 2 p, ⎞, ⎛ m, ⎞, K,  100⎟ %  ⎜,  100⎟ %,  100% = ⎜, ⎝, ⎠, ⎝, ⎠, p, m, K, , = 2 × 2% + 3%, K,  100% ⇒ 7%, K, , 28. The acceleration due to gravity is measured on the surface of earth by using a simple pendulum. If  and , are relative errors in the measurement of length and time period respectively, then percentage error in the, measurement of acceleration due to gravity is, , 1 ⎞, ⎛, (1) ⎜    ⎟  100, 2 ⎠, ⎝, , (2) ( – 2), , (3) (2 + ) × 100, , (4) ( + 2) × 100, , Sol. Answer (4), , T  2, , L, g, ,  T 2  4 2 L, g, , g, L, 2T,  100% ,  100% ,  100%, g, L, T, g,  100%  (  2)  100, g, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 27, , 29. A public park, in the form of a square, has an area of (100 ± 0.2) m2. The side of park is, (1) (10 ± 0.01) m, , (2) (10 ± 0.1) m, , (3) (10 ± 0.02) m, , (4) (10 ± 0.2) m, , Sol. Answer (1), , A  (100  0.2) m2, 100  l 2 ⇒ l  10 m, A 2l, , A, l, 0.2, l,  2, 100, 10, , , l  0.01 m, , So, length = (10 ± 0.01) m, 30. A physical quantity is represented by X = [MaLbT–c]. If percentage error in the measurement of M, L and T, are %, % and % respectively, then maximum percentage error in measurement of X should be (Given that, ,  and  are very small), (1) (a – b + c)%, , (2) (a + b + c)%, , (3) (a – b – c)%, , (4) (a + b – c)%, , Sol. Answer (2), X = [MaLbT–c], X, a M, b L, c T,  100% ,  100% ,  100% ,  100%, X, M, L, T, , , , X,  100%  (a  b  c  )%, X, , 1, second. The time of 20 oscillations of a pendulum is measured to be, 5, 25 seconds. The maximum percentage error in the measurement of time will be, , 31. The least count of a stop watch is, , (1) 0.1%, , (2) 0.8%, , (3) 1.8%, , (4) 8%, , Sol. Answer (2), Least count = T =, , 1, s = 0.2 s, 5, , T = 25 s, Percentage error =, , T, 0.2,  100% =,  100% = 0.8%, T, 25, , 32. A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this, data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the, distance and the time are e1 and e2 respectively, the maximum percentage error in the estimation of g is, (1) e2 – e1, , (2) e1 + 2e2, , (3) e1 + e2, , (4) e1 – 2e2, , Sol. Answer (2), g = LT–2, g, g L 2T, , ,  g  e1  2e2, g, L, T, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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28, , Units and Measurements, , Solution of Assignment, , SECTION - C, Previous Years Questions, 1., , If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of, surface tension will be, [AIPMT-2015], (1) [E–2 V–1 T–3], , (2) [E V–2 T–1], , (3) [E V–1 T–2], , (4) [E V–2 T–2], , Sol. Answer (4), 2., , If force (F), velocity (V) and time (T) are taken as fundamental units, then the dimensions of mass are, [AIPMT-2014], (1) [F V T–1], , (2) [F V T–2], , (3) [F V–1 T–1], , (4) [F V–1 T], , Sol. Answer (4), M = Fx Vy Tz, M = (MLT–2)x (LT–1)y (T)z, M = M x Lx+y T–2x–y+z, Equating powers of M, L and T both sides, x = 1, x + y = 0, –2x –y + z = 0, Solving equations x = 1, y = –1, z = 1, M = F V–1 T, 3., , In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4%, respectively. Quantity P is calculated as follows : P =, (1) 10%, , a3b2, . % error in P is, cd, , (2) 7%, , (3) 4%, , [NEET-2013], (4) 14%, , Sol. Answer (4), P=, , a3b2, cd, , P, ⎛ 3 a 2b c d ⎞, , , ,  100% = ⎜, ⎟  100% = 14%, ⎝ a, b, c, d ⎠, P, , 4., , The damping force on an oscillator is directly proportional to the velocity. The units of the constant of, proportionality are, [AIPMT (Prelims)-2012], (1) kgs–1, , (2) kgs, , (3) kgms–1, , (4) kgms–2, , Sol. Answer (1), F v, , 5., , , , F = bv, , , , b, , F, kgms2, =, = kgs–1, v, ms 1, , The dimensions of (00)–½ are, (1) [L–½ T½], , (2) [L½ T–½], , [AIPMT (Prelims)-2011 & (Mains)-2012 ], (3) [L–1 T], , (4) [L T–1], , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 29, , Sol. Answer (4), Speed of light c , , 1, 0 0, ,  c  (0 0 )1/2, , So, dimensional formula of (0 0 )1/2, 6., , The density of a material in CGS system of units is 4 g/cm3. In a system of units in which unit of length is, 10 cm and unit of mass is 100 g, the value of density of material will be, [AIPMT (Mains)-2011], (1) 400, , (2) 0.04, , (3) 0.4, , (4) 40, , Sol. Answer (4), Density, n1u1 = n2u2, , , 7., , 4g, cm3, ,  n2 , , 100 g, 103 cm3, , n2  40, , A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this, data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the, distance and the time are e1 and e2 respectively, the percentage error in the estimation of g is, [AIPMT (Mains)-2010], (1) e2 – e1, , (2) e1 + 2e2, , (3) e1 + e2, , (4) e1 – 2e2, , Sol. Answer (2), 8., , The dimension of, , 1,  E2, where 0 is permittivity of free space and E is electric field, is, 2 0, [AIPMT (Prelims)-2010], , (1) ML2 T–2, , (2) ML–1 T–2, , (3) ML2 T–2, , (4) MLT–1, , Sol. Answer (2), Energy density =, 9., , 1, E, 1, ML2 T 2, 1 2, 2, 2, = 0E ,  [ML T ]   0 E, 3, 2, V, 2, L, , If the dimensions of a physical quantity are given by Ma Lb Tc, then the physical quantity will be, [AIPMT (Prelims)-2009], (1) Velocity if a = 1, b = 0, c = –1, , (2) Acceleration if a = 1, b = 1, c = –2, , (3) Force if a = 0, b = –1, c = –2, , (4) Pressure if a = 1, b = –1, c = –2, , Sol. Answer (4), Pressure = [ML–1T–2], 10. Which two of the following five physical parameters have the same dimensions?, (a) Energy density, , (b) Refractive index, , (c) Dielectric constant, , (d) Young’s modulus, , [AIPMT (Prelims)-2008], , (e) Magnetic field, (1) (a) and (e), , (2) (b) and (d), , (3) (c) and (e), , (4) (a) and (d), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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30, , Units and Measurements, , Solution of Assignment, , Sol. Answer (4), Refractive index and dielectric constant are dimensional constant, Energy density =, , ML2 T 2, 3, , L, , Young's modulus =, ,  [ML1T 2 ], , MLT 2, L2, ,  [ML1T 2 ], , So, (d) & (a), 11. If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of, the sphere will be, [AIPMT (Prelims)-2008], (1) 2%, , (2) 4%, , (3) 6%, , (4) 8%, , Sol. Answer (3), Volume of sphere =, , , , 4, R 3, 3, , V, R,  100% = 3 ,  100%, V, R, , = 3 × 2%, , , , V,  100%  6%, V, , 12. Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and, of current I, would be, [AIPMT (Prelims)-2007], (1) [ML2T–3I–2], , (2) [ML2T–3I–1], , (3) [ML2T–2], , (4) [ML2T–1I–1], , Sol. Answer (1), V = IR  R , , V W ML2 T 2, , , , I, qI, AT  A, , R  [ML2 T 3 A 2 ], , 13. The velocity v of a particle at time t is given by, v  at , , b, , where a, b and c are constants, The dimensions, t c, , of a, b and c are respectively :, (1) [LT–2], [L] and [T], , (2) [L2], [T] and [LT2], , [AIPMT (Prelims)-2006], (3) [LT2], [LT] and [L], , (4) [L], [LT] and [T2], , Sol. Answer (1), v  at , , b, t c, , By the principle of homogeneity,, c = t = [T], at = v  a = [LT–2], b,  LT 1  b = [L], T, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 31, , 14. The ratio of the dimensions of Planck’s constant and that of the moment of inertia is the dimension of, [AIPMT (Prelims)-2005], (1) Frequency, , (2) Velocity, , (3) Angular momentum, , (4) Time, , Sol. Answer (1), h ML2 T 1, ,  [T 1]  Frequency, I, ML2, , 15. The pair of quantities having same dimensions is, (1) Young's modulus and Energy, , (2) Impulse and Surface Tension, , (3) Angular momentum and Work, , (4) Work and Torque, , Sol. Answer (4), Work = Force × Displacement, W  [ML2 T 2, , Torque = Perpendicular distance × Force = [ML2T–2], 16. The dimensions of 0 are, 1, 1⎤, ⎡, , (1) ⎢M1 L 2 T 2 ⎥, ⎢⎣, ⎥⎦, , 1, 1, ⎡,  ⎤, (2) ⎢M1L2 T 2 ⎥, ⎢⎣, ⎥⎦, , (3) [L–1T], , (4) [M1L1T–2A–2], , (3) [ML0T–2], , (4) [M1L0T–2], , (3) [ML–2T–2], , (4) [M–1L–1], , Sol. Answer (4), , 1, 0 0, , ,  LT 1, , 1,  L2 T 2, 0 0, ,  0 ,  0 , , , 1, L T 2  0, 2, , 1, L T 2, 2, , [ML3 T 4 A 2 ], ,  0  [MLT 2 A 2, , 17. What is the dimension of surface tension?, (1) [ML1T0], , (2) [ML1T–1], , Sol. Answer (3, 4), Surface tension =, , F, MLT 2, =, = [MT–2], L, L, , 18. Which of the following has the dimensions of pressure?, (1) [MLT–2], , (2) [ML–1T–2], , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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32, , Units and Measurements, , Solution of Assignment, , Sol. Answer (2), Pressure =, , Force, MLT 2, =,  [ML–1T–2], Area, L2, , P  [ML1T 2, , 19. Percentage errors in the measurement of mass and speed are 2% and 3% respectively. The error in the, estimate of kinetic energy obtained by measuring mass and speed will be, (1) 8%, , (2) 2%, , (3) 12%, , (4) 10%, , Sol. Answer (1), KE , , 1, MV 2, 2, , , , M, 2V, K,  100% ,  100% = 2% + 2 × 3%,  100% =, M, V, K, , , , K,  100%  8%, K, , 20. Which of the following is a dimensional constant?, (1) Relative density, , (2) Gravitational constant, , (3) Refractive index, , (4) Poisson’s ratio, , Sol. Answer (2), Dimensional constant [G] = [M–1L3T–2], 21. The dimensions of RC is, (1) Square of time, , (2) Square of inverse time (3) Time, , (4) Inverse time, , Sol. Answer (3), RC = Time, 22. The dimensions of impulse are equal to that of, (1) Pressure, , (2) Linear momentum, , (3) Force, , (4) Angular momentum, , Sol. Answer (2), Impulse = p  [MLT–1], 23. The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in, the measurement of mass and lengths are 3% and 2% respectively, the maximum error in the measurement, of density would be, (1) 12%, , (2) 14%, , (3) 7%, , (4) 9%, , Sol. Answer (4), Density =, , Mass, Volume, , m, 3 l, d,  100% ,  100%,  100% =, m, l, d, , = 3% + 3 × 2%, , d,  100%  9%, d, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Units and Measurements, , 33, , a ⎞, , ⎛, 24. An equation is given here ⎜ P  2 ⎟  b where P = Pressure, V = Volume and  = Absolute temperature., V, V ⎠, ⎝, If a and b are constants, the dimensions of a will be, (1) [ML–5 T–1], , (2) [ML5 T1], , (3) [ML5 T–2], , (4) [M–1 L5T2], , Sol. Answer (3), , a ⎞, , ⎛, ⎜P  2 ⎟  b, V, V ⎠, ⎝, Dimensionally,, P, , a, V2, , ML–1T–2 × L6 = a, , , a  [ML5 T 2 ], , 25. Which of the following dimensions will be the same as that of time?, (1), , L, R, , (2), , C, L, , R, L, , (3) LC, , (4), , (3) [ML2T–2A–1], , (4) [ML2T–1A3], , Sol. Answer (1), L,  Time, R, , 26. The dimensional formula of magnetic flux is, (1) [M0L–2T2A–2], , (2) [ML0T–2A–2], , Sol. Answer (3),  = BA =, , =, , F, A, qv, MLT 2, AT  LT 2, , [F  qvB ],  L2, , = [ML2T–2A–1], 27. Which pair do not have equal dimensions?, (1) Energy and torque, , (2) Force and impulse, , (3) Angular momentum and Planck’s constant, , (4) Elastic modulus and pressure, , Sol. Answer (2), Force = [MLT–2], Impulse = Force × Time  [MLT–1], 28. The dimensions of Planck’s constant equals to that of, (1) Energy, , (2) Momentum, , (3) Angular momentum, , (4) Power, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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34, , Units and Measurements, , Solution of Assignment, , Sol. Answer (3), E = h, ML2 T 2, T 1, ,  h  h  [ML2 T 1 ], , Angular momentum = mvr = MLT–1L, L  [ML2 T 1 ], , 29. The dimensions of universal gravitational constant are, (1) [M–1L3T–2], , (2) [ML2T–1], , (3) [M–2L3T–2], , (4) [M–2L2T–1], , Sol. Answer (1), Gravitational constant = [M–1L3T–2], , SECTION - D, Assertion-Reason Type Questions, 1., , A : Shake and light year, both measure time., R : Both have dimension of time., , Sol. Answer (4), Shake  Unit of time, Light year  Unit of length, 2., , A : Displacement gradient is a dimensionless quantity., R : Displacement is dimensionless quantity., , Sol. Answer (3), Displacement gradient =, , Displacement,  Dimensionless, Length, , But displacement is not dimensionless., 3., , A : Absolute error in a physical quantity can be positive, negative or zero., R : Absolute error is the difference in measured value and true value of physical quantity., , Sol. Answer (4), Absolute error is always positive as it is true value  measured value, 4., , A : A unitless physical quantity must be dimensionless., R : A pure number is always dimensionless., , Sol. Answer (2), If a quantity doesnot have units so definitely it will be dimensionless but reverse is not true., Pure number  also dimensionless., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, 5., , Units and Measurements, , 35, , A : Absolute error is unitless and dimensionless., R : All type of errors are unitless and dimensionless., , Sol. Answer (4), Absolute error is not dimensionless rather it will having dimensions of the measured quantity., 6., , A : Higher is the accuracy of measurement, if instrument have smaller least count., R : Smaller the percentage error, higher is the accuracy of measurement., , Sol. Answer (2), Higher accuracy means higher precisions., So, error will be very smaller., Low least count means low error and hence high accuracy., 7., , A : The maximum possible error in a reading is taken as least count of the measuring instrument., R : Error in a measurement cannot be greater than least count of the measuring instrument., , Sol. Answer (3), The assertion is true as least count is the maximum possible error in the measurement., But the error can be greater than least count it will depend upon power of quantity., 8., , A : In a measurement, two readings obtained are 20.004 and 20.0004. The second measurement is more, precise., R : Measurement having more decimal places is more precise., , Sol. Answer (1), The precisions is decided by the more number of decimal places so, 20.0004 is more precise., 9., , A : Out of the measurements A = 20.00 and B = 20.000, B is more accurate., R : Percentage error in B is less than the percentage error in A., , Sol. Answer (1), Out of 20.00 and 20.000, The second measurement is more precise and more accurate also. The percentage error in second reading, is less., , 0.01, 1,  100 , = 0.05%, 20.00, 20, 0.001,  100  0.0005%, 20.000, 10. A : When we change the unit of a measurement of a quantity, its numerical value changes., R : The product of numerical value of the physical quantity and unit for a quantity remain constant., , Sol. Answer (1), Numerical value × Unit = constant, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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36, 11., , Units and Measurements, , Solution of Assignment, , A : All physically correct equations are dimensionally correct., R : All dimensionally correct equations are physically correct., , Sol. Answer (3), If an equation is physically correct it has to be dimensionally correct also., But the reverse is not true., 12. A : Physical relations involving addition and subtraction cannot be derived by dimensional analysis., R : Numerical constants cannot be deduced by the method of dimensions., , Sol. Answer (2), Those equations carrying multiplication and divisions of physical quantities can be derived but not valid for, addition or subtraction., 13. A : If displacement y of a particle executing simple harmonic motion depends upon amplitude a angular, frequency  and time t then the relation y = a sint cannot be dimensionally achieved., R : An equation cannot be achieved by dimensional analysis; if it contains dimensionless expressions., , Sol. Answer (1), Assertion and reason is correct and correctly explains assertion., 14. A : An exact number has infinite number of significant digits., R : A number, which is not a measured value has infinite number of significant digits., , Sol. Answer (2), An exact number contains infinite number of significant figures., 15. A : A dimensionless quantity may have unit., R : Two physical quantities having same dimensions, may have different units., , Sol. Answer (2), Dimensionless quantity may have unit. for example, angle., Also two quantities having same dimensions may have different units., Work  ML2T–2  Joule, Torque  ML2T–2  Nm, ,   , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Chapter, , 3, , Motion in a Straight Line, Solutions, SECTION - A, Objective Type Questions, 1., , A body in one dimensional motion has zero speed at an instant. At that instant, it must have, (1) Zero velocity, , (2) Zero acceleration, , (3) Non-zero velocity, , (4) Non-zero acceleration, , Sol. Answer (1), Magnitude of velocity = Speed, So, if the speed is zero then it must have zero velocity also., 2., , A particle is moving along a circle such that it completes one revolution in 40 seconds. In 2 minutes 20 seconds,, the ratio, , displaceme nt, , is, , distance, , (1) 0, , (2), , 1, 7, , (3), , 2, 7, , (4), , 1, 11, , Sol. Answer (4), T = 40 s, If t = 2 minute 20 second, = 2 × 60 + 20, = 140 