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1, , Physics / XII (2021-22), , STUDENT SUPPORT MATERIAL, (ASSERTION REASONING AND CASE BASED QUESTIONS), , CLASS-XII, , PHYSICS, , Session 2021-22

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1, , Physics / XII (2021-22), , Course Structure :Term 1 and Term 2, Term 1, UNIT, Unit 1, Unit 2, Unit 3, Unit 4, , Chapters, Chapter 1: Electric Charges and Fields, Chapter 2: Electric Potential and Capacitance, Chapter 3: Current Electricity, Chapter 4: Moving Charges and Magnetism, Chapter 5: Magnetism and Matter, Chapter 6: Electromagnetic Induction, Chapter 7: Alternating Current, , Total, , Marks, 17, , 18, , 35, , Term 2, UNIT, Unit 5, Unit 6, Unit 7, Unit 8, Unit 9, , Chapters, Chapter 8: Electromagnetic Waves, Chapter 9: Ray Optics & Optical Instruments, Chapter 10: Wave Optics, Chapter 11: Dual nature of Radiation & Matter, Chapter 12: Atoms, Chapter 13: Nuclei, Chapter 14: Semiconductor-Electronics:, Materials, Devices and Simple Circuits, , Total, , Marks, 17, , 11, 7, , 35

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1, , Physics / XII (2021-22), , PHYSICS, Class-XII, , INDEX, S.No, , Particulars, , Page No., , 1, , SYLLABUS, , 04, , 2, , DELETED TOPICS FOR SESSION 2020-21, , 05, , 3, , UNIT-1 (ELECTROSTATICS), , 06-14, , 4, , UNIT-2 (CURRENT ELECTRICITY), , 15-35, , 5, , UNIT-3 (MAGNETIC EFFECTS OF CURRENT, AND MAGNETISM), , 36-51, , 6, , UNIT-4 (ELECTROMAGNETIC INDUCTION AND, ATERNATING CURRENTS), , 52-58, , 7, , UNIT-5 (ELECTROMAGNETIC WAVES), , 59-67, , 8, , UNIT-6 (OPTICS), , 68-79, , 9, , UNIT-7 (DUAL NATURE OF RADIATION AND, MATTER), , 80-83, , 10, , UNIT-8 (ATOMS AND NUCLEI), , 84-97, , 11, , UNIT-9 (ELECTRONIC DEVICES), , 98-102, , 12, , CBSE SAMPLE PAPER 2020-21, , 103-112

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1, , Physics / XII (2021-22), , SYLLABUS, TERM 1 & 2(Combined), , Session 2021-22, PHYSICS CLASS XII, TERM 1, Unit–I, , Marks, , Electrostatics, Chapter– 1: Electric Charges and Fields, Chapter– 2: Electrostatic Potential and Capacitance, , Unit-II, , 17, , Current Electricity, Chapter– 3: Current Electricity, , Unit-III, , Magnetic Effects of Current and Magnetism, Chapter– 4: Moving Charges and Magnetism, Chapter– 5: Magnetism and Matter, , Unit-IV, , 18, , Electromagnetic Induction and Alternating Currents, Chapter– 6: Electromagnetic Induction, Chapter– 7: Alternating Current, , TERM 2, Unit–V, , Electromagnetic Waves, Chapter– 8: Electromagnetic Waves, , Unit–VI, , Optics, , 17, , Chapter– 9: Ray Optics and Optical Instruments, Chapter– 10: Wave Optics, Unit–VII, , Dual Nature of Radiation and Matter, Chapter– 11: Dual Nature of Radiation and Matter, , Unit–VIII Atoms and Nuclei, , 11, , Chapter– 12: Atoms, Chapter– 13: Nuclei, Unit–IX, , Electronic Devices, Chapter– 14: Semiconductor, Electronics: Materials,, Devices and Simple Circuits, , 7