s, So, it has completed 3, , 1, revolution., 2, , B, , A, , Distance travelled = 3 × 2R + R, = 7R, Displacement = 2R, |Displacement|, Distance, , =, , 2R, 7R, , =, , 2, 1, =, 22, 11, 7, 7, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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38, 3., , Motion in a Straight Line, , Solution of Assignment, , Consider the motion of the tip of the second hand of a clock. In one minute (R be the length of second hand),, its, (1) Displacement is 2R, , (2) Distance covered is 2R, , (3) Displacement is zero, , (4) Distance covered is zero, , Sol. Answer (3), The second hand of the clock in minute covers an angle of, 360º and the initial and final positions are same., , B, A, , So, Displacement  0, 4., , The position of a body moving along x-axis at time t is given by x = (t2 – 4t + 6) m. The distance travelled, by body in time interval t = 0 to t = 3 s is, (1) 5 m, , (2) 7 m, , (3) 4 m, , (4) 3 m, , Sol. Answer (1), x = t2 – 4t + 6, dx,  2t  4, dt, , At t = 2 s, particle is at rest and reverses its position so,, x |t 0 6 m, x |t 2 s 2 m, x |t 3 s 3 m, , 4m, 1m, , Distance = (4 + 1) m = 5 m, Displacement = 3 m, 5., , If a particle is moving along straight line with increasing speed, then, (1) Its acceleration is negative, , (2) Its acceleration may be decreasing, , (3) Its acceleration is positive, , (4) Both (2) & (3), , Sol. Answer (2), If the speed of body is increasing then acceleration is in the direction of velocity., It may be positive or negative., If acceleration is in negative direction then acceleration is increasing but in negative side, so it will be called, as decreasing., 6., , At any instant, the velocity and acceleration of a particle moving along a straight line are v and a. The speed, of the particle is increasing if, (1) v > 0, a > 0, , (2) v < 0, a > 0, , (3) v > 0, a < 0, , (4) v > 0, a = 0, , Sol. Answer (1), For increasing speed both velocity (v) and acceleration (a) are in the same direction., 7., , A particle moves along x-axis with speed 6 m/s for the first half distance of a journey and the second half, distance with a speed 3 m/s. The average speed in the total journey is, (1) 5 m/s, , (2) 4.5 m/s, , (3) 4 m/s, , (4) 2 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 39, , Sol. Answer (3), If a body travels equal distance with speed v1 and v2 then average speed is given by, , v av , 8., , 2v1v 2, 263, ,  4 ms 1, v1  v 2, 63, , If magnitude of average speed and average velocity over a time interval are same, then, (1) The particle must move with zero acceleration, (2) The particle must move with non-zero acceleration, (3) The particle must be at rest, (4) The particle must move in a straight line without turning back, , Sol. Answer (4), The magnitude of average speed and average velocity can only be equal if object moves in a straight line without, turning back. In that condition distance will be equal to displacement., 9., , If v is the velocity of a body moving along x-axis, then acceleration of body is, (1), , dv, dx, , (2) v, , dv, dx, , (3) x, , du, dx, , (4) v, , dx, dv, , Sol. Answer (2), , a, , dv dv dx, , dt, dx dt, , a, , vdv, dx, , ⎛ dx, ⎞,  velocity⎟, ⎜⎝, ⎠, dt, , 10. If a body is moving with constant speed, then its acceleration, (1) Must be zero, , (2) May be variable, , (3) May be uniform, , (4) Both (2) & (3), , Sol. Answer (2), Acceleration is the rate of change of velocity. The magnitude of velocity (i.e., speed) is constant but it may, change in direction. So, acceleration may be variable due to change in direction., 11. When the velocity of body is variable, then, (1) Its speed may be constant, , (2) Its acceleration may be constant, , (3) Its average acceleration may be constant, , (4) All of these, , Sol. Answer (4), If velocity is changing they may change in magnitude or direction or both., (i) So, if velocity is changing in direction only the magnitude is constant so speed is constant., (ii) If only direction of velocity is changing and magnitude is constant then acceleration will also be constant, in magnitude (in case of uniform circular motion)., (iii) Average acceleration may be constant., , aav , , v 2  v1, t 2  t1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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40, , Motion in a Straight Line, , Solution of Assignment, , 12. An object is moving with variable speed, then, (1) Its velocity may be zero, , (2) Its velocity must be variable, , (3) Its acceleration may be zero, , (4) Its velocity may be constant, , Sol. Answer (2), If speed is changing then velocity must change., 13. The position of a particle moving along x-axis is given by x = 10t – 2t2. Then the time (t) at which it will, momently come to rest is, (1) 0, , (2) 2.5 s, , (3) 5 s, , (4) 10 s, , Sol. Answer (2), x = 10t – 2t2, v, , dx,  10  4t, dt, , v = 0, at the time of coming to rest, so, 10 – 4t = 0, , t  2.5 s, 14. A car moves with speed 60 km/h for 1 hour in east direction and with same speed for 30 min in south direction., The displacement of car from initial position is, (2) 30 3 km, , (1) 60 km, , (3) 30 5 km, , (4) 60 2 km, , Sol. Answer (3), Displacement of car =, , 60 km, , O, , 602  302  30 5 km, , A, , V = 60 km/h, t=1h, , Distance OA = Speed × Time, ⇒ 60 × 1 h = 60 km, , 30 km, , B, 15. A person travels along a straight road for the first, , t, 2t, time with a speed v1 and for next, time with a speed, 3, 3, , v2. Then the mean speed v is given by, (1) v , , v 1  2v 2, 3, , (2), , 1, 1, 2, , , v 3v 1 3v 2, , (3) v , , 1, 2v 1v 2, 3, , (4) v , , 3v 2, 2v 1, , Sol. Answer (1), , v av , , Distance Speed  Time, , , Time, Time, ,  v, av, , v1 2v 2, , 3  v1  2v 2,  3, 1, 3, , v1 , , , , t, 2t,  v2 , 3, 3, t 2t, , 3 3, , v av , , v1  2v 2, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 41, , x  t  7 , then, , 16. If the displacement of a particle varies with time as, , (1) Velocity of the particle is inversely proportional to t, (2) Velocity of the particle is proportional to t2, (3) Velocity of the particle is proportional to, , t, , (4) The particle moves with constant acceleration, Sol. Answer (4), x  t 7, , , , x  (t  7)2,  t 2  49  14t, , (squaring), , dx,  2t  14, dt, , v  2t  14  v  t, Acceleration :, a, , dv, dt, , a  2 ms2  constant, 17. A boat covers certain distance between two spots in a river taking t1 hrs going downstream and t2 hrs going, upstream. What time will be taken by boat to cover same distance in still water?, (1), , t1  t 2, 2, , (2) 2(t2 – t1), , (3), , 2t1t 2, t1  t 2, , (4), , t1t 2, , tstill , , d, v, , t still , , 2t1t 2, t1  t 2, , Sol. Answer (3), For upstream, Speed  v – u, , (where v  man and u  water), , For downstream, Speed  v + u, t up , , t2 , , d, v u, , t down , , d, v u, ,  d  (v  u ) t2, , t1 , ...(i), , , , d, v u, , d, v u, d  (v  u ) t1, , ...(ii), , On equating (i) and (ii), (v – u) t2 = (v + u) t1,  vt2 – ut2 = vt1 + ut1,  v(t2 – t1) = u(t1 + t2),  u, , v (t2  t1 ), t2  t1, , v (t2  t1 ) ⎞, ⎛, ⎛t t t t ⎞, So, d  ⎜ v , t2  vt2 ⎜ 1 2 2 1 ⎟, ⎟, t, t, t1  t2, , ⎝, ⎝, ⎠, 1, 2 ⎠, 2t t, d,  12, v t1  t 2  Remember as shortcut, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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42, , Motion in a Straight Line, , Solution of Assignment, , 18. A particle starts moving with acceleration 2 m/s2. Distance travelled by it in 5th half second is, (1) 1.25 m, , (2) 2.25 m, , (3) 6.25 m, , (4) 30.25 m, , Sol. Answer (2), (distance travelled in 5th half second), , S2.5 – S2 = ?, S2.5  ut , , 1 2, at, 2, ,  S2.5 , , 1,  2  (2.5)2  6.25 m (∵ u  0), 2, , S2 , , a = 2 ms–2, 1, t=0, , 1, 24  4 m, 2, , 2, t, , 1, 2, , 3, t=1, , 4, t, , 3, 2, , 5, t=2, , t, , 5, 2, , So, S2.5  S2  2.25 m, 19. The two ends of a train moving with constant acceleration pass a certain point with velocities u and 3u. The, velocity with which the middle point of the train passes the same point is, (1) 2u, , (2), , Sol. Answer (3), , 3, u, 2, , (3), , 5u, , (4), , 10 u, , Final velocity, Initial velocity, , Velocity at the mid-point =, , v 2  u2, 2, , (When acceleration is constant), Given, v = 3u, u = u, , 9u 2  u 2, 10u 2, , 2, 2, , So, v mid , , v mid  5u 2 , , 5u  v mid, , 20. The initial velocity of a particle is u (at t = 0) and the acceleration a is given by t3/2. Which of the following, relations is valid?, (1) v = u + t3/2, , (2) v  u , , 3 t 3, 2, , (3) v  u , , 2 5/2, t, 5, , (4) v = u + t5/2, , Sol. Answer (3), a  t 3/2, v, , t, , u, , 0, , (acceleration is a function of time), , ∫ dv  ∫ adt, , , , v, , t, , u, , 0, , 3/2, ∫ dv  ∫ t dt, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, ,  v, , v, u, , Motion in a Straight Line, , 43, , t, , t 3/2  1, , 3, 1, 2, 0, , 2,  (v  u )    5, ,  t 5/2  0, , 2 5/2,  v  u  5 t, , , , v u, , 2 5/2, t, 5, , Note : The equations of kinematics are valid only for constant acceleration, here a is a function of 't' so, we didn't apply those equations., 21. A train starts from rest from a station with acceleration 0.2 m/s2 on a straight track and then comes to rest, after attaining maximum speed on another station due to retardation 0.4 m/s2. If total time spent is half an, hour, then distance between two stations is [Neglect length of train], (1) 216 km, , (2) 512 km, , (3) 728 km, , (4) 1296 km, , Sol. Answer (1), , 1  2, T, 2,   Acceleration, Shortcut : S , ,   Deceleration, , (magnitude only), , T  Time of journey, S  Distance travelled, Given,, ,  = 0.2 ms–2,  = 0.4 ms–2, T = half an hour = 30 × 60 s = 1800 s, , S, , 1 ⎛ 0.2  0.4 ⎞, 2, ⎜, ⎟  (1800), 2 ⎝ 0.2  0.4 ⎠, ,  S = 216000 m, , , S  216 km, , 22. The position x of particle moving along x-axis varies with time t as x = Asin (t) where A and  are positive, constants. The acceleration a of particle varies with its position (x) as, (1) a = Ax, , (2) a = – 2x, , (3) a = A x, , (4) a = 2 x A, , Sol. Answer (2), x = Asin t, dx,  A cos t, dt, , , , , d2x, dt 2, ,   A 2 sin t, , a  2 x, , (∵ A sin t  x ), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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44, , Motion in a Straight Line, , Solution of Assignment, , 23. A particle moves in a straight line and its position x at time t is given by x2 = 2+ t. Its acceleration is given, by, (1), , 2, x3, , (2) , , 1, 4x 3, , (3) , , 1, 4x 2, , (4), , 1, x2, , Sol. Answer (2), x2 = t + 2 , , 1, x, , 2, , , , 1, t2, , , , x  t2, , , , 1, dx 1,  (t  2) 2, dt 2, , , , , dx 1,  (t  2) 2, dt 2, , , , ....(i), , 1, , 1, , d2x, dt 2, , 1, ,  1, 1 ⎛ 1⎞, ⎜⎝  ⎟⎠ (t  2) 2, 2 2, , , , 3, 1, , 1, ,  a =  (t  2) 2 = , 4(t  2), 4, , , , a, , 1, (t , , 1, 2) 2, , = , , 1 1 1, , , 4 x2 x, , 1, 4x3, , 24. A body is projected vertically upward direction from the surface of earth. If upward direction is taken as positive,, then acceleration of body during its upward and downward journey are respectively, (1) Positive, negative, , (2) Negative, negative, , (3) Positive, positive, , (4) Negative, positive, , Sol. Answer (2), Whether body move upwards or downwards, the earth tries to pull it downwards only., Hence during both the motion g will negative., , u, , g, , So, negative, negative, 25. A particle start moving from rest state along a straight line under the action of a constant force and travel, distance x in first 5 seconds. The distance travelled by it in next five seconds will be, (1) x, , (2) 2 x, , (3) 3 x, , (4) 4 x, , Sol. Answer (3), Body starts from rest and moves with a constant acceleration, then the distance travelled in equal time intervals, will be in the ratio of odd number. (Galileo's law of odd number), x1 : x2  1 : 3, x : x2  1 : 3, , , x, 1, , x2 3, , , , x2  3 x, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 45, , 26. A body is projected vertically upward with speed 40 m/s. The distance travelled by body in the last second, of upward journey is [take g = 9.8 m/s2 and neglect effect of air resistance], (1) 4.9 m, , (2) 9.8 m, , (3) 12.4 m, , (4) 19.6 m, , Sol. Answer (1), As the motion under gravity is symmetric, so distance travelled in last second of ascent is equal to first second, of descent., (1st second), , t=1s, , v=0, , 1,  x2  ut  g  12, 2, , x1, , x2 , , 1,  9.8  12 (∵ u  0), 2, , , , x2  4.9 m, , x2, , This distance is constant for every body thrown with any speed., 27. A body is moving with variable acceleration (a) along a straight line. The average acceleration of body in time, interval t1 to t2 is, a[t 2  t1 ], (1), 2, , (2), , a[t 2  t1 ], 2, , (3), , t2, , t2, , t1, , t1, , ∫ a dt, , t 2  t1, , (4), , ∫ a dt, , t 2  t1, , Sol. Answer (4), t2, , ∫ a dt, t1, Change in velocity,  aav , Average acceleration =, t2  t1, Time, 28. A body is projected vertically upward with speed 10 m/s and other at same time with same speed in downward, direction from the top of a tower. The magnitude of acceleration of first body w.r.t. second is {take g = 10 m/s2}, (2) 10 m/s2, , (1) Zero, , (3) 5 m/s2, , (4) 20 m/s2, , Sol. Answer (1), The acceleration of first body, a1 = 10 ms–2, a2 = 10 ms–2, arel = a1 – a2 = 10 ms–2 – 10 ms–2 = 0, 29. The position of a particle moving along x-axis given by x = (–2t3 + 3t2 +5)m. The acceleration of particle at, the instant its velocity becomes zero is, (1) 12 m/s2, , (2) –12 m/s2, , (3) –6 m/s2, , (4) Zero, , Sol. Answer (3), , x  ( 2t 3  3t 2  5) m, , , , dx,  6t 2  6t  v, dt, , d2x, dt 2, ,  12t  6, , (for v = 0, 6t = 6t2 t = 1 s), , a t 1 s  12  6  6 ms 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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46, , Motion in a Straight Line, , Solution of Assignment, , 30. A car travelling at a speed of 30 km/h is brought to rest in a distance of 8 m by applying brakes. If the same, car is moving at a speed of 60 km/h then it can be brought to rest with same brakes in, (1) 64 m, , (2) 32 m, , (3) 16 m, , (4) 4 m, , Sol. Answer (2), , ds , , u2, ⇒ ds  u 2, 2a, , u   2u, , d  (2u )2, , d, u2, , , d, 4, 8, ,  d   32, 31. A particle is thrown with any velocity vertically upward, the distance travelled by the particle in first second of, its decent is, (1) g, , (2 ), , g, 2, , g, 4, , (3), , (4) Cannot be calculated, , Sol. Answer (2), , s, , g, 1, g  12  s , 2, 2, , 32. A body is thrown vertically upwards and takes 5 seconds to reach maximum height. The distance travelled, by the body will be same in, (1) 1st and 10th second, , (2) 2nd and 8th second, , (3) 4th and 6th second, , (4) Both (2) & (3), , Sol. Answer (1), The motion under gravity is a symmetric motion and the time taken to go up is same as time taken to come, back to the initial position., , t=5s, 4s, , 1st second, , 6s, , 3s, , 7s, , 2s, , 8s, , 1s, , 9s, 0, , 10th second, , 10 s, , So, clearly the distance travelled in 1st second is same as that travelled in 10th second., 33. A ball is dropped from a bridge of 122.5 metre above a river. After the ball has been falling for two seconds,, a second ball is thrown straight down after it. Initial velocity of second ball so that both hit the water at the, same time is, (1) 49 m/s, , (2) 55.5 m/s, , (3) 26.1 m/s, , (4) 9.8 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 47, , Sol. Answer (3), , 1,  h   gt 2, 2,  122.5 , , 1st ball, , 1,  9.8 t 2, 2, ,  t 2  25  t  5 s, Another ball is dropped after 2 second so it took only (5 – 2) = 3 s, 122.5  u(3) , , 1,  9.8  32, 2, ,  122.5 = 3u + 4.9 × 9,  3u = 78.4, , , u  26.1 s, , 34. A balloon starts rising from ground from rest with an upward acceleration 2 m/s2. Just after 1 s, a stone is dropped, from it. The time taken by stone to strike the ground is nearly, (1) 0.3 s, , (2) 0.7 s, , (3) 1 s, , (4) 1.4 s, , Sol. Answer (2), u = 0, a = 2 ms–2, The velocity of object after one second, v = u + at, , , s, , 1,  2  12  1 m, 2, , v  2 ms1, , Now after separating from the balloon it will move under the effect of gravity alone.,  h  vt , , 1,  9.8  t 2, 2, ,  –1 = 2t – 4.9t2,  4.9t2 – 2t – 1 = 0, , , t  0.7 s, , 35. A boy throws balls into air at regular interval of 2 second. The next ball is thrown when the velocity of first, ball is zero. How high do the ball rise above his hand? [Take g = 9.8 m/s2], (1) 4.9 m, , (2) 9.8 m, , (3) 19.6 m, , (4) 29.4 m, , Sol. Answer (3), T=2s, v=0, , 2u, u, 2T ,  2,  u = 19.6, g, 9.8, H, , u2, 19.6  19.6, =, 2g, 2  9.8, , , , H  19.6 m, , 2s, , H, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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48, , Motion in a Straight