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1, , Physics / XII (2021-22), , DELETED TOPICS, (for Session 2021-22), PHYSICS, CLASS XII, S.No, 01, 02, 03, 04, , 05, 06, 07, , 08, 09, , 10, , Name of the Chapter, Electric charges and, fields, Current Electricity, Moving Charges and, Magnetism, Magnetism and Matter, , Alternating Current, Electromagnetic Waves, Ray Optics and Optical, Instruments, , Dual Nature of radiation, and matter, Nuclei, , Semiconductor, Electronics: Materials,, Devices and Simple, Circuits, , Deleted Topics, uniformly charged thin spherical shell (field inside and, outside)., Carbon resistors, colour code for carbon resistors;, series and parallel combinations of resistors, Cyclotron, magnetic field intensity due to a magnetic dipole, (bar magnet) along its axis and perpendicular to its, axis, torque on a magnetic dipole (bar magnet) in a, uniform magnetic field;, Para-, dia- and ferro - magnetic substances, with, examples. Electromagnets and factors affecting their, strengths, permanent magnets., power factor, wattless current, Basic idea of displacement current, Reflection of light, spherical mirrors,(recapitulation), mirror formula ,, Scattering of light - blue colour of sky and reddish, appearance of the sun at sunrise and sunset., resolving power of microscope and astronomical, telescope, polarisation, plane polarised light,, Brewster's law, uses of plane polarised light and, Polaroids., Davisson-Germer experiment, Radioactivity, alpha, beta and gamma, particles/rays and their properties; radioactive, decay law, half life and mean life, binding energy per nucleon and its variation with, mass number, Zener diode and their characteristics, zener, diode as a voltage regulator

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1, , Physics / XII (2021-22), , UNIT-I ELECTROSTATISC, Assertion (A) & Reason(R), For question numbers 1 to 20, two statements are given-one labelled Assertion (A), and the other labelled Reason (R). Select the correct answer to these questions from, the codes (a), (b), (c) and (d) as given below., a) Both A and R are true and R is the correct explanation of A, b) Both A and R are true but R is NOT the correct explanation of A, c) A is true but R is false, d) A is false and R is also false, 1. Assertion (A): The electrostatics force increases with decrease the distance between the, charges., Reason (R): The electrostatic force of attraction or repulsion between any two stationary, point charges is inversely proportional to the square of the distance between them., Answer: A, 2. Assertion(A): The Coulomb force between two points charges depend upon the dielectric, constant of the intervening medium., Reason(R): Coulomb’s force varies inversely with the dielectric constant of medium., Answer: A, 3. Assertion(A): The charge given to a metallic sphere does not depend on whether it is, hollow or solid, Reason(R): The charge resides only at the surface of conductor., Answer: A, 4. Assertion (A): A comb run through one’s dry hair attracts small bits of paper., Reason(R): Molecules in the paper gets polarized by the charged comb resulting in net, force of attraction, Answer: A, 5. Assertion(A): A proton is placed in a uniform electric field, it tend to move along the, direction of electric field., Reason(R): A proton is placed in a uniform electric field it experiences a force., Answer: B, 6. Assertion(A): Electric field at the surface of a charged conductor is always normal to the, surface at every point., Reason(R): Electric field gives the magnitude & direction of electric force, (⃗𝐹⃗⃗→) experienced by any charge placed at any point., Answer: B

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1, , Physics / XII (2021-22), 7. Assertion(A): The potential inside a hollow spherical charged conductor is zero., Reason(R): Inside the hollow spherical conductor electric field is constant., Answer: D, 8. Assertion(A): Electric filed lines not form closed loops., Reason(R): Electric filed lines are always normal to the surface of a conductor., Answer: B, 9. Assertion(A): No work is done in moving a test charge from one point to another over an, equipotential surface., Reason(R): Electric field is always normal to the equipotential surface at every point, Answer: B, 10. Assertion(A): No work is done in moving a point charge 𝑄 around a circular arc of radius ′𝑟′, at the Centre of which another point charge ′𝑞′ is located., Reason(R): No work is done in moving a test charge from one point to another over an, equipotential surface., Answer: A, 11. Assertion(A): A metal plate is introduced between the plates of a charged parallel plate, capacitor, its capacitance increased., Reason(R): A metal plate is introduced between the plates of a charged parallel plate, capacitor, the effective separation between the plates is decreased., Answer: A, 12. Assertion(A): In the presence of external electric field the net electric field within the, conductor becomes zero., Reason(R): In the presence of external electric field the free charge carriers move and, charge distribution in the conductor adjusts itself., Answer: A, 13. Assertion (A): Sensitive instruments can protect from outside electrical influence by, enclosing them in a hollow conductor., Reason(R): Potential inside the cavity is zero., Answer: C, 14. Assertion(A): Earthing provides a safety measure for electrical circuits and, appliances., Reason(R): When we bring a charged body in contact with the earth, all the, excess charge on the body disappears by causing a momentary current to, pass to the ground through the connecting conductor., Answer: A

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1, , Physics / XII (2021-22), 15. Assertion(A): The total amount of charge on a body equal to 4X10-19 C is not, possible., Reason(R): Experimentally it is established that all free charges are integral, multiples of a basic unit of charge denoted by e. Thus, charge q on a body is, always given by q = ne, Answer: A, 16. Assertion(A): The net force on a dipole in a uniform electric dipole is zero., Reason(R): Electric dipole moment is a vector directed from –q to +q., Answer: B, 17. Assertion(A): Electrostatic forces are conservative in nature., Reason(R): Work done by electrostatic force is path dependent., Answer: C, 18. Assertion(A) The field intensity in between such sheets having equal and opposite uniform, surface densities of charge become constant., Reason(R): The field intensity does not depend upon the distance between the thin sheet., Answer: A, 19. Assertion(A): Work done by the electrostatic force in bringing the unit positive Charge form, infinity to the point P is positive., Reason(R): The force on a unit positive test charge is attractive, so that the electrostatic, force and the displacement (from infinity to P) are in the same direction., Answer: A, 20. Assertion(A): The interior of a conductor can have no excess charge in the static situation, Reason(R): Electrostatic potential is constant throughout the volume of the conductor and, has the same value (as inside) on its surface., Answer: B, , CASE STUDY BASED QUESTIONS, Q.1 The electric field inside, the cavity is zero, whatever, be the size and shape of the, cavity and whatever be the, charge on the conductor, and the external fields in, which it might be placed., The electric field inside a, charged spherical shell is, zero. But the vanishing of, electric field in the (chargefree) cavity of a conductor

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1, , Physics / XII (2021-22), is, as mentioned above, a very general result. A related result is that even if the conductor, is charged or charges are induced on a neutral conductor by an external field, all charges, reside only on the outer surface of a conductor with cavity., The proofs of the results noted in Fig. are omitted here, but we note their important, implication. Whatever be the charge and field configuration outside, any cavity in a, conductor remains shielded from outside electric influence: the field inside the cavity is, always zero. This is known as electrostatic shielding. The effect can be made use of in, protecting sensitive instruments from outside electrical influence., (1) A metallic shell having inner radius R1 and outer radii R2 has a point charge Q kept, inside cavity. Electric field in the region R 1 < r < R2 where r is the distance from the, centre is given by, (a) depends on the value of r, (b) Zero, (c) Constant and nonzero everywhere, (d) None of the above, (2) The electric field inside the cavity is depend on, (a) Size of the cavity, (b) Shape of the cavity, (c) Charge on the conductor, (d) None of the above, (3) Electrostatic shielding is based, (a) electric field inside the cavity of a conductor is less than zero, (b) electric field inside the cavity of a conductor is zero, (c ) electric field inside the cavity of a conductor is greater than zero, (d) electric field inside the cavity of a plastic is zero, (4), (a), (b), (c), (d), , During the lightning thunderstorm, it is advised to stay, inside the car, under trees, in the open ground, on the car, , (5) Which of the following material can be used to make a Faraday cage (based on, electrostatic shielding), (a) Plastic, (b) Glass, (c) Copper, (d) Wood, Answer:, , 1. b, , 2. d, , 3. b, , 4. a, , 5. c