Line, , Solution of Assignment, , 36. A ball projected from ground vertically upward is at same height at time t1 and t2. The speed of projection of, ball is [Neglect the effect of air resistance], (1) g [t 2  t1 ], , (2), , g [t1  t 2 ], 2, , g [t 2  t1 ], 2, , (3), , (4) g [t1  t2 ], , Sol. Answer (2), t1 + t2 = total time of flight, t1 + t2 = 2T, , T , , u, t1  t2, , also T , g, 2, , u t1  t2, , g, 2, ,  u, , 1, g (t1  t 2 ), 2, , 37. For a body moving with uniform acceleration along straight line, the variation of its velocity (v) with position, (x) is best represented by, , v, , (2), , (1), , O, , x, , v, , v, , v, , (3), , (4), , x, , O, , x, , O, , Sol. Answer (3), , O, , x, , v, , For uniform acceleration, a  constant, v2 = u2 + 2as, , , v2  x, , (∵ u  rest), , x, , 38. The position-time graph for a particle moving along a straight line is shown in figure. The total distance travelled, by it in time t = 0 to t = 10 s is, x (m), , 10, , 2, (1) Zero, , 4, , (2) 10 m, , 6, , 8, , 10, , t (s), , (3) 20 m, , (4) 80 m, , Sol. Answer (3), , x (m), The total distance travelled from 0 to 2 s is 10 m, 2 s to 8 s  Zero distance, , 10, , and from 8 s to 10 s  10 m, So, distance = 10 + 0 + 10 = 20 m, , 0, , 2, , 4, , 6, , 8, , 10, , t (s), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 49, , Motion in a Straight Line, , 39. The position-time graph for a body moving along a straight line between O and A is shown in figure. During, its motion between O and A, how many times body comes to rest?, , x, A, , t, , O, (1) Zero, , (2) 1 time, , (3) 2 times, , (4) 3 times, , Sol. Answer (3), As there are two extremes in the graph one is maxima and other is minima. At both maxima and minima the, slope is zero. So, it comes to rest twice., 40. Which one of the following graph for a body moving along a straight line is possible?, , Speed, , Speed, , (2), , (1), , t, , O, , Time, , Speed, , (3), , t, , O, , (4), , t, , O, , O, , Position, , Sol. Answer (4), This graph is possible., 41. A body is projected vertically upward from ground. If we neglect the effect of air, then which one of the following, is the best representation of variation of speed (v) with time (t)?, , v, , v, , (2), , (1), , t, , O, , v, , v, , (3), , t, , O, , (4), , t, , O, , t, , O, , Sol. Answer (2), The speed of an object is directly proportional to time v  t ., 42. Which one of the following time-displacement graph represents two moving objects P and Q with zero relative, velocity?, d, , d, Q, P, , (1), O, , P, , (2), t, , P, Q, , (3), O, , d, , d, , P, Q, , t, O, , Q, , (4), t, , O, , t, , Sol. Answer (2), Zero relative velocity means that both of them have same slope., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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50, , Motion in a Straight Line, , Solution of Assignment, , vA, 43. The displacement-time graph for two particles A and B is as follows. The ratio v is, B, , Y, Displacement, , B, , 15°, , A, , 15°, , 45°, , X, , t, (2) 1 : 3, , (1) 1 : 2, , 3 :1, , (3), , Sol. Answer (4), , (4) 1 : 3, , Y, , B, , The slope of line A is tan30º and B = tan60º, 15°, , 1, v A tan30º, , , v B tan 60º, , 3 1  v :v  1: 3, , A, B, 3 3, , 15°, , 45°, , A, , X, , 44. For the acceleration-time (a-t) graph shown in figure, the change in velocity of particle from t = 0 to t = 6 s, is, 2, a(m/s ), , 4, 2, , 4, , t(s), , 6, , –4, (1) 10 m/s, , (2) 4 m/s, , (3) 12 m/s, , Sol. Answer (2), , 2, , Area under a-t graph gives change in velocity., 1, 1, So, v   4  4   2  4  8  4, 2, 2, , v  4 ms1, , a(m/s ), 4, , 2, , 4, , 6, , t(s), , –4, , 45. Figure shows the graph of x-coordinate of a particle moving, along x-axis as a function of time. Average velocity during, t = 0 to 6 s and instantaneous velocity at t = 3 s respectively,, will be, (1) 10 m/s, 0, , (2) 60 m/s, 0, , (3) 0, 0, , (4) 0, 10 m/s, , Sol. Answer (3), , (4) 8 m/s, , x, (m), 20, 10, , 0 1 2 3 4 5 6, , t (s), , From 0 to 6 s  Displacement = 0, so, average velocity = 0, at t = 3 s, the displacement = 0, so v = 0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 51, , 46. Position-time graph for a particle is shown in figure. Starting from t = 0, at what time t, the average velocity, is zero?, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, , (1) 1 s, , x (m), , t (s), 1 2 3 4 5 6 7 8 9 10 1112 13, , (2) 3 s, , (3) 6 s, , (4) 7 s, , Sol. Answer (3), If we look at the graph very carefully at t = 0, x = 6 m, The average velocity will be zero if it comes back to the initial position., It is evident that at t = 6 s, x = 6 m, So, vav at t = 6 s is zero., 47. The velocity versus time graph of a body moving in a straight line is as, shown in the figure below, , v (m/s), , (1) The distance covered by the body in 0 to 2 s is 8 m, (2) The acceleration of the body in 0 to 2 s is 4 ms–2, , 4, 2, 0, , (3) The acceleration of the body in 2 to 3 s is 4 ms–2, , 1, , 2, , (4) The distance moved by the body during 0 to 3 s is 6 m, , 3, t (s), , v (m/s), , Sol. Answer (4), Distance covered = Area under v-t graph =, , Acceleration t  0 to 2 s , , 1, 34  6 m, 2, , 4, , 40,  2 ms 2, 2, , 0, , 1m 2, , 3, , t (s), , 48. Acceleration-time graph for a particle is given in figure. If it starts motion at t = 0, distance travelled in 3 s, will be, a, 2, , (m/s ), 2, 0, , 1, , 3, , 2, , t (s), , –2, , (1) 4 m, , (2) 2 m, , (3) 0, , (4) 6 m, , v, , Sol. Answer (1), Draw the v-t graph from a-t graph., Area under v-t graph =, , 2 m/s, , 1,  2  (3  1), 2, , =4m, , 1s, , 2s, , 3s, , t, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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52, , Motion in a Straight Line, , Solution of Assignment, , 49. Figure shows the position of a particle moving on the x-axis as a, function of time, , (m), , x, , 20, , (1) The particle has come to rest 4 times, (2) The velocity at t = 8 s is negative, , 10, , (3) The velocity remains positive for t = 2 s to t = 6 s, (4) The particle moves with a constant velocity, , 0, , Sol. Answer (1), , 2, , 4, , 6, , 8, , t (s), , The particle has come to rest four times., 50. A particle moves along x-axis in such a way that its x-co-ordinate varies with time according to the equation, x = 4 – 2t + t2. The speed of the particle will vary with time as, , (1) O, , (2), , O, , (3) O, , (4) O, , Sol. Answer (1), , dx,  2  2t, dt, v = 2t – 2  Straight line, x = 4 – 2t + t2 , , Slope  Positive, Intercept  Negative, 51. Two balls are projected upward simultaneously with speeds 40 m/s and 60 m/s. Relative position (x) of second, ball w.r.t. first ball at time t = 5 s is [Neglect air resistance]., (1) 20 m, , (2) 80 m, , (3) 100 m, , (4) 120 m, , Sol. Answer (3), , 1, arel t 2, 2, = (60 – 40) 5, , Srel  Urel t ,  Srel, , , (arel = 0), , Srel  100 m, , x (m), , 52. The position (x) of a particle moving along x-axis varies with time (t), as shown in figure. The average acceleration of particle in time interval, t = 0 to t = 8 s is, (1) 3 m/s2, , (2) –5 m/s2, , (3) – 4 m/s2, , (4) 2.5 m/s2, , 40, , 0, , 2 4 6, , 8, , t (s), , Sol. Answer (2), t = 0 to t = 2, , t = 6 to t = 8, , v = 20 m/s, , v = – 20 m/s, , aavg , , v 20  20 40, , ,  5 ms2, t, 8, 8, , aavg  5 ms 2, , x (m), 40, , 0, , 2 4 6, , 8, , t (s), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 53, , 53. The position (x)-time (t) graph for a particle moving along a straight line is shown in figure. The average speed, of particle in time interval t = 0 to t = 8 s is, , x (m), 20, 10, , 0, (1) Zero, , 2, , 4, , 6, , (2) 5 m/s, , t (s), 8, (3) 7.5 m/s, , (4) 9.7 m/s, , Sol. Answer (2), , v, , Distance 40, ,  5 ms 1, Time, 8, , 54. A ball is dropped from a height h above ground. Neglect the air resistance, its velocity (v) varies with its height, above the ground as, (1), , 2g (h  y ), , (2), , 2gh, , (3), , 2gy, , (4), , 2g (h  y ), , Sol. Answer (1), , (h – y), , h, , y, , Reference, Ground, , v  2g (h  y ), 55. A train of 150 m length is going towards North at a speed of 10 m/s. A bird is flying at 5 m/s parallel to the, track towards South. The time taken by the bird to cross the train is, (1) 10 s, , (2) 15 s, , Sol. Answer (1), Time =, , , (3) 30 s, , (4) 12 s, , 5 m/s Bird, , 150, 150, , 10  5 15, , Train, , T  10 s, , 150 m, 10 ms–1, , 56. Two cars are moving in the same direction with a speed of 30 km/h. They are separated from each other by, 5 km. Third car moving in the opposite direction meets the two cars after an interval of, 4 minutes. The speed of the third car is, (1) 30 km/h, , (2) 25 km/h, , (3) 40 km/h, , (4) 45 km/h, , Sol. Answer (4), , 30 kmh–1, , 30 kmh–1, , A, , B, , v, C, , 5 km, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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54, , Motion in a Straight Line, , Solution of Assignment, , The distance of 5 km is in between A and B is covered by C in 4 minute with relative velocity (v + 30)., So, drel  v rel  t, , 4, 60, = v + 30, ,  5 km  (v  30) ,  75 kmh–1, , , v  45 kmh1, , 57. Two cars A and B are moving in same direction with velocities 30 m/s and 20 m/s. When car A is at a distance, d behind the car B, the driver of the car A applies brakes producing uniform retardation of 2 m/s2. There will, be no collision when, (1) d < 2.5 m, , (2) d > 125 m, , (3) d > 25 m, , (4) d < 125 m, , Sol. Answer (3), v2 = u2 + 2ad,  0=, , (10)2, , a = –2 ms–2, , – 2 × 2 × drel, , , , 100,  drel, 4, , , , drel  25 m, , B, , A, –1, , 20 ms, , d, , –1, , 30 ms, , 58. Two trains each of length 100 m moving parallel towards each other at speed 72 km/h and, 36 km/h respectively. In how much time will they cross each other?, (1) 4.5 s, , (2) 6.67 s, , (3) 3.5 s, , (4) 7.25 s, , Sol. Answer (2), When two trains are moving in opposite direction then, , v rel  (20  10)  30 ms1, t, , 200,  6.67 s, 30, , 59. A ball is dropped from the top of a building of height 80 m. At same instant another ball is thrown upwards, with speed 50 m/s from the bottom of the building. The time at which balls will meet is, (1) 1.6 s, , (2) 5 s, , (3) 8 s, , (4) 10 s, , Sol. Answer (1), t, , , , h, 80, , v rel 50, , t  1.6 s, , 60. A particle move with velocity v1 for time t1 and v2 for time t2 along a straight line. The magnitude of its average, acceleration is, (1), , v 2  v1, t1  t 2, , (2), , v 2  v1, t1  t 2, , (3), , v 2  v1, t 2  t1, , (4), , v1  v 2, t1  t 2, , Sol. Answer (2), aavg , , v 2  v1, Change in velocity, t1  t2 =, Time interval, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 55, , SECTION - B, Objective Type Questions, 1., , If average velocity of particle moving on a straight line is zero in a time interval, then, (1) Acceleration of particle may be zero, (2) Velocity of particle must be zero at an instant, (3) Velocity of particle may be never zero in the interval, (4) Average speed of particle may be zero in the interval, , Sol. Answer (2), If average velocity = zero, then displacement is zero it means particle takes a turn in the opposite direction, and at the time of turning back velocity has to be zero., 2., , A car moving with speed v on a straight road can be stopped with in distance d on applying brakes. If same, car is moving with speed 3v and brakes provide half retardation, then car will stop after travelling distance, (1) 6 d, , (2) 3 d, , (3) 9 d, , (4) 18 d, , Sol. Answer (3), , u2, 2a, , ds , , ds  u 2 ⇒, , ds u 2, , ds u  2, , u   3v, , u=v, So,, , ds, v2,  2, ds 9v, ,  ds  9ds, , 3., , ds  9d, , The initial velocity of a particle moving along x-axis is u (at t = 0 and x = 0) and its acceleration a is given, by a = kx. Which of the following equation is correct between its velocity (v) and position (x)?, (1) v2 – u2 = 2kx, , (2) v2 = u2 + 2kx2, , (3) v2 = u2 + kx2, , (4) v2 + u2 = 2kx, , Sol. Answer (3), a = kx and, v, , vdv, a, dx, x, , x, ,  ∫ vdv  ∫ adx  ∫ kxdx, u, , , , 0, , 2 v, , v, 2, , u, , , , 0, , 2 x, , kx, 2, , 0, ,  v2 – u2 = kx2  v 2  u 2  kx 2, 4., , The velocity v of a body moving along a straight line varies with time t as v = 2t2 e–t, where v is in, m/s and t is in second. The acceleration of body is zero at t =, (1) 0, , (2) 2 s, , (3) 3, , (4) Both (1) & (2), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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56, , Motion in a Straight Line, , Solution of Assignment, , Sol. Answer (2), v = 2t2 e–t, dv,  2[t 2e  t  ( 1)  e  t  2t ], dt, Put, a = 0,, , 2t 2e  t  4te  t  0, , , 5., , 2t 2  4t  0  t2 = 2t  t  2 s, , The relation between position (x) and time (t) are given below for a particle moving along a straight line. Which, of the following equation represents uniformly accelerated motion? [where  and  are positive constants], (1) x  t  , , (2) x    t, , (3) xt  , , (4) t    x, , Sol. Answer (4), For uniformly accelerated motion,, 2, , 2, , v = u + 2as, , 1 2, at, 2, , s  ut , , or, , Constant, , Constant, , 1 2, at  ut, 2, Or the maximum power of t has to be two., x, , So, 4 ., 6., , The velocity v of a particle moving along x-axis varies with its position (x) as v   x ; where  is a constant., Which of the following graph represents the variation of its acceleration (a) with time (t)?, a, , a, , (1), , (2), t, , O, , a, , a, , (3), O, , t, , (4), O, , t, , O, , t, , Sol. Answer (3), v x, , Squaring both sides v 2   2 x, Comparing above equation with 3rd equation of kinematics., , v 2  u 2  2ax,  2 x  2ax, , , , a, , 2, 2, , Constant  not a function of time, a, So,, , t, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 7., , Motion in a Straight Line, , 57, , The velocity (v) of a particle moving along x-axis varies with its position x as shown in figure. The acceleration, (a) of particle varies with position (x) as, v (m/s), 4, , 0, , (1) a2 = x + 3, , x (m), , 2, , (2) a = 2x2 + 4, , (3) 2a = 3x + 5, , (4) a = 4x – 8, , Sol. Answer (4), , a, , vdv, dx, , dv,  slope, dx, , , , v 4 04, , x 0 20, , 4,  2, 2, , , , v  u 4, , x, 2, , Intercept = + 4, , , , v – 4 = – 2x, , a  Negative, , , , v = – 2x + 4, , , , dv,  2, dx, , , , a, , , , a  4x  8, , So,, , So, a , , 8., , Relation between v and x, , vdv, dx, , vdv,  ( 2 x  4)( 2), dx, , A ball is dropped from an elevator moving upward with acceleration ‘a’ by a boy standing in it. The acceleration, of ball with respect to [Take upward direction positive], (1) Boy is – g, , (2) Boy is – (g + a), , (3) Ground is – g, , (4) Both (2) & (3), , Sol. Answer (4), Upward direction  Positive, Negative direction  Negative, If a person is observing from ground then, for, him the acceleration of ball is in the, downward direction., , ELEVATOR, 2, , aball G  aball  aground  g  0, , u=0, g, , a (Positive), 1, , abG  g, , abG = Acceleration of ball w.r.t. ground., , aball boy  aball  aboy  g  a, , a bb   ( g  a ) , abb = Acceleration of ball w.r.t. boy., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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58, 9., , Motion in a Straight Line, , Solution of Assignment, , The velocity (v)-time (t) graph for a particle moving along x-axis is shown in the figure. The corresponding position, (x)- time (t) is best represented by, v, , t, , O, , x, , x, , x, , (1), , (2), O, , t, , x, , (3), t, , O, , (4), t, , O, , Sol. Answer (1), , O, , t, , v, , The graph of v-t can be converted into the x-t (parabolic) graph., , x, , t, , t, 10. The speed-time graph for a body moving along a straight line is shown in figure. The average acceleration of, body may be, , speed (m/s), 20, , 0, (2) 4 m/s2, , (1) 0, , 5, , t(s), , 10, , (3) – 4 m/s2, , (4) All of these, , Sol. Answer (4), The acceleration from zero to 5 s is, , 0  20 20, ,  4 ms2, 50, 5, From 5 s to 10 s, a, , a, , 20  0,  4 ms2, 10  5, , a, , Total change in velocity, Time, , , , speed (m/s), 20 m/s, , 0, , 5, , 10 s, , t, , 20  20,  0 ms2, 10  0, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 59, , 11. The acceleration (a)-time (t) graph for a particle moving along a straight starting from rest is shown in figure., Which of the following graph is the best representation of variation of its velocity (v) with time (t)?, a, , T, , O, v, , v, , (1), O, , t, , T, , t, , (2), O, , v, , v, , T, , (3) O, , t, , t, , T, , (4) O, , T, , t, , Sol. Answer (1), From the graph it is evident that the acceleration is decreasing with time., Also, a  –t,  a = –kt, , a, (decreasing with time), , To find velocity,, , t, , T, , dv,  kt, dt, , ∫ dv  ∫ ktdt, v  t 2 or graph of velocity should be parabolic with a decreasing slope., , 12. A ball is thrown upward with speed 10 m/s from the top of the tower reaches the ground with a speed, 20 m/s. The height of the tower is [Take g = 10 m/s2], (1) 10 m, , (2) 15 m, , (3) 20 m, , (4) 25 m, , Sol. Answer (2), , v  u 2  2gh,  (–20)2 = 102 + 2 × 10 × h, , , 300,  h  h  15 m, 2  10, , 13. A ball dropped from the top of tower falls first half height of tower in 10 s. The total time spend by ball in air, is [Take g = 10 m/s2], (1) 14.14 s, , (2) 15.25 s, , (3) 12.36 s, , (4) 17.36 s, , Sol. Answer (1), , u=0, , H, 1,  ut  g  102, 2, 2, , H, 2, ,  H = g × 102, , , 1, H   gt 2, 2, , t = 10 s, , (Full journey), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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60, , Motion in a Straight Line, , g  102 , , Solution of Assignment, , 1 2, gt, 2, ,  t2 = 200,  t  10 2 s,  t = 10 × 1.414 s, = 14.14 s  t, 14. An object thrown vertically up from the ground passes the height 5 m twice in an interval of 10 s. What is its, time of flight?, (1), , 28 s, , (2), , 86 s, , 104 s, , (3), , (4), , 72 s, , Sol. Answer (3), h=5m, , (given), , t2 – t1 = 10 s, T  Time taken to reach the highest point., , t1  T  T 2 , , T, , 2h, 2h, 2, , t2  T  T , g, g, , t1, , 2h, 2h, t2  t1  T  T , T  T 2 , g, g, , t2, , 2, , 2,  10  2 T , , h, , 25, 10, ,  5  T2 1 , , 25 = T2 – 1, T2 = 26, , , , T  26, , Total time of flight  2T  2 26 , , 4  26  104 s, , 15. A ball is projected vertically upwards. Its speed at half of maximum height is 20 m/s. The maximum height, attained by it is [Take g = 10 ms2], (1) 35 m, , (2) 15 m, , (3) 25 m, , Sol. Answer (4), , v=0, , ⎛H⎞, v B2  v A2  2g ⎜ ⎟, ⎝ 2⎠,  0  400  2  10 , , , 40 m  H, , (4) 40 m, , H, 2, , B, , H, 2, , 20 m/s, H, 2 A, H, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 61, , 16. A particle starts with initial speed u and retardation a to come to rest in time T. The time taken to cover first half of, the total path travelled is, (1), , ⎛, 1 ⎞, ⎟⎟, (2) T ⎜⎜1 , 2⎠, ⎝, , T, 2, , (3), , T, 2, , a, u, , Sol. Answer (2), Retardation  a, Initial velocity  u, , (4), , d, 2, , A, , 3T, 4, , C, B, , d, Time = T, , (I) For total journey, , (II) For half journey, , d, 1,  ut  at 2, 2, 2, , v = u + at, , ...