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1, , Physics / XII (2021-22), 2. When a glass rod is rubbed with silk, the rod acquires one kind of charge and the silk, acquires the second kind of charge. This is true for any pair of objects that are rubbed to, be electrified. Now if the electrified glass rod is brought in contact with silk, with which it, was rubbed, they, no longer attract, each other. They, also do not attract, or repel other light, objects as they did, on, being, electrified., Thus, the charges, acquired, after, rubbing are lost, when the charged bodies are brought in contact. What can you conclude from these, observations? It just tells us that unlike charges acquired by the objects neutralise or nullify, each other’s effect. Therefore, the charges were named as positive and negative by the, American scientist Benjamin Franklin. We know that when we add a positive number to a, negative number of the same magnitude, the sum is zero. This might have been the, philosophy in naming the charges as positive and negative. By convention, the charge on, glass rod or cat’s fur is called positive and that on plastic rod or silk is termed negative. If, an object possesses an electric charge, it is said to be electrified or charged. When it has, no charge it is said to be electrically neutral., (1) When you charge a balloon by rubbing it on your hair this is an example of what, method of charging?, (a) Friction, (b)Conduction, (c)Grounding, (d)Induction, (2) Neutral atoms contain equal numbers of positive, (a)Electrons and Protons, (b)Protons and Electrons, (c)Neutrons and Electrons, (d)Protons and Neutrons, , and negative, , ., , (3) Which particle in an atom can you physically manipulate?, (a)protons, (b)electrons, (c)neutrons, (d)you can't manipulate any particle in an atom, (4) If a negatively charged rod touches a conductor, the conductor will be charged by, what method?, (a) Friction, (b)Conduction, (c)Induction, (d)Convection

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1, , Physics / XII (2021-22), (5) A negatively charged rod is touched to the top of an electroscope, which on is, correct in the given figure, (a) A, (b) B, (c) C, (d) D, Answer:, , 1. a, , 2. b, , 3. b, , 4. b, , 5. C, , 3. For electrostatics, the concept of electric field is convenient, but not really necessary., Electric field is an elegant way of characterizing the electrical environment of a system of, charges. Electric field at a point in the space around a system of charges tells you the force, a unit positive test charge would experience if placed at that point (without disturbing the, system). Electric field is a characteristic of the system of charges and is independent of the, test charge that you place at a point to determine the field. The term field in physics, generally refers to a quantity that is defined at every point in space and may vary from, point to point. Electric field is a vector field, since force is a vector quantity., (1) Which of the following statement is correct? The electric field at a point is, (a) always continuous., (b) continuous if there is a charge at that point., (c) discontinuous only if there is a negative charge at that point., (d) discontinuous if there is a charge at that point., (2) The force per unit charge is known as, (a) electric flux, (b) electric field, (c) electric potential, (d) electric current, (3) The SI unit of electric field is, (a) N/m, (b) N-m, (c) N/C, (d) N/C2, (4) The magnitude of electric field intensity E is such that, an electron placed in it, would experience an electrical force equal to its weight is given by, (a) mge, (b) mg/e, (c) e/mg, (d) e²g/m², (5) At a particular point, Electric field depends upon, (a) Source charge Q only, (b) Test Charge q0 only., (c)Both q and q0, (d)Neither Q nor q0, Answer:, , 1. b, , 2. b, , 3. c, , 4. b, , 5. a

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1, , Physics / XII (2021-22), 4. Dielectric with, polar, molecules, also develops a, net dipole moment, in an external, field, but for a, different reason., In the absence of, any external field,, the, different, permanent dipoles, are oriented, randomly due to, thermal agitation;, so, the total dipole, moment is zero., When, an external field is, applied,, the, individual dipole, moments tend to alignwith the field. When summed overall the molecules, there is then a, net dipole moment in the direction of the external field, i.e., the dielectric is polarised. The, extent of polarisation depends on the relative strength of two factors: the dipole potential, energy in the external field tending to align the dipoles mutually opposite with the field and, thermal energy tending to disrupt the alignment. There may be, in addition, the ‘induced, dipole moment’ effect as for non-polar molecules, but generally the alignment effect is, more important for polar molecules. Thus in either case, whether polar or non-polar, a, dielectric develops a net dipole moment in the presence of an external field. The dipole, moment per unit volume is called polarization., (1) The best definition of polarisation is, (a) Orientation of dipoles in random direction, (b) Electric dipole moment per unit volume, (c) Orientation of dipole moments, (d)Change in polarity of every dipole, (2) Calculate the polarisation vector of the material which has 100 dipoles per unit volume, in a volume of 2 units., (a) 200, (b) 50, (c) 0.02, (d) 100, (3) The total polarisation of a material is the, (a) Product of all types of polarisation, (b) Sum of all types of polarisation, (c)Orientation directions of the dipoles, (d)Total dipole moments in the material