(iii), , 0 = u – aT, , , u = aT, , ...(i), , 1 2, aT, 2, Dividing by 2 on both sides, d  uT , , d uT 1 aT 2, , , ....(ii), 2, 2 2 2, On comparing equation (i) & (iii), uT 1 aT 2, 1, ,  ut  at 2, 2 2 2, 2, Put u = aT, , , aT 2 aT 2, 1, ,  aTt  at 2, 2, 4, 2, , T2, t2,  Tt , 4, 2, Multiplying by 4 on both sides, , , T2 = 4Tt – 2t2  2t2 – 4Tt + T2 = 0, On solving this quadratic equation,, , t T , , 1 ⎞, ⎛, ⎟,  t  T ⎜⎝ 1 , 2⎠, 2, , T, , 17. Which of the following speed-time (v - t) graphs is physically not possible?, , v, , v, , v, (2), , (1), , t, , (3), , t, , t, , (4) All of these, , Sol. Answer (4), None of the graph is physically possible., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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62, , Motion in a Straight Line, , Solution of Assignment, , 18. A particle travels half of the distance of a straight journey with a speed 6 m/s. The remaining part of the distance, is covered with speed 2 m/s for half of the time of remaining journey and with speed 4 m/s for the other half of time., The average speed of the particle is, (1) 3 m/s, , (2) 4 m/s, , (3) 3/4 m/s, , (4) 5 m/s, , Sol. Answer (2), From C to B the time interval of travelling is same., So, v av, , v  v3 2  4,  2, ,  3 m/s, 2, 2, , A, , v1 = 6 m/s, , C, v2 2 m/s v3 4 m/s, t, , B, , t, , 6 m/s, , Now, A, , B, 3 m/s, Now, first half is covered with 6 ms–1 and second half with 3 ms–1. So when distances are same., v av , , 2v1v 2, 263, ,  4 ms 1, v1  v 2, 63, , v av  4 ms 1, , 19. The acceleration-time graph for a particle moving along x-axis is shown in figure. If the initial velocity of particle is, –5 m/s, the velocity at t = 8 s is, a (m/s2), , 10, 8, –10, (1) +15 m/s, , 2, , 4, , (2) +20 m/s, , 6, , t (s), , (3) –15 m/s, , (4) –20 m/s, , Sol. Answer (1), The area under a-t graph gives change in velocity., Given, u = – 5 m/s, , 1,  6  10 = 30 ms–1, 2, ,  Area on positive side =,  Area on negative side =, , 1,  2  10 = 10 ms–1, 2, , Net area = 30 – 10 = 20 ms–1, v = Area, v – (– 5) = 20, , , v  15 ms1, , 20. A body thrown vertically up with initial velocity 52 m/s from the ground passes twice a point at h height above at an, interval of 10 s. The height h is (g = 10 m/s2), (1) 22 m, , (2) 10.2 m, , (3) 11.2 m, , (4) 15 m, , Sol. Answer (2), Given, t2 – t1 = 10 s, , t2  t1 , , 2u 2  52, ,  10.4, g, 10, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 63, ,  2t2 = 20.4,  t2 = 10.2 s, t1 = 0.2 s, So, t1t2 , , 2h, g, , 0.2  10.2 , , 2h, 10, ,  1 × 10.2 = h  10.2 m  h, 21. A body falling from a vertical height of 10 m pierces through a distance of 1 m in sand. It faces an average, retardation in sand equal to (g = acceleration due to gravity), , O, 10 m, 1m, (1) g, , (2) 9 g, , (3) 100 g, , (4) 1000 g, , Sol. Answer (2), If the ball is dropped then x = 0, the velocity with which it will hit the sand will be given by, v2 – u2 = 2(–g) (– 9), v2 – 0 = 18 g, v 2  18 g, , O u=0, , ...(i), , v, , Now on striking sand, the body penetrates into sand, for 1 m and comes to rest. So, v  initial for sand, and final velocity = 0, , 9m, , 10 m, , 1m, , v  2  v 2  2(a)  ( 1),  – 18 g = – 2 a, , , a9g, , 22. When a particle is thrown vertically upwards, its velocity at one third of its maximum height is, 10 2 m/s. The maximum height attained by it is, , (1) 20 2 m, , (2) 30 m, , (3) 15 m, , v=0, , Sol. Answer (3), v 2  u 2  2g , , , , , 2H, 3, , 100  2  2  10 , , H  15 m, , (4) 12.8 m, , 2H, 3, 2H, 3, , v   10 2 ms 1, , 10 2 m/s, H, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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64, , Motion in a Straight Line, , Solution of Assignment, , 23. A body is dropped from a height H. The time taken to cover second half of the journey is, (1) 2, , 2H, g, , H, g, , (2), , H, g, , (3), ,  2  1, , (4), , 2H, 1, , g, ( 2  1), , Sol. Answer (3), The total time of journey, , s  ut ,  H, , 1 2, gt, 2, , 1 2, gT, 2, , u=0, , ...(i), , H, 1,  ut  gt 2  T , 2, 2, , , , , H 1 2,  gt, 2 2, 1 2, gT  gt 2, 2, ,  t, , t, , 2H, g, , H, T, , (T – t), , (∵ ut  0), , T, 2, ,  Second half time = T – t = T , , 1 ⎞, ⎛, = T ⎜⎝ 1 , ⎟ =, 2⎠, 2, , T, , 2H ⎛, 1 ⎞, 1, ⎜, ⎟ =, ⎝, g, 2⎠, , H, g, , , , 2  1, , ⎛5⎞, 24. A stone dropped from the top of a tower is found to travel ⎜ ⎟ of the height of the tower during the last second of, ⎝9⎠, its fall. The time of fall is, , (1) 2 s, , (2) 3 s, , (3) 4 s, , Sol. Answer (2), , u=0, , Let the total height of tower = H, , h, , Total time of journey = t, Time taken to cover the, So, st  st  1, , 5h, is = last second, 9, , 5h, , 9, , , , 1 2 1, 5 1, gt  g (t  1)2   gt 2, 2, 2, 9 2, , , , 1, 1, 5, g (t 2  t 2  1  2t )  gt 2 , 2, 2, 9, ,  (2t  1) , , (4) 5 s, , (t – 1) s, , t, 5h, 9, , 1 2⎤, ⎡, ⎢⎣∵ h  2 gt ⎥⎦, , 5 2, t, 9, ,  18t – 9 = 5t2,  5t2 – 18t + 9 = 0,  5t2 – 15t – 3t + 9 = 0,  5t (t – 3) – 3 (t – 3 ) = 0,  (5t – 3) (t – 3) = 0, t, , 3, , t 3s, 5, , (t , , 3, , doesn't satisfy the given criterion, so we neglect it), 5, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , (1) Uniform acceleration, , t2,  20 . The body is undergoing, 10, (2) Uniform retardation, , (3) Non-uniform acceleration, , (4) Zero acceleration, , 65, , 25. The velocity of a body depends on time according to the equation v , , Sol. Answer (3), , v, , t2,  20, 10, , To find acceleration find, So, a , , dv, dt, , dv 2t, , 0, dt 10, , t, ⇒ at, 5, ∵ a is a function of time so it is not constant, rather it is non-uniform., ,  a, , 26. The displacement (x) - time (t) graph of a particle is shown in figure. Which of the following is correct?, , x, , (1) Particle starts with zero velocity and variable acceleration, (2) Particle starts with non-zero velocity and variable acceleration, (3) Particle starts with zero velocity and uniform acceleration, (4) Particle starts with non-zero velocity and uniform acceleration, Sol. Answer (1), , O, , t, , From the graph it is clear that the x is a function of time and speed/velocity is also changing. So, if velocity, is changing then definitely the acceleration also changes with time. So, at t = 0, x = 0, so v = 0 but it is, function of time and hence non-uniform., 27. A stone thrown upward with a speed u from the top of a tower reaches the ground with a velocity 4u. The height of, the tower is, (1), , 15u 2, 2g, , (2), , 7u 2, 2g, , (3), , 16u 2, g, , (4) Zero, , Sol. Answer (1), , v  u 2  2gh, , (4u )2  u 2  2gh, 15u 2, 16u 2  u 2, h  h, 2g, 2g, 28. If magnitude of average speed and average velocity over an interval of time are same, then, (1) Particle must move with zero acceleration, (2) Particle must move with uniform acceleration, (3) Particle must be at rest, (4) Particle must move in a straight line without turning back, Sol. Answer (4), Particle should have same distance and displacement in order to have final average speed and average velocity, which is only possible only in case of an object moving on a straight line without turning back., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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66, , Motion in a Straight Line, , Solution of Assignment, , 29. A body is dropped from a certain height h (h is very large) and second body is thrown downward with velocity of 5, m/s simultaneouly. What will be difference in heights of the two bodies after 3 s?, (1) 5 m, , (2) 10 m, , (3) 15 m, , (4) 20 m, , Sol. Answer (3), , urel  u1  u2  0  ( 5)  5 ms1, t=3s, , arel  a1  a2  g  ( g )  0 ms2, srel  urel t , , 1, arel t 2, 2, ,  srel  5  3  15 m, , (∵ arel  0), , So, srel  15 m, 30. Ball A is thrown up vertically with speed 10 m/s. At the same instant another ball B is released from rest at height, h. At time t, the speed of A relative to B is, (1) 10 m/s, , (2) 10 – 2 gt, , (3), , 10 2  2gh, , (4) 10 – gt, , Sol. Answer (1), , v A  10 ms1  10t, v B  0  10t, v AB  v A  v B  10  (10t )  ( 10t )  10  10t  10t = 0, , , , v AB  10 ms 1, , 31. A man moves in an open field such that after moving 10 m on a straight line, he makes a sharp turn of 60º to, his left. The total displacement after 8 such turn is equal to, (1) 12 m, , (2) 15 m, , (3) 17.32 m, , (4) 14.14 m, , Sol. Answer (3), 60º, , After 8 such turns object is at 'B'., 60º, , Displacement = AB, , 60º, , Two vectors are at 60º, 102  102  2  102 , , , , B, , 1  10 3 m, 2, , 60º, , 10 m, 60º, , 17.32 m  AB, , 60º, , A, , 10 m, , 32. A body starts from origin and moves along x-axis so that its position at any instant is x = 4t2 – 12t where t, is in second and v in m/s. What is the acceleration of particle?, (1) 4 m/s2, , (2) 8 m/s2, , (3) 24 m/s2, , (4) 0 m/s2, , Sol. Answer (2), x  4t 2  12t, , v, a, , dx,  8t  12, dt, d2x, dt 2, , 8, ,  a  8 ms2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 67, , 33. Position time graph of a particle moving along straight line is shown which is in the form of semicircle starting, from t = 2 to t = 8 s. Select correct statement, , x(m), , (0, 0) 2, , 5, , 8, , t(s), , (1) Velocity of particle between t = 0 to t = 2 s is positive, (2) Velocity of particle is opposite to acceleration between t = 2 to t = 5 s, (3) Velocity of particle is opposite to acceleration between t = 5 to t = 8 s, (4) Acceleration of particle is positive between t1 = 2 s to t2 = 5 s while it is negative between t1 = 5 s to, t2 = 8 s, , x(m), , Sol. Answer (2), (i) From 0 to 2 s the velocity = 0 as displacement, is zero., (ii) From 2 to 5 s velocity is decreasing but nature, is positive, but acceleration is negative., , (0, 0), , So, v and a have opposite nature., , 2, , 5, , 8, , t(s), , 34. Two bodies starts moving from same point along a straight line with velocities v 1 = 6 m/s and, v2 = 10 m/s, simultaneously. After what time their separation becomes 40 m?, (1) 6 s, , (2) 8 s, , (3) 12 s, , (4) 10 s, , Sol. Answer (4), , srel  urel t , , 1, arel t 2, 2, , arel = 0,,  40  (10  6)  t, , , 40,  t  t  10 s, 4, , SECTION - C, Previous Years Questions, 1., , A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = x–2n,, where and n are constants and x is the position of the particle. The acceleration of the particle as a function of, x, is given by, [AIPMT-2015], (1) –2n2 e–4n + 1, , (2) –2n2 x–2n – 1, , (3) –2n2 x–4n – 1, , (4) –22 x–2n + 1, , Sol. Answer (3), 2., , A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next, 5 seconds and the next 5 seconds respectively. The relation between h1, h2 and h3 is, [NEET-2013], (1) h1 , , h2 h3, , 3, 5, , (2) h2 = 3h1 and h3 = 3h2, , (3) h1 = h2 = h3, , (4) h1 = 2h2 = 3h3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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68, , Motion in a Straight Line, , Solution of Assignment, , Sol. Answer (1), When a body starts from rest and under the effect of constant acceleration then the distance travelled by the, body in final time intervals is in the ratio of odd number i.e., 1 : 3 : 5 : 7, So, h1 : h2 : h3  1 : 3 : 5, h1 1 h1 1, , , h2 3 , h3 5, , 3., ,  h1 , , h3, h2, , h1 , 3, 5, , So, h1 , , h2 h3, , 3, 5, , The motion of a particle along a straight line is described by equation x = 8 + 12t – t3 where x is in metre, and t in second. The retardation of the particle when its velocity becomes zero, is [AIPMT (Prelims)-2012], (1) 6 ms–2, , (2) 12 ms–2, , (3) 24 ms–2, , (4) Zero, , Sol. Answer (2), x = 8 + 12t – t3, , dx,  12  3t 2, dt, If v = 0, then 12 – 3t2 = 0, 4 = t2  t  2 s, a, , d2x, dt 2, ,  6t, , a t  2 s ⇒ 12 ms2, | a |  12 ms 2, , 4., , A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 ms–2, the velocity with, which it hits the ground is, [AIPMT (Prelims)-2011], (1) 5.0 m/s, , (2) 10.0 m/s, , (3) 20.0 m/s, , (4) 40.0 m/s, , Sol. Answer (3), s  ut , , , , 1 2, gt, 2, , 20  , , 1,  10  t 2, 2, , u=0, , (∵ u  0), , 20 m, ,  40 = 10t2, , , t2s, , v = u – gt, , , v  20 ms1, , , , | v |  20 ms 1, , (∵ u  0), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 5., , Motion in a Straight Line, , 69, , A particle covers half of its total distance with speed v1 and the rest half distance with speed v2. Its average, speed during the complete journey is, [AIPMT (Mains)-2011], (1), , v 12v 22, v12  v 22, , (2), , v1v 2, (3) v  v, 1, 2, , v1  v 2, 2, , 2v1v 2, (4) v  v, 1, 2, , Sol. Answer (4), As the distances are same so,, v av , , 6., , A, , v1, , v2, , B, , 2v1v 2, v1  v 2, , A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown, downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v?, (Take g = 10 m/s2), [AIPMT (Prelims)-2010], (1) 60 m/s, , (2) 75 m/s, , (3) 55 m/s, , (4) 40 m/s, , Sol. Answer (2), , u=0, , As the ball meet at t = 18 s, , v, , So, it means both of them covered the same distance 'h'., , h, , But the time of travel is different, 1st body  t, 2nd body  (t – 6)  as theorem after 6 s., 1st body, , 2nd body, , 1,  h   gt 2, 2, , h  v (t  6) , , h, , 1 2, gt, 2, , ...(i), , h  v (t  2) , , 1, g (t  6)2, 2, , 1, (t  2)2, 2, , ...(ii), , Equating (i) and (ii), we get, v = 75 m/s, For fitst body, t = 18 s, For second body, t = (18 – 6) = 12 s, , h, , 1,  10  (18)2  5  324, 2, , h = 1620 m, For second body, , 1, × 10 (18 – 6)2, 2, 1620 = v × 12 + 5 × 144, 1600 = v × (18 – 6) +, , 1620  720, v, 12, , 900, v, 12, , , v  75 ms1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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70, 7., , Motion in a Straight Line, , Solution of Assignment, , A particle moves a distance x in time t according to equation x = (t + 5)–1. The acceleration of particle is, proportional to, [AIPMT (Prelims)-2010], (1) (Velocity)3/2, , (2) (Distance)2, , (3) (Distance)–2, , (4) (Velocity)2/3, , Sol. Answer (1), x = (t + 5)–1, , v, , d n, ⎡, n 1 ⎤, ⎢⎣∵ dx ( x )  nx ⎥⎦, , dx,  ( 1)(t  5)2, dt, , v = – (t + 5)–2, a, , dv,  ( 1)( 2)(t  5)3, dt, , a  2(t  5), , 3, ,  2(t  5), , 2, ,  (t  5), , 1, , 1, ⎡, 1, 1 ⎤, 2 , ⇒, v, ⎢∵ v , ⎥, 2, t  5 ⎥⎦, (t  5), ⎢⎣, , 1, ,  2(v )  v 2, 3, , a  2v 2, 3, , a  (velocity) 2, , 8., , A bus is moving with a speed of 10 ms–1 on a straight road. A scooterist wishes to overtake the bus in, 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase, the bus?, [AIPMT (Prelims)-2009], (1) 40 ms–1, , (2) 25 ms–1, , (3) 10 ms–1, , (4) 20 ms–1, , Sol. Answer (4), , 10 ms–1, , T = 100 s, Srel = 1000 m, Srel = Urel t, , (∵ arel  0), , 1000 = (v – 10) × 100, , 1 km, , v  20 ms1, 9., , A particle starts its motion from rest under the action of a constant force. If the distance covered in first, 10 seconds is S1 and that covered in the first 20 seconds is S2, then, [AIPMT (Prelims)-2009], (1) S2 = 3S1, , (2) S2 = 4S1, , (3) S2 = S1, , (4) S2 = 2S1, , Sol. Answer (2), u = 0, a  Constant, S1 , , 1, 1, a (10)2 , S2  a (20)2, 2, 2, , S1, 102, 100, , , 2, S2 (20), 400, S2  4S1, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 71, , 10. A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle, is around the point, [AIPMT (Prelims)-2008], Distance, , D, C, , S, A, , B, Time t, , (1) A, , (2) B, , (3) C, , (4) D, , Sol. Answer (3), Maximum instantaneous velocity will be at that point which has maximum slope., As clear from the graph 'C' has maximum slope., 11. A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms–1 to 20 ms–1, while passing through a distance 135 m in t second. The value of t is, [AIPMT (Prelims)-2008], (1) 9, , (2) 10, , (3) 1.8, , (4) 12, , Sol. Answer (1), Using 3rd equation, we first find acceleration,, v2 – u2 = 2as, 202 – 102 = 2a × 135, , , 300, a, 2  135, ,  a, , 10, 20, ms2  a, , 9, 18, ,  v = u + at,  20  10 ,  10 , , , 10, t, 9, , 10, t, 9, , t 9s, 4, ms–2, in the third, 3, [AIPMT (Prelims)-2008], , 12. The distance travelled by a particle starting from rest and moving with an acceleration, second is, (1), , 19, m, 3, , (2) 6 m, , (3) 4 m, , (4), , 10, m, 3, , Sol. Answer (4), , S, , n th, , u, , a, (2n  1), 2, , n = 3, (given), a , , S, , n th, , u, , a, (2n  1), 2, ,  Sn th  0 , , , 4, ms2, 3, , 4 1, 2,  (2  3  1) =  5, 3 2, 3, , 10, m  S rd, 3, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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72, , Motion in a Straight Line, , Solution of Assignment, , t ⎞, ⎛, 13. A particle moving along x-axis has acceleration f, at time t, given f  f0 ⎜1  ⎟ , where f0 and T are constants., ⎝ T⎠, The particle at t = 0 has zero velocity. When f = 0, the particle’s velocity (vx) is, (1), , 1, f0T, 2, , (2) f0T, , (3), , 1, f0T 2, 2, , [AIPMT (Prelims)-2007], (4) f0T 2, , Sol. Answer (1), 14. A car moves from x to y with a uniform speed vu and returns to y with a uniform speed vd. The average speed, for this round trip is, [AIPMT (Prelims)-2007], , (3), , 2v u v d, (2) v  v, d, u, , vu  v d, 2, , (4), , vuvd, , vd  vu, (4) v  v, d, u, , Sol. Answer (4), Repeated., 15. The position x of a particle with respect to time t along x-axis is given by x = 9t2 – t3, where x is in metres, and t in seconds. What will be the position of this particle when it achieves maximum speed along the positive, x-direction?, [AIPMT (Prelims)-2007], (1) 24 m, , (2) 32 m, , (3) 54 m, , (4) 81 m, , Sol. Answer (3), x = 9t2 – t3, dx,  18t  3t 2, dt, ,  v = 18t – 3t2, To find the maxima of speed,, dv,  18  6t, dt, dv, 0, Put,, dt, , , , , 18 – 6t = 0, , t 3s, , So, the positions of particle at t = 3 = ?, x t  3 s  9(32 )  33, , x  54 m, 16. A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle, from O is given by x = 40 + 12t – t3. How long would the particle travel before coming to rest?, [AIPMT (Prelims)-2006], (1) 24 m, , (2) 40 m, , (3) 56 m, , (4) 16 m, , Sol. Answer (4), , x  40  12t  t 3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 73, , The particle will come to rest when v = 0,, v, , dx,  12  3t 2, dt, ,  v=0, , , , 12 = 3t2, , , , t2 = 4  t  2 s, , So, the distance travelled by object is 2 s., , x t  0  40 m ,, x t  2 s  40  12  2  8 = 40 + 24 – 8 = 40 + 16  56 m, Distance travelled = (56 – 40) = 16 m, 17. Two bodies, A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m,, respectively. The ratio of the time taken by them to reach the ground is, [AIPMT (Prelims)-2006], (1), , 5, 4, , (2), , 12, 5, , (3), , 5, 12, , (4), , 4, 5, , Sol. Answer (4), , T , , 2H, ⇒T  H, g, , T1,  T , 2, , H1, H2, , , , T1, , T2, , , , T1 4, , T2 5, , 16 4, , 25 5, , (Given, H1 = 16 m, H2 = 25 m), , 18. The displacement x of a particle varies with time t as x = ae–t + bet, where a, b,  and  are positive, constants. The velocity of the particle will, [AIPMT (Prelims)-2005], (1) Go on decreasing with time, , (2) Be independent of , , (3) Drop to zero when  and , , (4) Go on increasing with time, , Sol. Answer (4), x  ae t  bet, dx,  a(  ) e t  b () et, dt, , v  bet  ae t, As we increase time et increases and e–t decreases., So, v keeps on increasing with time., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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74, , Motion in a Straight Line, , Solution of Assignment, , 19. A ball is thrown vertically upward. It has a speed of 10 m/s when it has reached one half of its maximum height., [AIPMT (Prelims)-2005], How high does the ball rise? (Taking g = 10 m/s2), (1) 15 m, , (2) 10 m, , (3) 20 m, , (4) 5 m, , Sol. Answer (2), 20. The displacement 'x' (in meter) of a particle of mass 'm' (in kg) moving in one dimension under the action of a, force, is related to time 't' (in sec) by t  x  3 . The displacement of the particle when its velocity is zero, will, be, (1) 2 m, , (2) 4 m, , (3) 0 m (zero), , (4) 6 m, , Sol. Answer (3), t, , x 3, , x  x  (t  3)2  t 2  9  6t, , (t  3) , , dx,  2t  6, dt, ,  v, If v = 0,, , 2t – 6 = 0, , , At, t = 3 s,, , t 3s, x=?, , , , x  (t  3)2  (3  3)2, , x0, , Speed (ms–1), , 21. The speed-time graph of a particle moving along a solid curve is shown below. The distance traversed by the, particle from t = 0 to t = 3 is, , 2, 1.5, 1, 1, , 2, Time (second), , (1), , 9, m, 2, , (2), , 9, m, 4, , (3), , 3, , 10, m, 3, , (4), , 10, m, 5, , Sol. Answer (2), , Speed, 1.5 ms–1, –1, , 1 ms, , t=0, , 1s, , 2s, , t=3s, , t, , Area under the speed-time graph gives distance., Area =, , 1, 9,  3  1.5 , m, 2, 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 75, , Displacement, , 22. The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle, is negative at the point, , C, , D, , F, , E, Time, , (1) E, , (2) F, , (3) C, , (4) D, x, , Sol. Answer (1), The angle made by the tangent at point 'C' is obtuse hence, tan QE = negative, so slope = negative, , E, , hence, velocity is also negative., , QE, , t, , 23. Two bodies A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively., The ratio of the time taken by them to reach the ground is, , 4, 5, Sol. Answer (1), , (2), , (1), , 24., , 5, 4, , (3), , 12, 5, , (4), , 5, 12, , t ⎞, ⎛, A particle moving along x-axis has acceleration f at time t given by f  f0 ⎜1  ⎟ , where f0 and T are, ⎝ T⎠, constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when, f = 0, the particle’s velocity (vx) is, , (1), , 1, f0T 2, 2, , (2) f0T2, , (3), , 1, f0T, 2, , (4) f0T, , Sol. Answer (3), , t⎞, ⎛, f  f0 ⎜ 1  ⎟, ⎝ T⎠, f  Acceleration, f0  Initial acceleration, Initial/lower limit of time = 0, u = 0, Upper limit of time = T, v = ?, , dv, a, , dt, vx, , t, , 0, , 0, , ∫ dv  ∫ adt, , T, , t⎞, ⎛, dv  ∫ f0 ⎜ 1  ⎟ dt, ⎝ T⎠, , ∫, , 0, , v, , vx, , 0, , vx, 0, ,  f0 t, , T, 0, , f t2,  0, T 2, , T, 0, , v x  0  f0 (T  0) , ,  v x  f0T , , , vx , , f0, (T 2  0), 2T, , 1, f0T, 2, , 1, f0T, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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76, , Motion in a Straight Line, , Solution of Assignment, , 25. A bus is moving with a speed of 10 ms–1 on a straight road. A scooterist wishes to overtake the bus in 100, s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the, bus?, (1) 10 ms–1, , (2) 20 ms–1, , (3) 40 ms–1, , (4) 25 ms–1, , Sol. Answer (2), Repeated, 26. A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown, downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v?, (Take g = 10 m/s2), (1) 60 m/s, , (2) 75 m/s, , (3) 55 m/s, , (4) 40 m/s, , Sol. Answer (2), Repeated, 27. The velocity of train increases uniformly from 20 km/h to 60 km/h in 4 hour. The distance travelled by the, train during this period is, (1) 160 km, , (2) 180 km, , (3) 100 km, , (4) 120 km, , Sol. Answer (1), v2 – u2 = 2as, v = u + at, 60 = 20 + a × 4, 40 = 4a, , a  10 km/h2, 602 – 202 = 2 × 10 × s, 3600  400, s, 20, ,  s  160 km, 28. A particle moves along a straight line such that its displacement at any time t is given by s = (t3 – 6t2, – 3t + 4) metres. The velocity when the acceleration is zero is, (1) 3 m/s, , (2) 42 m/s, , (3) –9 m/s, , (4) –15 m/s, , Sol. Answer (4), s = t3 – 6t – 3t + 4, v, , ds,  3t 2  12t  3, dt, , a, , dv,  6t  12, dt, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Put a = 0 , , Motion in a Straight Line, , 77, , 6t – 12 = 0, , t2s, v t  2 s  3(2)2  12(2)  3, = 12 – 24 – 3, = – 12 – 3, , v  15 ms1, 29. A car accelerates from rest at a constant rate  for some time after which it decelerates at a constant rate,  and comes to rest. If total time elapsed is t, then maximum velocity acquired by car will be, , (1), , ( 2 –  2 ) t, , , (2), , ( 2   2 ) t, , , (   ) t, , , (3), , (4), , t, , , Sol. Answer (4), v max , , v, , t, , , , , v max, In ABC, tan   slope  t, 1, , In BCD, 1   , , A, , , , v max, T  t1, , t1  T  t1, ,  t1 , , B, , vmax, , , t1 C, T, , D, , t, , v max    t1, , T, , , v max , , T, , , 30. The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap, at instant the first drop touches the ground. How far above the ground is the second drop at that instant?, (Take g = 10 ms–2), (1) 3.75 m, , (2) 4.00 m, , (3) 1.25 m, , (4) 2.50 m, , Sol. Answer (1), x = 3x = 5 m,  4x = 5 m, , x  1.25 m, So, second drop is at 3x, , 5m, , x, , 3rd drop, 2nd drop, , 3x, 1st drop, ,  3 × 1.25 = 3.75 m above ground., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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78, , Motion in a Straight Line, , Solution of Assignment, , 31. The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with an, initial velocity v0. The distance travelled by the particle in time t will be, (1) v 0 t , , 1 2, bt, 3, , (2) v 0 t , , 1 2, bt, 2, , (3) v 0 t , , 1 3, bt, 6, , (4) v 0 t , , 1 3, bt, 3, , Sol. Answer (3), a = bt, u = v0, a, , dv, dt, , v, , t, , v0, , 0, , ∫ dv  ∫ adt, v, , ∫ dv  ∫ btdt, , v0, , t, , v  v0 , , bt 2, 2, , v  v0 , , b 2, (t  0), 2, , v  v0 , , 1 2, bt, 2, , Now, v , , dx, dt, , x, , t, , 0, , 0, , 0, ,  ∫ dx  ∫ vdt, x, , t, , 1 2⎞, ⎛, ∫ dx  ∫ ⎜⎝ v 0  2 bt ⎟⎠ dt, 0, 0, , x  v 0t , , 1 3, bt, 6, , 32. If a car at rest accelerates uniformly to a speed of144 km/h in 20 s, it covers a distance of, (1) 1440 cm, , (2) 2980 cm, , (3) 20 m, , (4) 400 m, , Sol. Answer (4), u = 0, a  constant, v = 144 km/h–1 = 144 , , 5, = 40 ms–1, 18, , 1 2 1, at   2  400, 2, 2, , t = 20 s, , s, , v = u + at, , s  400 m, , 40 = a × 20, , a  2 ms2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Straight Line, , 79, , 33. The position x of a particle varies with time, (t) as x = at2 – bt3. The acceleration will be zero at time t equal to, , (1), , a, 3b, , (2) Zero, , (3), , 2a, 3b, , (4), , a, b, , Sol. Answer (1), x = at2 – bt3, , v, , dx,  2at  3bt 2, dt, , a, , dv,  2a  6bt, dt, , Put a = 0, to find 't', 2a = 6bt, , t, , a, 3b, , 34. Motion of a particle is given by equation, s = (3t3 + 7t2 + 14t + 8) m, The value of acceleration of the particle at t = 1 s is, (1) 10 m/s2, , (2) 32 m/s2, , (3) 23 m/s2, , (4) 16 m/s2, , Sol. Answer (2), s = 3t3 + 7t2 + 14t + 8, , v, , a, , ds,  9t 2  14t  14, dt, , d 2s, dt 2, ,  18t  14, , a t  1 s  18  14, a t  1 s  32 ms2, 35. If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent, is, (1) ut, , (2), , 1 2, gt, 2, , (3) ut , , 1 2, gt, 2, , (4) (u + gt)t, , Sol. Answer (2), As the motion is symmetric the distances covered during the last t seconds of ascent is same as that travelled, during 1st t seconds of descent., At highest point, v = 0, 1, s   gt 2, 2, , , , s, , 1 2, gt, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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80, , Motion in a Straight Line, , Solution of Assignment, , 36. A man throws balls with the same speed vertically upwards one after the other at an interval of, 2 second. What should be the speed of the throw so that more than two balls are in the sky at any time?, (Given g = 9.8 m/s2), (1) More than 19.6 m/s, , (2) At least 9.8 m/s, , (3) Any speed less than 19.6 m/s, , (4) Only with speed 19.6 m/s, , Sol. Answer (1), For move than two ball in air, time of flight should be, Total time of flight , , 4, , 2u, g, , 2u, g, , 2  9.8  u, , u  19.6 ms1, 37. A particle is moving such that its position coordinates (x, y) are, (2 m, 3 m) at time t = 0,, (6 m, 7 m) at time t = 2 s and, (13 m, 14 m) at time t = 5 s, , Average velocity vector (Vav ) from t = 0 to t = 5 s is, (1), , 1, (13iˆ  14 ˆj ), 5, , (2), , 7 ˆ ˆ, (i  j ), 3, , (3) 2(iˆ  ˆj ), , (4), , 11 ˆ ˆ, (i  j ), 5, , Sol. Answer (4), , Vav, ,  , rf – ri, 13 – 2 iˆ  14 – 3 jˆ  11 iˆ  jˆ, , =, 5, t,  5 – 0, , , , , , SECTION - D, Assertion - Reason Type Questions, 1., , A : It is not possible to have constant velocity and variable acceleration., R : Accelerated body cannot have constant velocity., , Sol. Answer (1), 2., , A : The direction of velocity of an object can be reversed with constant acceleration., R : A ball projected upward reverse its direction under the effect of gravity., , Sol. Answer (2), 3., , A : When the velocity of an object is zero at an instant, the acceleration need not be zero at that instant., R : In motion under gravity, the velocity of body is zero at the top-most point., , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 4., , Motion in a Straight Line, , 81, , A : A body moving with decreasing speed may have increasing acceleration., R : The speed of body decreases, when acceleration of body is opposite to velocity., , Sol. Answer (1), 5., , A : For a moving particle distance can never be negative or zero., R : Distance is a scalar quantity and never decreases with time for moving object., , Sol. Answer (1), 6., , A : If speed of a particle is never zero than it may have zero average speed., R : The average speed of a moving object in a closed path is zero., , Sol. Answer (4), 7., , A : The magnitude of average velocity in an interval can never be greater than average speed in that interval., R : For a moving object distance travelled  | Displacement |, , Sol. Answer (1), 8., , A : The area under acceleration-time graph is equal to velocity of object., R : For an object moving with constant acceleration, position-time graph is a straight line., , Sol. Answer (4), 9., , A : The motion of body projected under the effect of gravity without air resistance is uniformly accelerated, motion., R : If a body is projected upwards or downwards, then the direction of acceleration is downward., , Sol. Answer (2), 10. A : The relative acceleration of two objects moving under the effect of gravity ,only is always zero, irrespective of, direction of motion., R : The acceleration of object moving under the effect of gravity have acceleration always in downward direction, and is independent from size and mass of object., Sol. Answer (1), 11. A : In the presence of air resistance, if the ball is thrown vertically upwards then time of ascent is less than the, time of descent., R : Force due to air friction always acts opposite to the motion of the body., Sol. Answer (1), , path length, , 12. A : The following graph can’t exist actually, , time, , R : Total path length never decreases with time., Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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82, , Motion in a Straight Line, , Solution of Assignment, , 13. A : The displacement (s) time graph shown in the figure represents an accelerated motion., , s, , t, R : Slope of graph increases with time., Sol. Answer (1), 14. A : Average velocity can be zero, but average speed of a moving body can not be zero in any finite time interval., R : For a moving body displacement can be zero but distance can never be zero., Sol. Answer (1), 15. A : For a particle moving in a straight line, its acceleration must be either parallel or antiparallel to velocity., R : A body moving along a curved path may have constant acceleration., Sol. Answer (2), ,   , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Chapter, , 4, , Motion in a Plane, Solutions, SECTION - A, Objective Type Questions, 1., , Which of the following is a vector?, (1) Current, , (2) Time, , (3) Acceleration, , (4) Volume, , Sol. Answer (3), Acceleration is a vector quantity., 2., , The change in a vector may occur due to, (1) Rotation of frame of reference, , (2) Translation of frame of reference, , (3) Rotation of vector, , (4) Both (1) & (3), , Sol. Answer (3), Change in a vector may occur due to rotation of vector and not due to rotation of frame of reference., 3., , Which one of the following pair cannot be the rectangular components of force vector of 10 N?, (1) 6 N & 8 N, , (2) 7 N &, , 51 N, , (3) 6 2 N & 2 7 N, , (4) 9 N & 1 N, , Sol. Answer (4), The vector magnitude =, , Ax 2  Ay 2, , Vector magnitude = 10, But (4) option gives the magnitude, , ⇒, , 92  12  82  10, , [by trial method check options], , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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84, 4., , Motion in a Plane, , Solution of Assignment, , The resultant of two vectors at an angle 150° is 10 units and is perpendicular to one vector. The magnitude, of the smaller vector is, (2) 10 3 units, , (1) 10 units, , (3) 10 2 units, , (4) 5 3 units, , Sol. Answer (2), , ⇒ R 2  A2  B 2, , .....(1), , R = 10, Also tan 30º =, , 1, 3, , , , Perpendicular, Base, , R, A, , B, , R, , 30°, , 150°, , A, , From equation (1) A  10 3, , 102  10 3   B 2, 2, , B = 20, 5., , Two vectors, each of magnitude A have a resultant of same magnitude A. The angle between the two vectors, is, (1) 30°, , (2) 60°, , (3) 120°, , (4) 150°, , Sol. Answer (3), , , , , | A || B || R |, R =, , A2  B 2  2 AB cos , , A2 = A2 + A2 + 2A2cos, –A2 = 2A2cos, cos  = , , 6., , 1, ⇒   120º, 2, , , , , Let  be the angle between vectors A and B . Which of the following figures correctly represents the angle, ?, , B, , A, (1), , B, , , , (2), , B, , , , A, , (3), , , , B, A, , (4), , , A, , Sol. Answer (3), To find angle between vectors, they will be joined either head to head or tail to tail., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 7., , Motion in a Plane, , 85, , A is a vector of magnitude 2.7 units due east. What is the magnitude and direction of vector 4 A ?, , (1) 4 units due east, , (2) 4 units due west, , (3) 2.7 units due east, , (4) 10.8 units due east, , Sol. Answer (4), , A  2.7 iˆ, , , Vector 4A, ⇒ 4(2.7iˆ)  10.8iˆ or 10.8 units due east., 8., , Two forces of magnitude 8 N and 15 N respectively act at a point. If the resultant force is 17 N, the angle, between the forces has to be, (1) 60°, , (2) 45°, , (3) 90°, , (4) 30°, , Sol. Answer (3), R=, , A2  B 2  2 AB cos , , A = 8, B = 15, R = 17, 172 = 82 + 152 + 2 × 8 × 15 × cos , 289 = 64 + 225 + 240 cos , , ⇒ 289 = 289 + 24cos , 24cos = 0, cos = 0 ⇒ = 90º, 9., , A particle is moving in a circle of radius r having centre at O, with a constant speed v. The magnitude of change, in velocity in moving from A to B is, v, B, v, , O, , (1) 2v, , (2) 0, , 60°, , (3), , A, , 3v, , (4) v, , Sol. Answer (4), , , ,  V   2V sin, , ,   2  V  sin ⎛ 60º ⎞, 1, ⎜, ⎟  2  V  ⇒ V | V |, ⎝ 2 ⎠, 2, 2, , 10. Two forces of 10 N and 6 N act upon a body. The direction of the forces are unknown. The resultant force on, the body may be, (1) 15 N, , (2) 3 N, , (3) 17 N, , (4) 2 N, , Sol. Answer (1), The resultant of two vectors always lie between (A + B) & (A – B)., So the resultant of 10 N & 6 N should lie between 16 N & 4 N., So answer is 15 N., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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86, , Motion in a Plane, , Solution of Assignment, , 11. The vector OA where O is origin is given by OA  2iˆ  2 ˆj . Now it is rotated by 45° anticlockwise about O. What, will be the new vector?, (1) 2 2 ĵ, , (3) 2 iˆ, , (2) 2 ĵ, , Sol. Answer (1), , OA  2iˆ  2 jˆ, , OA  4  4 ⇒ 2 2, , (4) 2 2 iˆ, , P(2, 2), , On rotating by an angle of 45º anticlockwise it will lie along y-axis., , So A  2 2 jˆ, , 45°, , 12. A car moves towards north at a speed of 54 km/h for 1 h. Then it moves eastward with same speed for same, duration. The average speed and velocity of car for complete journey is, (1) 54 km/h, 0, , (2) 15 m/s,, , 15, 2, , m/s, , (3) 0, 0, , (4) 0,, , 54, 2, , km/h, , Sol. Answer (2), 54 Km, 54 Kmh, , d = 54 Km, , N, , B, , –1, , W, , t = 1h, , S, , A, , Displacement , , E, , 54 2, Km, , Distance = 2 × 54 = 108 Km, Average speed=, , 108, 5,  54 Kmh1 ,  15 ms1, 2, 18, , Average velocity , , disp. 54 2, 5, 15, , ⇒ 27 2 , ⇒, m/s, time, 2, 18, 2, , 13. If the sum of two unit vectors is also a unit vector, then magnitude of their difference and angle between the, two given unit vectors is, (1), , 3 , 60, , (2), , 3 , 120 , , (3), , 2, 60, , (4), , 2 , 120 , , Sol. Answer (2), , ,  , 2, 2, R  A  B  A  B  2 AB cos , , , , A  B  R 1, 1 = 1 + 1 + 2 × 1 × 1 × cos , cos  = , , 1, ⇒   120º, 2, , ,  , 2, 2, R  A  B  A  B  2 AB cos120º,  , ⎛ 1⎞,  12  12  2  1 1 ⎜  ⎟  3  A  B, ⎝ 2⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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88, , Motion in a Plane, , Solution of Assignment, , 16. A train is running at a constant speed of 90 km/h on a straight track. A person standing at the top of a boggey, moves in the direction of motion of the train such that he covers 1 meters on the train each second. The speed, of the person with respect to ground is, (1) 25 m/s, , (2) 91 km/h, , (3) 26 km/h, , (4) 26 m/s, , Sol. Answer (4), VT = 90 Kmh–1 = 90 , , 5,  25 ms1, 18, , Vm = ?, d = speed × time, dnet = Vnet × t, 1 = (Vm – 25) × 1, Vm = 26 ms–1, 17 . Figure shows two ships moving in x-y plane with velocities VA and VB. The ships move such that B always remains, , VA, north of A. The ratio V is equal to, B, y, , N, , VB, , , S, , B, , A, (1) cos, , E, , W, , x, , VA, , (2) sin, , (3) sec, , (4) cosec, , Sol. Answer (1), If ship B is always north of ship A then, their horizontal component should be equal, so,, VA = VBcos , ⇒, , VA,  cos , VB, , 18. Four persons P, Q, R and S are initially at the four corners of a square of side d. Each person now moves, with a constant speed v in such a way that P always moves directly towards Q, Q towards R, R towards S,, and S towards P. The four persons will meet after time, (1), , d, 2v, , (2), , d, v, , (3), , 3d, 2v, , (4) They will never meet, , Sol. Answer (2), T , , d, v rel, , v rel  v  v cos 90º, , P, , d, v, , v, , =v–0, =v, d, T , v, , S, , v, Q, , v, , R, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 89, , 19. A person, reaches a point directly opposite on the other bank of a flowing river, while swimming at a speed, of 5 m/s at an angle of 120° with the flow. The speed of the flow must be, (1) 2.5 m/s, , (2) 3 m/s, , (3) 4 m/s, , (4) 1.5 m/s, , Sol. Answer (1), For drift to be zero, u = v sin 30º, = 5, , v, , 1, 2, , 30º, , v cos 30, 120º, , u, , v sin 30, , = 2.5 ms–1, , 20. A body of mass 1 kg is projected from ground at an angle 30º with horizontal on a level ground at a speed 50 m/s., The magnitude of change in momentum of the body during its flight is (g = 10 m/s2), (1) 50 kg ms–1, , (2) 100 kg ms–1, , (3) 25 kg ms–1, , (4) Zero, , Sol. Answer (1), , ⇒ The change in momentum = 2mu sin ˆj, , u sin , , u = 50 ms, , –1, , , p  2mu sin , , , = 2 × 1 × 50 × sin 30º, , u cos , , , p = 50 Kg ms–1, , 21. A car with a vertical windshield moves in a rain storm at a speed of 40 km/hr. The rain drops fall vertically, with constant speed of 20 m/s. The angle at which rain drops strike the windshield is, (1) tan–1, , 5, 9, , (2) tan–1, , 9, 5, , (3) tan–1, , 3, 2, , (4) tan–1, , 2, 3, , Sol. Answer (1), , v, tan   m , vr, , 20  5, 9, 20, , Q, , vm, , , ⎛ 5⎞,   tan1 ⎜ ⎟, ⎝ 9⎠, , vr, , ⎛, ⎞, ⎛, ⎞, , 22. Two projectiles are projected at angles ⎜   ⎟ and ⎜   ⎟ with the horizontal, where   , with same speed., 4, 4, ⎝, ⎠, ⎝, ⎠, 4, , The ratio of horizontal ranges described by them is, (1) tan  : 1, , (2) 1 : tan2 , , (3) 1 : 1, , (4) 1 : 3, , Sol. Answer (3), The horizontal range is same when the angles of projection are complimentary to each other., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 91, , for t = 0, Vy = 4, , tan  , , Vy, , , , Vx, , 4, 3, ,  = 53º with horizontal, With vertical,  = 37º, 25. A particle is projected from ground with speed 80 m/s at an angle 30° with horizontal from ground. The magnitude, of average velocity of particle in time interval t = 2 s to t = 6 s is [Take g = 10 m/s2], (1) 40 2 m/s, , (2) 40 m/s, , (4) 40 3 m/s, , (3) Zero, , Sol. Answer (2), Average velocity of the projectile when it is at the same vertical height is : u cos ., , ⇒ 80 × cos 30º ⇒ 40 ms–1., , h, t=2, , t=6, , 26. A stone projected from ground with certain speed at an angle  with horizontal attains maximum height h1., When it is projected with same speed at an angle  with vertical attains height h2. The horizontal range of, projectile is, (1), , h1  h2, 2, , (3) 4 h1h2, , (2) 2h1h2, , (4) h1 + h2, , Sol. Answer (3), When the angles are complimentary the range is same,, h1 , , u 2 sin2 , ,, 2g, , h2 , , u 2 sin2 (90  ), 2g, , u, , , , u 2 sin2 , h1 , 2g, h2 , , u, , h1, R, , (90–), , h2, , R, , u 2 cos2 , 2g, 2, , u 4 sin2  cos2 , 1 1, ⎛ 2u sin  cos  ⎞, h1h2 , ⇒ ⎜, ⎟  4g  4, 2, g, 4g, ⎝, ⎠, , h1h2  R 2, , 1, 16, , ⇒ R 2  16 h1h2, R  4( h1h2 ), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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92, , Motion in a Plane, , Solution of Assignment, , 27. Two objects are thrown up at angles of 45° and 60° respectively, with the horizontal. If both objects attain same, vertical height, then the ratio of magnitude of velocities with which these are projected is, (1), , 5, 3, , (2), , 3, 5, , 2, 3, , (3), , (4), , 3, 2, , Sol. Answer (4), h1 = h2, u12 sin2 45º V22 sin2 60º, , 2g, 2g, , 3, 3, , 3, 2, 2, , , 2, 1, 2, V2, 2, u12, , V1, 3, , V2, 2, 28. For an object projected from ground with speed u horizontal range is two times the maximum height attained, by it. The horizontal range of object is, (1), , 2u 2, 3g, , (2), , 3u 2, 4g, , (3), , 3u 2, 2g, , (4), , 4u 2, 5g, , Sol. Answer (4), R = 24 also,, , H 1,  tan , R 4, 2, , H 1, 1 1,  ⇒  tan , R 2, 2 4, , tan   2 , , 2, , 2u 2 sin  cos , g, , R, , 2u 2 2, 1, ., , g, 5, 5, , R, , 4u 2, 5g, , 29. The velocity at the maximum height of a projectile is, on the horizontal plane is, (1), , 3u 2, 2g, , 1, , 5, , 2, , , 1, , P, B, , R, , +, , =, , (2), , 3u 2, 2g, , 3, times its initial velocity of projection (u). Its range, 2, , (3), , 3u 2, g, , (4), , u2, 2g, , Sol. Answer (1), uh  u cos , , uh, , u cos , , 3, u  u cos , 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , ⇒ cos  , , Motion in a Plane, , 93, , 3, 2, ,  = 30º, R =, , =, , u 2 sin 2, g, u 2 sin 60º, 3u 2, ⇒, R, g, 2g, , 30. A projectile is thrown into space so as to have a maximum possible horizontal range of 400 metres. Taking, the point of projection as the origin, the co-ordinates of the point where the velocity of the projectile is minimum, are, (1) (400, 100), , (2) (200, 100), , (3) (400, 200), , (4) (200, 200), , Sol. Answer (2), Rmax = 400 m, The velocity is minimum at the highest point, , ⇒ H, , R, 2, , (200, 100), 200 N, 400 m, , R = 4H, 400 = 4 × H, H = 100 m, 31. If the time of flight of a bullet over a horizontal range R is T, then the angle of projection with horizontal is, ⎛ gT 2, (1) tan 1 ⎜⎜, ⎝ 2R, , ⎞, ⎟, ⎟, ⎠, , 2, 1 ⎛ 2R, (2) tan ⎜⎜, ⎝ gT, , ⎞, ⎟, ⎟, ⎠, , ⎛ 2R, (3) tan 1 ⎜⎜ 2, ⎝g T, , ⎞, ⎟⎟, ⎠, , 1 ⎛ 2R ⎞, ⎟⎟, (4) tan ⎜⎜, ⎝ gT ⎠, , Sol. Answer (1), , T , , 2u sin , gT, ⇒u , 2sin , g, , R, , 2u 2 sin  cos , g, , R, , 2u sin ,  u cos , g, , R = T × u cos , R T , , R, , gT cos , 2sin , , gT 2 1, 2 tan , , tan  , , gT 2, 2R, , 2⎞, ⎛,   tan1 ⎜ gT ⎟, ⎝ 2R ⎠, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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94, , Motion in a Plane, , Solution of Assignment, , 32. In the graph shown in figure, which quantity associated with projectile motion is plotted along, y-axis, y-axis, , x-axis, t, (1) Kinetic energy, , (2) Momentum, , Sol. Answer (3), , (3) Horizontal velocity, , (4) None of these, , y-axis, , It is the horizontal component of, velocity that remains constant, throughout the motion as there is no, acceleration in that direction ax = 0,, ux = constant, , t, , x-axis, , 33. The equation of a projectile is y = ax – bx2. Its horizontal range is, (1), , a, b, , (2), , b, a, , (3) a + b, , (4) b – a, , Sol. Answer (1), y = ax – bx2, When the body lands then y = 0, x = R, 0 = aR – bR2, aR = bR, R, , y=0, , R, , a, b, , u, 30°, , 34. Figure shows a projectile thrown with speed u = 20 m/s at an angle, 30° with horizontal from the top of a building 40 m high. Then the, horizontal range of projectile is, (1) 20 3 m, , (2) 40 3 m, , (3) 40 m, , (4) 20 m, , 40 m, , Sol. Answer (2), Sy  uyT , , 1, g yT 2, 2, , –40 = 4 sin30T , , 1 2, gT, 2, , 1, 2, –40 = 20  T  5T, 2, –8 = 2T – T2, , uy = 4 sin 30º, u = 20 ms–1, 30º, ux = u cos 30º, , T2 – 2T – 8 = 0, 40 m, , T2 – 4T + 2T – 8 = 0, T = –2, 4, R  u cos T = 20 , , 3, 4, 2, , R  40 3 m, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 95, , 35. When a particle is projected at some angle to the horizontal, it has a range R and time of flight t1. If the same, particle is projected with the same speed at some other angle to have the same range, its time of flight is, t2, then, (1) t1  t 2 , , 2R, g, , (2) t1  t 2 , , R, g, , (3) t1t 2 , , 2R, g, , (4) t1t 2 , , R, g, , Sol. Answer (3), The angles has to be complimentary i.e., if 1  , 2  (90  ), , t1 , , 2u sin , 2u sin(90  ), , t2 , g, g, , t2 , , 2u cos , g, , t1t2 , , 2u sin  2u cos , , g, g, , t1t2 , , 2R, g, , 36. A projectile is thrown with velocity v at an angle  with horizontal. When the projectile is at a height equal to, half of the maximum height, the vertical component of the velocity of projectile is, (1) v sin  × 3, , (2), , v sin, 3, , v sin, , (3), , (4), , 2, , v sin, 3, , Sol. Answer (3), 2g ⎛ u 2 sin  ⎞, ⎜, ⎟, 2 ⎝ 2g ⎠, , v sin , , v B2  v 2 sin2  , , v 2 sin2 , v B2 , 2, vB , , vB, v, , , v sin , , v cos , , 2, , 37. In the given figure for a projectile, , u, , P, , , x1, ⎡ x1x 2 ⎤, (1) y  ⎢ x  x ⎥ tan θ, 2⎦, ⎣ 1, , ⎡ x1x 2 ⎤, (2) y  ⎢, ⎥ tan θ, ⎣ x1  x 2 ⎦, , x2, ⎡ 2 x1x 2 ⎤, (4) y  ⎢ x  x ⎥ tan θ, 2⎦, ⎣ 1, , ⎡ 2 x1x 2 ⎤, (3) y  ⎢, ⎥ cos θ, ⎣ x1  x 2 ⎦, , Sol. Answer (2), The equation of trajectory for point 'P' can be written as :, x1 ⎞, x⎞, ⎛, ⎛, ⎛ x  x2  x1 ⎞, = x1 tan  ⎜ 1, y = x tan  ⎜ 1  ⎟ = x1 tan  ⎜ 1 , ⎟, ⎟, x1  x2 ⎠, ⎝ R⎠, ⎝, ⎝ x1  x2 ⎠, , x1x2, y = x  x tan , 1, 2, , u, , , P, , x1, , y, , P, x2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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96, , Motion in a Plane, , Solution of Assignment, , 38. Two paper screens A and B are separated by distance 100 m. A bullet penetrates A and B, at points P and, Q respectively, where Q is 10 cm below P. If bullet is travelling horizontally at the time of hitting A, the velocity, of bullet at A is nearly, (1) 100 m/s, , (2) 200 m/s, , (3) 600 m/s, , (4) 700 m/s, , Sol. Answer (4), 10 cm ⇒ 10 × 10–2 m ⇒ 10–1 ⇒ 0.1 m, It is a case of horizontal projectile., So, ax = 0, ux = 4, uy = 0, ay = –g, R = 100m, T , , 2H, ⇒ Time of flight, g, , A, , B, , P, 10 cm, Q, 100 m, , R = uxT, 100 = u, , u, , 2  0.1, u 2, ⇒,  100, 100, 10, , 1000,  707 ms1, 2, , 39. A car is going round a circle of radius R1 with constant speed. Another car is going round a circle of radius, R2 with constant speed. If both of them take same time to complete the circles, the ratio of their angular, speeds and linear speeds will be, , (1), , R1 R1, ,, R2 R2, , (2) 1, 1, , (3) 1,, , R1, R2, , (4), , R1, ,1, R2, , Sol. Answer (3), The angular speed is given by, , , 2, T, , , , , T, 1, ⇒ 1  2, T,  2 T1, , if T1 = T2 ⇒ 1 = 2, So, ratio ⇒ 1 : 1, and linear speed v = R, V R, , V1 R1, , V2 R2, 40. A body revolves with constant speed v in a circular path of radius r. The magnitude of its average acceleration, during motion between two points in diametrically opposite direction is, (1) Zero, , (2), , v2, r, , (3), , 2v 2, r, , (4), , v2, 2r, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 97, , Sol. Answer (3), , aavg, , aavg, , ⎛ ⎞, 2v sin ⎜ ⎟, ⎝ 2⎠, , ⎛ r ⎞, ⎜⎝ ⎟⎠, v, , 2v 2 sin ⎛⎜ ⎞⎟, 2, ⎝, ⎠, , r, , 180º, , B, , A, , Here,  =  rad, aavg, , , 2v 2 sin ⎛⎜ ⎞⎟, ⎝2⎠, , r , , aavg , , 2v 2, r, , 41. An object of mass m moves with constant speed in a circular path of radius R under the action of a force of, constant magnitude F. The kinetic energy of object is, (1), , 1, FR, 2, , (2) FR, , (3) 2FR, , (4), , 1, FR, 4, , (4), , , rad/s, 1800, , Sol. Answer (1), 1 F v2, 1F, 1, 1, 2, 2, , v, FR, =, KE = mv =, 2 ⎞ =, 2, ⎛, 2, a, 2, 2, v, ⎜, ⎟, ⎝R ⎠, , 42. The angular speed of earth around its own axis is, (1), , , rad/s, 43200, , (2), , , rad/s, 3600, , (3), , , rad/s, 86400, , Sol. Answer (1), Angular speed =, , 2, T, , T  Time period of earth = 24 h, =, , 2, , , rad s1, 24  60  60 43200, , 43. A particle moves in a circle of radius 25 cm at two revolutions per second. The acceleration of the particle is, (in m/s2), (1) 2, , (2) 82, , (3) 42, , (4) 22, , Sol. Answer (3), a = r2, a=, , 25,  2  2 2, 100, , a = 42 m/s2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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98, , Motion in a Plane, , Solution of Assignment, , 44. A particle is revolving in a circular path of radius 25 m with constant angular speed 12 rev/min. Then the angular, acceleration of particle is, (1) 22 rad/s2, , (2) 42 rad/s2, , (3) 2 rad/s2, , (4) Zero, , Sol. Answer (4), , , , Angular acceleration is the rate of change of angular speed or angular velocity if  remains constant then, 0, , 45. Two particles are moving in circular paths of radii r1 and r2 with same angular speeds. Then the ratio of their, centripetal acceleration is, (1) 1 : 1, , (2) r1 : r2, , (4) r22 : r12, , (3) r2 : r1, , Sol. Answer (2), Centripetal acceleration is given by, , a, , v2,  r 2, r, , For same '', ac  r ⇒, , a1 r1, , a2 r2, , 46. A particle P is moving in a circle of radius r with uniform speed v. C is the centre of the circle and AB is, diameter. The angular velocity of P about A and C is in the ratio, (1) 4 : 1, , (2) 2 : 1, , (3) 1 : 2, , (4) 1 : 1, , Sol. Answer (3), P /C, , P, , d, , dt, , P / A , , 1 d, 2 dt, , P / A , , 1, P /C, 2, , , , /2, C, , B, , P / A 1,   1: 2, P /C 2, , 47. A car is moving at a speed of 40 m/s on a circular track of radius 400 m. This speed is increasing at the, rate of 3 m/s2. The acceleration of car is, (1) 4 m/s2, , (2) 7 m/s2, , (3) 5 m/s2, , (4) 3 m/s2, , Sol. Answer (3), v = 40 ms–1, r = 400 m, aT = 3 ms–2, , ac , , V 2 40  40, ,  4 ms2, r, 400, , a  aC2  aT2, a=, , 42  32  5 ms2, , a = 5 ms–2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 99, , 48. Four particles A, B, C and D are moving with constant speed v each. At the instant shown relative velocity of A, with respect to B, C and D are in directions, B, , A, , C, , D, (1), , (2), , (3), , (4), , Sol. Answer (1), vC, , vA, , (i), , vA, , –vC, , ,  , v AC  v A  vC ⇒, , vA – vC, , v A  vC, , vB, , , , , , , ⇒ v AB  v A  v B ⇒ v A  ( v A ), , (ii) vA, , v B, vA, , v A  ( vB ), vA, , (iii), vD, , , , , , , , , , , , ⇒ vAD  vA  vD  vA  (vD ), , v A  vD, vA, , ⇒, , v D, 49. The ratio of angular speeds of minute hand and hour hand of a watch is, (1) 6 : 1, , (2) 12 : 1, , (3) 60 : 1, , (4) 1 : 60, , Sol. Answer (2), mh = Angular speed of minute hand, hh = Angular speed of hour hand, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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100, , Motion in a Plane, , Solution of Assignment, , mh =, , 2, 2, , rad s1, 60m 60  60, , hh, , =, , 2, 2, , rad s1, 12 h 12  60  60, , mh, , hh, , 2, 60  60  1  12, 2, 1 1, 12  60  60, , mh : hh ⇒ 12 : 1, 50. If  is angle between the velocity and acceleration of a particle moving on a circular path with decreasing speed,, then, (1)  = 90°, , (2) 0° <  < 90°, , (3) 90° <  < 180°, , (4) 0°    180°, , Sol. Answer (3), V, , ac, a, aT, ,  between v & Q is, 90º <  < 180º, 51. If speed of an object revolving in a circular path is doubled and angular speed is reduced to half of original, value, then centripetal acceleration will become/remain, (1) Same, , (2) Double, , (3) Half, , (4) Quadruple, , Sol. Answer (1), ac = r2 = (r)(), ac = v, , ⎛⎞, ac = (2v) ⎜ ⎟  v   ac, ⎝2⎠, 52. An object is projected from ground with speed u at angle  with horizontal. the radius of curvature of its, trajectory at maximum height from ground is, u 2 sin 2, g, , (1), , (2), , u 2 cos2 , g, , (3), , u 2 sin2 , g, , (4), , u 2 sin2 , 2g, , Sol. Answer (2), ac , , v2, r, , v 2 u 2 cos2 , r  ,, ac, g, , r , , u cos , , = 90º, g, , u 2 cos2 , g, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 101, , SECTION - B, Objective Type Questions, 1., , Two particles A and B start moving with velocities 20 m/s and 30 2 m/s along x-axis and at an angle 45º with, x-axis respectively in xy-plane from origin. The relative velocity of B w.r.t. A, (1) (10iˆ  30 ˆj ) m/s, , (2) (30 iˆ  10 jˆ) m/s, , (3) (30 iˆ  20 2 ˆj )m/s, , (4) (30 2 iˆ  10 2 ˆj ) m/s, , Sol. Answer (1), vA = 20 m/s, , v B  30 2 m/s along 45º with x-axis, , v B  v B cos 45º iˆ  v B sin 45º jˆ  30iˆ  30 jˆ, , , , v BA  v B  v A  30iˆ  30 jˆ  20iˆ, , , vB, , 45º, , , v BA  10iˆ  30 jˆ, , 2., , A particle is projected at angle  with horizontal from ground. The slop (m) of the trajectory of the particle, varies with time (t) as, m, , m, , m, , (1), , t, , O, , (2), , t, , O, , (3), O, , m, , t, , (4), O, , t, , Sol. Answer (1), Slope of trajectory, , tan  , , u sin   gt, u cos , , So, m , , u sin , gt, , u cos  u cos , , m  tan  , , g, t, u cos , , y, ,  y = a – bx, , x, , Therefore,, , m, , O, , t, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 5., , Motion in a Plane, , 103, , A particle projected from ground moves at angle 45º with horizontal one second after projection and speed is, minimum two seconds after the projection. The angle of projection of particle is [Neglect the effect of air resistance], (1) tan–1(3), , (3) tan1( 2), , (2) tan–1(2), , (4) tan–1(4), , Sol. Answer (2),  = 45º, t = 1 s, tan  , , Vy, Uy, , tan 45º , , , , u sin   gt, u cos , , u sin   g  1, ⇒ u cos   u sin   g, u cos , , also, Vy = 0, after 1st (as speed is minimum), u sin   g  2  0, , , , u sin   2g, , ...(i), , so, u cos   2g  g, u cos   g, , so,, , ...(ii), , (i) u sin  2g, , , (ii) u cos , g, ,  tan   2,   tan 1(2), , 6., , A ball is projected from ground at an angle 45º with horizontal from distance d1 from the foot of a pole and just, after touching the top of pole it the falls on ground at distance d2 from pole on other side, the height of pole is, (1) 2 d1d 2, , (2), , d1  d 2, 4, , 2 d1 d 2, (3) d  d, 1, 2, , (4), , d1 d 2, d1  d 2, , Sol. Answer (4), Repeated., tan   tan   tan , , º, 45, =, , , , y, y, ,  tan 45º, d1 d 2, , d1, , ⎛ dd ⎞, y⎜ 1 2 ⎟, ⎝ d1  d 2 ⎠, , 7., , , d2, , A particle is projected with speed u at angle  with horizontal from ground. If it is at same height from ground, at time t1 and t2, then its average velocity in time interval t1 to t2 is, (1) Zero, , (2) u sin , , (3) u cos , , (4), , 1, u cos , 2, , Sol. Answer (3), When projectile is at same height, average velocity = u cos., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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104, 8., , Motion in a Plane, , Solution of Assignment, , A particle is projected from ground at an angle  with horizontal with speed u. The ratio of radius of curvature, of its trajectory at point of projection to radius of curvature at maximum height is, (1), , 1, sin  cos , , (2) cos2 , , 2, , (3), , 1, sin3 , , (4), , 1, cos3 , , Sol. Answer (4), At the point of projection, rA , , u2, g cos , 2, , u, , 2, , u cos , rH , g, , A, , H, , u = cos, , , , u2, rA, r, 1, g cos ,  2, ,  A, 2, 3, Ratio, rH, cos  rH, u cos , g, 9., , An object of mass 10 kg is projected from ground with speed 40 m/s at an angle 60º with horizontal. The rate, of change of momentum of object one second after projection in SI unit is [Take g = 9.8 m/s2], (1) 73, , (2) 98, , (3) 176, , (4) 140, , Sol. Answer (2), Force =, , p, , force remains constant = mg, t, ,  10 × 9.8, , , , 98 N, , At t = 1, particle is at its maximum height., 10. An object is projected from ground with speed 20 m/s at angle 30º with horizontal. Its centripetal acceleration, one second after the projection is [Take g = 10 m/s2], (1) 10 m/s2, , (2) Zero, , (3) 5 m/s2, , (4) 12 m/s2, , Sol. Answer (1), Centripetal acceleration =, , v2,  g  10 ms2, r, , 11. A particle is moving on a circular path with constant speed v. It moves between two points A and B, which, subtends an angle 60º at the centre of circle. The magnitude of change in its velocity and change in magnitude, of its velocity during motion from A to B are respectively, (1) Zero, Zero, , (2) v, 0, , (3) 0, v, , (4) 2v, v, , Sol. Answer (2), , v  2v sin, , , 2, , Change in magnitude of velocity = 0, , ⎛ 60 ⎞,  2v  sin ⎜ ⎟, ⎝ 2⎠, | v |  v, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 105, , 12. A particle is moving with constant speed v in xy plane as shown in figure. The magnitude of its angular velocity, about point O is, y, , v, , (0,b), , v, , (1), , 2, , a b, , (2), , 2, , x, , (a,0), , O, v, b, , (3), , vb, (a  b 2 ), , Sol. Answer (3), , v cos, r,  v, , v sin   r , , (0, b), , v sin   a 2  b 2  , v, , b, , 2, , a b, , 2, , vb, (a 2  b 2 ), , 2, , a b, , 2, , , , r, O, , b, , a, , , , (4), , 2, , v, a, , v  r , , v sin, , (a, 0), , 13. A particle is moving in xy-plane in a circular path with centre at origin. If at an instant the position of particle is, 1 ˆ ˆ, (i  j ), then velocity of particle is along, given by, 2, (1), , 1 ˆ ˆ, (i  j ), 2, , (2), , 1, 2, , ( jˆ  iˆ), , (3), , 1 ˆ ˆ, (i  j ), 2, , (4) Either (1) or (2), , Sol. Answer (4), ,  1, 1 ˆ, r  iˆ , j, 2, 2,  , v  r  0 as velocity is always tangential to the path., 1, (v x iˆ  v y jˆ)  (iˆ  ˆj )  0, 2, , vx  vy  0, , v  v x2  v y2, , vy  , or, , ⇒, , v x  v y, , or, , v y  v x, , , , 2v x  v, , , , vx , , v, 2, , v, 2, , vx  , , v, 2, , , vy , , v, 2, , So, possible value of v  v x iˆ  v y ˆj , , v ˆ v ˆ, v ˆ v ˆ, i , j or, i, j, 2, 2, 2, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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106, , Motion in a Plane, , Solution of Assignment, , 14. A particle is moving eastwards with a speed of 6 m/s. After 6 s, the particle is found to be moving with same speed, in a direction 60° north of east. The magnitude of average acceleration in this interval of time is, , N, , W, , 6 m/s, 60°, , 6 m/s, , E, , S, (1) 6 m/s2, , (2) 3 m/s2, , (3) 1 m/s2, , (4) Zero, , Sol. Answer (3), , , , ⎛ 60 ⎞, | v |  2v sin  2v sin ⎜ ⎟  2v sin30º  v, ⎝ 2⎠, 2, , 6 ms–1, 60º, 6 m/s, , , | v |  6 ms1, , aav , , , ( v ), t, , t = 6 s, so, aav , , 6,  1 ms2, 6, , 15. What is the path followed by a moving body, on which a constant force acts in a direction other than initial velocity, (i.e. excluding parallel and antiparallel direction)?, (1) Straight line, , (2) Parabolic, , (3) Circular, , (4) Elliptical, , Sol. Answer (2), The path will be parabolic., 16. Two stones are thrown with same speed u at different angles from ground in air. If both stones have same range, and height attained by them are h1 and h2, then h1 + h2 is equal to, (1), , u2, g, , (2), , u2, 2g, , (3), , u2, 3g, , (4), , u2, 4g, , Sol. Answer (2), If range is same then, one angle is  and other angle is (90 – ),  h1 , , h1 , , u 2 sin2 , u 2 sin2 (90   ), , h2 , 2g, 2g, u 2 sin2 , u 2 cos2 , , h2 , 2g, 2g, , So, h 1  h 2 ⇒, , h1  h2 , , u 2 sin2  u 2 cos2  u 2, , , (sin2   cos2 ), 2g, 2g, 2g, , u2, 2g, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 107, , 17. When a force F acts on a particle of mass m, the acceleration of particle becomes a. Now if two forces of, magnitude 3F and 4F acts on the particle simultaneously as shown in figure, then the acceleration of the, particle is, 4F, , 90º, m, (1) a, , (2) 2a, , 3F, (3) 5a, , (4) 8a, , Sol. Answer (3), F, a, m, , Fnet  32  42  5 F, , 4F, , So, Fnet = ma, 5F = ma,  a , , 90º, , 5F, m, , m, , 3F, , a   5a, 18. Consider the two statements related to circular motion in usual notations,  , , A. In uniform circular motion  , v and a are always mutually perpendicular,  , , B. In non-uniform circular motion,  , v and a are always mutually perpendicular, (1) Both A and B are true, , (2) Both A and B are false, , (3) A is true but B is false, , (4) A is false but B is true, , Sol. Answer (3), , v, , a, , , , Only first statement is correct.,   mutually perpendicular to v and a., 19. Which of the following quantities remains constant during uniform circular motion?, (1) Centripetal acceleration, , (2) Velocity, , (3) Momentum, , (4) Speed, , Sol. Answer (4), Speed remains constant., 20. A projectile is projected with speed u at an angle  with the horizontal. The average velocity of the projectile, between the instants it crosses the same level is, (1) u cos , , (2) u sin , , (3) u cot , , (4) u tan , , Sol. Answer (1), Repated., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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108, , Motion in a Plane, , Solution of Assignment, , 21. A ball is thrown at an angle  with the horizontal. Its horizontal range is equal to its maximum height. This is, possible only when the value of tan  is, (1) 4, , (2) 2, , (3) 1, , (4) 0.5, , Sol. Answer (1), H 1,  tan , R 4,  H = R, given,, , tan   4, , ,   tan 1(4), , 22. A ball is projected from a point O as shown in figure. It will strike the ground after (g = 10 m/s2), 10 m/s, 30°, O, 60 m, , (1) 4 s, , (2) 3 s, , (3) 2 s, , (4) 5 s, , Sol. Answer (1), , sy  u xT , , 1, ayT 2, 2, , 60  10 sin30º T , , –1, , 10 ms, , 1 2, gT, 2, , , , 60  5T  5T 2, , 60 m, , T 2 T  2  0, , T 4s, 23. A particle is thrown with a velocity of u m/s. It passes A and B as shown in figure at time t1 = 1 s and t2 = 3 s. The, value of u is (g = 10 m/s2), y, , u, 30º, O, , (1) 20 m/s, , (2) 10 m/s, , A, , B, x, , (3) 40 m/s, , (4) 5 m/s, , Sol. Answer (3), , t1  t2 , , 2u sin , g, , 2u  sin30º, 10, 20 × 2 = u, 1 3 , , , , u  40 ms1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 109, , 24. Which one of the following statements is not true about the motion of a projectile?, (1) The time of flight of a projectile is proportional to the speed with which it is projected at a given angle of, projection, (2) The horizontal range of a projectile is proportional to the square root of the speed with which it is projected, (3) For a given speed of projection, the angle of projection for maximum range is 45°, (4) At maximum height, the acceleration due to gravity is perpendicular to the velocity of the projectile, Sol. Answer (2), R, , u 2 sin 2,  R  u2, g, , 25. Out of the two cars A and B, car A is moving towards east with a velocity of 10 m/s whereas B is moving towards, north with a velocity 20 m/s, then velocity of A w.r.t. B is (nearly), (1) 30 m/s, , (2) 10 m/s, , (3) 22 m/s, , (4) 42 m/s, , Sol. Answer (3), , , , v AB  v A  v B, –1, , vB = 20 ms, , v AB  v A2  v B2, , vA = 10 ms–1, , , | v AB |  102  202  100  400  500  22 ms 1, , 26. A projectile is thrown with speed 40 ms–1 at angle  from horizontal. It is found that projectile is at same height, at 1 s and 3 s. What is the angle of projection?, –1 ⎛ 1 ⎞, ⎟, (2) tan ⎜, ⎝ 3⎠, , –1 ⎛ 1 ⎞, ⎟, (1) tan ⎜, ⎝ 2⎠, , –1, (3) tan, ,  3, , –1, (4) tan, ,  2, , Sol. Answer (2), , tan  , , vy, vx, , Also, t1  t2 , 4, , 2u sin , g, , 2  40  sin , 10, , sin  , , 1, ⇒   30º, 2, , So, tan   tan30º ⇒, , 1, 3, , ⎛ 1 ⎞,   tan1 ⎜, ⎝ 3 ⎟⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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110, , Motion in a Plane, , Solution of Assignment, , SECTION - C, Previous Years Questions, 1., , A ship A is moving Westwards with a speed of, 10 km h–1 and a ship B 100 km South of A, is moving, –1, Northwards with a speed of 10 km h . The time after which the distance between them becomes shortest,, is, [AIPMT-2015], (1) 10 2 h, , (2) 0 h, , (3) 5 h, , (4) 5 2 h, , Sol. Answer (3), 2., , A projectile is fired from the surface of the earth with a velocity of 5 ms–1 and angle  with the horizontal. Another, projectile fired from another planet with a velocity of 3 ms–1 at the same angle follows a trajectory which is identical, with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet, is (in ms–2) is (Given g = 9.8 ms–2), [AIPMT-2014], (1) 3.5, , (2) 5.9, , (3) 16.3, , (4) 110.8, , Sol. Answer (1), Since trajectory is same, so range and maximum height both will be identical from earth and planet. So, equating maximum height (Answer can be obtained by equating range also), 2, 2, ue2 sin2  u p sin , , 2ge, 2g p, , 2.5, 9, , 9.8 g p, , gp = 3.5 m/s2, 3., , A particle is moving such that its position coordinates (x, y) are, (2 m, 3 m) at time t = 0,, (6 m, 7 m) at time t = 2 s and, (13 m, 14 m) at time t = 5 s., , , Average velocity vector (v av ) from t = 0 to t = 5 s is, , (1), , , , 1, 13iˆ  14 ˆj, 5, , , , (2), , 7 ˆ ˆ, (i  j ), 3, , [AIPMT-2014], (3) 2(iˆ  ˆj ), , (4), , 11 ˆ ˆ, (i  j ), 5, , Sol. Answer (4), 4., , The velocity of a projectile at the initial point A is (2iˆ  3 jˆ) m/s. Its velocity (in m/s) at point B is, [NEET-2013], , Y, , A, (1) 2iˆ  3 jˆ, , (2) 2iˆ  3 jˆ, , B, , X, , (3) 2iˆ  3 ˆj, , (4) 2iˆ  3 jˆ, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 111, , Sol. Answer (2), The change is only in the y-component, So, v f  2iˆ  3 ˆj, 5., , ∵ ax  0, , The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile, is, [AIPMT (Prelims)-2012], (2)  = 45º, , (1)  = tan–1(2), , ⎛ 1⎞, (3)   tan–1 ⎜ ⎟, ⎝4⎠, , (4)  = tan–1(4), , Sol. Answer (4), H=R, tan = 4, , H 1, ⎛, ⎞,  tan ⎟, ⎜⎝∵, ⎠, R 4, ,   tan 1(4), , 6., , A particle has initial velocity, , , , , ,  2i  3 j , , , , and acceleration 0.3i  0.2 j . The magnitude of velocity after, , , , 10 s will be, (1) 5 units, , , , [AIPMT (Prelims)-2012], (2) 9 units, , (3) 9 2 units, , (4) 5 2 units, , Sol. Answer (4), , , , u  3iˆ  3 ˆj , a  0.3iˆ  0.2 ˆj, t = 10 s,   , v  u  at, , , v  2iˆ  3 ˆj  (0.3iˆ  0.2 ˆj )  10,  2iˆ  3 ˆj  3iˆ  2 ˆj, , v  5iˆ  5 jˆ, v  52  52  50, , v ⇒ 5 2 ms1, 7., , A particle moves in a circle of radius 5 cm with constant speed and time period 0.2  s. The acceleration of, the particle is, [AIPMT (Prelims)-2011], (1) 5 m/s2, , (2) 15 m/s2, , (3) 25 m/s2, , (4) 36 m/s2, , Sol. Answer (1), r = 5 cm, v = ?, T = 0.2  s, , T , , 2, 20, ⇒ ,  10 rad s 1, , 0.2, , a  r  2  5  10 2  100, , a  5 ms2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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112, 8., , Motion in a Plane, , Solution of Assignment, , A body is moving with velocity 30 m/s towards east. After 10 s its velocity becomes 40 m/s towards north., The average acceleration of the body is, [AIPMT (Prelims)-2011], (1) 5 m/s2, , (2) 1 m/s2, , (3) 7 m/s2, , (4), , 7 m/s2, , Sol. Answer (1), aav , , , | v |, t, , , , v  (v 2  v1 )  v 22  v12, , –1, , 40 ms, , (∵   90º ), ,  5 302  402  50 ms1, so, aav , , 9., , –1, , 30 ms, , 50, 10, , 5 ms 2  aav, , A missile is fired for maximum range with an initial velocity of 20 m/s. If g = 10 m/s2, the range of the, missile is, [AIPMT (Prelims)-2011], (1) 20 m, , (2) 40 m, , (3) 50 m, , (4) 60 m, , Sol. Answer (2), For maximum range  = 45º, v = 20 ms–1, R, , u 2 20  20, , g, 10, , [∵   45º ], , R  40 m, 10. A particle of mass m is released from rest and follows a parabolic path as shown. Assuming that the, displacement of the mass from the origin is small, which graph correctly depicts the position of the particle, as a function of time?, [AIPMT (Prelims)-2011], v(x), , m, (x), , 0, , (1), , 0, , x(t), , x(t), , x(t), t, , (2), , 0, , t, , (3), , 0, , x(t), , t, , (4), , t, , 0, , Sol. Answer (2), 11. A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as, seen from the point of projection is, [AIPMT (Mains)-2011], , ⎛ 3⎞, (1) tan ⎜⎜, ⎟⎟, ⎝ 2 ⎠, , (2) 45°, , (3) 60°, , 1, (4) tan, , 1, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 113, , Sol. Answer (4), , 1, H, tan  , 4, R, 4H = R, , B, ...(i), , H, , H 2H, , In ABC, tan  , R, R, 2, , tan  , , A, , , , 45º, , R, 2, , C, , 2H 1, , 4H 2, , 1,    tan1 ⎛⎜ ⎞⎟, ⎝ 2⎠, , 12. Six vectors, â through fˆ have the magnitudes and directions indicated in the figure. Which of the following, statements is true?, [AIPMT (Prelims)-2010], , b, , a, d, ,   , (1) b  c  f, ,   , (2) d  c  f, , c, f, , e, ,   , (3) d  e  f, ,   , (4) b  e  f, , Sol. Answer (3), , , d, , f, , , e,   , d e  f, , 13. The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is, [AIPMT (Mains)-2010], (1) 60º, , (2) 15º, , (3) 30º, , (4) 45º, , Sol. Answer (1), uH  u cos , , u,  u  cos , 2, cos  , , 1, 2, ,   60º, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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114, , Motion in a Plane, , Solution of Assignment, , 14. A particle moves in x-y plane according to rule x = a sin t and y = a cost. The particle follows, [AIPMT (Mains)-2010], (1) An elliptical path, (2) A circular path, (3) A parabolic path, (4) A straight line path inclined equally to x and y-axes, Sol. Answer (2), x  a sin t ⇒ x 2  a 2 sin2 t, , y  a cos t ⇒ y 2  a2 cos2 t, , x 2  y 2  a2 (sin2 t  cos2 t ), x 2  y 2  a 2  equation of circle., , 15. A particle has initial velocity (3iˆ  4 ˆj ) and has acceleration (0.4iˆ  0.3 ˆj ) . Its speed after 10 s is, [AIPMT (Prelims)-2010], (1) 7 units, , (2) 7 2 units, , (3) 8.5 units, , (4) 10 units, , Sol. Answer (2), 16. A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle, lands on the level ground the magnitude of the change in its momentum will be, [AIPMT (Prelims)-2008], (1) Zero, , (2) 2 mv, , (3), , mv, 2, , (4) mv 2, , Sol. Answer (4), , p  2mv sin ˆj, , 1, | p |  2mv sin   2mv , 2, , | p |  2 mv, 17. A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane. Its coordinates at a later, time are ( 3,3) . The path of the particle makes with the x-axis an angle of, , [AIPMT (Prelims)-2007], , (1) 0°, , (4) 60°, , (2) 30°, , (3) 45°, , Sol. Answer (4), , tan  , , P ( 3, 3), , P, 3, , B, 3, ,   60º, , , 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , 18., , Motion in a Plane, , 115, ,  ,  , A and B are two vectors and  is the angle between them, if A  B  3( A  B ) , the value of  is, [AIPMT (Prelims)-2007], (1) 90°, , (2) 60°, , (3) 45°, , (4) 30°, , Sol. Answer (2), 19. For angles of projection of a projectile at angles (45°– ) and (45° + ), the horizontal ranges described by, the projectile are in the ratio of, [AIPMT (Prelims)-2006], (1) 1 : 1, , (2) 2 : 3, , (3) 1 : 2, , (4) 2 : 1, , Sol. Answer (1), Repeated., 20. A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The, average velocity and average speed for each circular lap respectively is, [AIPMT (Prelims)-2006], (1) 0, 0, , (2) 0, 10 m/s, , (3) 10 m/s, 10 m/s, , (4) 10 m/s, 0, , Sol. Answer (2), T = 62.8 s, r = 100 m, , T , , 2, , , , , 2 2  3.14  10, , T, 62.8  100, , 0, 10, , m, ,   0.1 rad s1, v = r, v = 100 × 0.1, , v  10 ms1, Average velocity = 0, Average speed = 10 ms 1, , 21. The vectors A and B are such that: | A  B |  | A – B | . The angle between the two vectors is, [AIPMT (Prelims)-2006], (1) 90°, , (2) 60°, , (3) 75°, , (4) 45°, , Sol. Answer (1),  ,  , | AB|  | AB|, A2  B 2  2 AB cos  , , A2  B 2  2 AB cos , , Squaring both the sides,, , , A2  B 2  2 AB cos   A2  B 2  2 AB cos , , 4 AB cos   0,  cos   0, , , ,   90º, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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116, , Motion in a Plane, , Solution of Assignment, , 22. If a vector 2iˆ  3 ˆj  8kˆ is perpendicular to the vector 4 ˆj  4iˆ  kˆ , then the value of  is, [AIPMT (Prelims)-2005], (1) –1, , (2), , 1, 2, , 1, 2, , (3) , , (4) 1, , Sol. Answer (3), 23. A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the, stone makes 22 revolutions in 44 s, what is the magnitude and direction of acceleration of the stone?, [AIPMT (Prelims)-2005], , 2, ms–2 and direction along the radius towards the centre, 4, (2) 2 ms–2 and direction along the radius away from centre, , (1), , (3) 2 ms–2 and direction along the radius towards the centre, (4) 2 ms–2 and direction along the tangent to the circle, Sol. Answer (3),  = 22 r/s  22 × 2, , 22  2, ⇒  rad s 1, 44 s, a = r2, , a  1   2 ms2, , centripetal acceleration., , 24. Two boys are standing at the ends A and B of a ground, where AB = a. The boy at B starts running in a, direction perpendicular to AB with velocity v1. The boy at A starts running simultaneously with velocity v and, catches the other boy in a time t, where t is, [AIPMT (Prelims)-2005], a, , (1), , v, , 2, , (2), ,  v12, , a2, v, , 2, , (3), ,  v 12, , a,  v  v1 , , (4), , a, v, ,  v1 , , C, , Sol. Answer (2), The distance travelled by body at B,, , v, , v1, , = Speed × t, , A, , = v1t, , a, , C, , So, BC = v1t, similarly, AC = vt, Applying pythagoras in ABC,, 2 2, , v t , , v12t 2, , a, , t, , vt, , v1t, , 2, , (v 2  v12 )t 2  a2, t2 , , B, , A, , a, , B, , a2, v 2  v12, , a, 2, , v  v12, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 117, , , ,   , 25. If the angle between the vectors A and B is , the value of the product (B  A)  A is equal to, [AIPMT (Prelims)-2005], (1) BA2 cos, , (2) BA2 sin, , (3) BA2 sincos, , (4) Zero, , Sol. Answer (4), 26. A boat is sent across a river with a velocity of 8 km/h. If the resultant velocity of the boat is, 10 km/h, then velocity of the river is, (1) 8 km/h, , (2) 10 km/h, , (3) 12.8 km/h, , Sol. Answer (4), , (4) 6 km/h, , u, , vr  v 2  u2, vr = 10 kmh–1, v = 8 kmh–1, , v, , u=?, ,  , vr  v  u, , 100  82  uR2,  uR2  36, , , uR  6 km/h, , , , 27. Which of the following is correct relation between an arbitrary vector A and null vector 0 ?,     ,     ,     , (1) A  0  A  0  A, (2) A  0  A  0  A, (3) A  0  A  0  0, (4) None of these, , Sol. Answer (1), Knowledge based., 28. An object is being thrown at a speed of 20 m/s in a direction 45° above the horizontal. The time taken by, the object to return to the same level is, (1) 20/g, , (2) 20 g, , (3) 20 2/g, , (4) 20 2g, , Sol. Answer (3), u = 20 ms–1,  = 45º, , T , , 2u sin , g, , T , , 2u, 1, 2, u, , , g, 2 10, , T , , 2  20 1, 20 2, ⇒, T, g, g, 2, , 29. A body is whirled in a horizontal circle of radius 20 cm. It has an angular velocity of 10 rad/s. What is its linear, velocity at any point on circular path?, (1) 20 m/s, , (2), , 2 m/s, , (3) 10 m/s, , (4) 2 m/s, , Sol. Answer (4), v  r   20  10 2  10, , v  2 ms1, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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118, , Motion in a Plane, , Solution of Assignment, , 30. Identify the vector quantity among the following., (1) Distance, , (2) Angular momentum, , (3) Heat, , (4) Energy, , Sol. Answer (2), Angular momentum is an axial vector., 31. Two particles A and B are connected by a rigid rod AB. The rod slides along perpendicular rails as shown here., The velocity of A to the left is 10 m/s. What is the velocity of B when angle  = 60°?, , B, , , , (1) 10 m/s, , (2) 9.8 m/s, , A, , (3) 5.8 m/s, , Sol. Answer (3), vB = cos30º, , v A cos 60º  v B cos30º, , 10 , , vB, vB = sin30º, , 30º, , 1, 3,  vB , 2, 2, , (4) 17.3 m/s, , 30º, , vA = cos60º, , vB , , 60º, , 10, , vA, vA = sin60º, , 3, , 32. The speed of a boat is 5 km/h in still water. It crosses a river of width 1.0 km along the shortest possible, path in 15 minute. The velocity of the river water (in km/h) is, (1) 3, , (2) 1, , (3) 4, , (4) 5, , Sol. Answer (1), Repeated., v = 5 kmh–1, d = 1.0 km, t = 15 min, 33. Two racing cars of masses m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are, such that each makes a complete circle in the same time t. The ratio of the angular speeds of the first to, the second car is, (1) r1 : r2, , (2) m1 : m2, , (3) 1 : 1, , (4) m1 m2 : r1 r2, , Sol. Answer (3), If time is same then,, 1 :  2 ⇒ 1 : 1, , 2 ⎤, ⎡, ⎢∵   T ⎥, ⎣, ⎦, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 119, , 34. A person aiming to reach exactly opposite point on the bank of a stream is swimming with a speed of, 0.5 m/s at an angle of 120° with the direction of flow of water. The speed of water in the stream is, (1) 0.25 m/s, , (2) 0.5 m/s, , (3) 1.0 m/s, , (4) 0.433 m/s, , Sol. Answer (1), v, , v sin30º = u, , 0.5 , , 30º, , 1,  u ⇒ u  0.25 ms 1, 2, , 120º, , v sin30º, , u, , 35. Two projectiles of same mass and with same velocity are thrown at an angle 60° and 30° with the horizontal,, then which will remain same, (1) Time of flight, , (2) Range of projectile, , (3) Maximum height acquired, , (4) All of these, , Sol. Answer (2), Range is same for complimentary angles., 36. Two particles having mass M and m are moving in a circular path having radius R and r. If their time periods, are same, then the ratio of their angular velocities will be, , (1), , r, R, , (2), , R, r, , (3) 1, , (4), , R, r, , Sol. Answer (3), Repeated., 37. If | A  B |  | A |  | B | then angle between A and B will be, (1) 90°, , (2) 120°, , (3) 0°, , (4) 60°, , Sol. Answer (2),  , , , | AB|  | A|  |B|, ,  , | AB| , , A2  B 2  2AB cos , , A2  A2  A2  2 A2 cos , , , 1,  cos  ⇒   120º, 2, , ⎛ 20 ⎞, ⎟ m with constant tangential acceleration. If the velocity of the, 38. A particle moves along a circle of radius ⎜, ⎝  ⎠, particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is, , (1) 40 m/s2, , (2) 640 m/s2, , (3) 160 m/s2, , (4) 40 m/s2, , Sol. Answer (1), , r , , 20, m, , , aT  constant, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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120, , Motion in a Plane, , Solution of Assignment, , v = 80 ms–1,  = 4 rad, v = r, 80 , , 20,   ⇒   4 rad s1, , ,  = 0,, , 2  02  2, 4  4  2    4, , , ,   2 rad s2, , a  r , , 20,  2, , , a  40 ms2, 39. The vector sum of two forces is perpendicular to their vector differences. In that case, the forces, (1) Are equal to each other, , (2) Are equal to each other in magnitude, , (3) Are not equal to each other in magnitude, , (4) Cannot be predicted, , Sol. Answer (2),    , ( A  B)  ( A  B)  0, A2  B 2  AB  BA  0, , A2  B 2, , , AB, , , so, | A |  | B |, 40. A wheel has angular acceleration of 3.0 rad/s2 and an initial angular speed of 2.00 rad/s. In a time of 2 s it, has rotated through an angle (in radian) of, (1) 10, , (2) 12, , (3) 4, , (4) 6, , Sol. Answer (1),  = 3 rad s–2, 0 = 2 rad s–1, t=2s,  = 0 + t, =2+3×2, , https://t.me/NEET_StudyMaterial, ,   8 rad s1, , 2  02  2  , 64  4  2  3  , , 60,   ⇒   10 rad s 1, 6, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solution of Assignment, , Motion in a Plane, , 121, , SECTION - D, Assertion - Reason Type Questions, 1., ,  ,  , , A : If A  B , then | A  B |  | A  B | .,  ,  ,  , R : If A  B , then ( A  B ) is perpendicular to A  B ., , Sol. Answer (3), 2., , , , A : The addition of two vectors P and Q is commutative.,    , R : By triangle law of vector addition we can prove P  Q  Q  P ., , Sol. Answer (1), 3., , A : A vector cannot be divided by other vector., R : A vector can be divided by a scalar., , Sol. Answer (2), 4., , A : At the highest point the velocity of projectile is zero., R : At maximum height projectile comes to rest., , Sol. Answer (4), 5., , A : Horizontal range of a projectile is always same for angle of projection  with horizontal or  with vertical., R : Horizontal range depends only on angle of projection., , Sol. Answer (4), 6., , A : Horizontal motion of projectile without effect of air is uniform motion., R : Without air effect the horizontal acceleration of projectile is zero., , Sol. Answer (1), 7., , A : Path of a projectile with respect to another projectile is straight line., R : Acceleration of a projectile with respect to another projectile is zero., , Sol. Answer (1), 8., , A : In the case of ground to ground projection of a projectile from ground the angle of projection with horizontal is,  = 30º. There is no point on its path such that instantaneous velocity is normal to the initial velocity., R : Maximum deviation of the projectile is 2 = 60º., , Sol. Answer (1), 9., , A : Three vectors having magnitudes 10, 10 and 25 cannot produce zero resultant., R : If three vectors are producing zero resultant, then sum of magnitude of any two is more than or equal to, magnitude of third and difference is less than or equal to the magnitude of third., , Sol. Answer (1), 10. A : Uniform circular motion is accelerated motion still speed remains unchanged., R : Instantaneous velocity is always normal to instantaneous acceleration in uniform circular motion., Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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122, , Motion in a Plane, , Solution of Assignment, , 11. A : When a body moves on a curved path with increasing speed, then angle between instantaneous velocity, and acceleration is acute angle., R : When the speed is increasing, its tangential acceleration is in the direction of instantaneous velocity., Sol. Answer (1), 12. A : A uniform circular motion have non uniform acceleration., R : The direction of acceleration of a particle in uniform circular motion changes continuously., Sol. Answer (1), 13. A : Angular displacement is vector quantity only for small values., R : The direction of angular displacement is perpendicular to plane of rotation of object., Sol. Answer (2), ,   , , https://t.me/NEET_StudyMaterial, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456