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1, , Physics / XII (2021-22), (4) Dipoles are created when dielectric is placed in, (a) Magnetic Field, (b) Electric field, (c) Vacuum, (d) Inert Environment, (5) Identify which type of polarisation depends on temperature., (a)Electronic, (b)Ionic, (c)Orientational, (d) Interfacial, Answer:, , 1. b, , 2. a, , 3. b, , 4. b, , 5. C, , 5. Figure (a) shows an uncharged metallic sphere on an insulating metal stand. If we Bring, a negatively charged rod close to the metallic sphere, as shown in Fig. (b). As the rod is, brought close to the sphere, the free electrons in the sphere move away due to repulsion, and start piling up at the farther, end. The near end becomes, positively charged due to deficit, of electrons., This, process, of, charge, distribution stops when the net, force on the free electrons, inside the metal is zero. Now if, we Connect the sphere to the, ground by a conducting wire., The electrons will flow to the, ground while the positive, charges at the near end will, remain held there due to the, attractive force of the negative, charges on the rod, as shown in Fig. (c). Disconnect the sphere from the ground. The, positive charge continues to be held at the near end Fig.(d). if we remove the electrified, rod. The positive charge will spread uniformly over the sphere as shown in Fig. (e). In this, experiment, the metal sphere gets charged by the process of induction and the rod does, not lose any of its charge., (1) What do you call the process of charging a conductor by bringing it near another, Charged object?, (a) Induction, (b) Polarisation, (c) neutralization, (d) conduction

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1, , Physics / XII (2021-22), (2) The negatively charged balloon is brought near the two cans. What happens?, , (a) The negative charges in Can B move towards the balloon, (b)The negative charges in Can A move away from the balloon, (c)The positive charges in Can B move towards the balloon, (d)The positive charges in Can A move away from the balloon, (3) Transferring a charge without touching is, (a)Conduction, (b) Induction, (c)Grounding, (d)Newtons 3rd law, (4) Due to electrostatic induction in aluminum rod due to charged plastic rod, the total, charge on the aluminum rod is, (a)Zero, (b)Positive, (c) Negative, (d) Dual, (5) If we bring charged plastic rod near-neutral aluminum rod, then rods will, (a)Repel each other, (b) Attract each other, (c)Remain their position, (d)Exchange charges, Answer:, , 1. a, , 2. b, , 3. b, , 4. a, , 5. b

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1, , Physics / XII (2021-22), , UNIT-II CURRENT ELECTRICITY, Assertion (A) & Reason(R), Q1:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: A, Q2:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: A, Q3:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I.

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1, , Physics / XII (2021-22), C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: D, Q4:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: A, Q5:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: D

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Physics / XII (2020-21), Q6:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: A, Q7:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: C, Q8:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: A

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Physics / XII (2020-21), Q9:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: D, Q10:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: B, Q11:

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Physics / XII (2020-21), A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: A, Q12:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: D, Q13:, , A) If both Statement I and Statement II are true and the statement II is the correct, explanation of the statement I., B) If both Statement I and Statement II are true but statement II is not the correct, explanation of the statement I., C) If Statement I is true but Statement II is false., D) If the Statement I and Statement II both are false., E) If Statement I is false but Statement II is true., Answer: D

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Physics / XII (2020-21), Q14:, Statement 1: The possibility of an electric bulb fusing is higher at the time of switching ON, and OFF, Statement 2: Inductive effects produce a surge at the time of switch ON and OFF, A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: A, Q15:, Statement 1: The 200 W bulbs glow with more brightness than 100 W bulbs., Statement 2: A 100 W bulb has more resistance than a 200 W bulb., A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: A, Q16:, Statement 1: Fuse wire must have high resistance and low melting point., Statement 2: Fuse is used for small current flow only., A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: C, Q17:, Statement 1: Two electric bulbs of 50 and 100 W are given. When connected in series 50, W bulb glows more but when connected parallel 100 W bulb glows more., Statement 2: In series combination, power is directly proportional to the resistance of the, circuit. But in parallel combination, power is inversely proportional to the resistance of the, circuit.

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Physics / XII (2020-21), A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: A, Q18:, Statement 1: Two bulbs of same wattage, one having a carbon filament and the other, having a metallic filament are connected in series. Metallic bulbs will glow more brightly, than a carbon filament bulb., Statement 2: Carbon is a semiconductor., A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: D, Q19:, Statement 1: An electric bulb is first connected to a dc source and then to an ac source, having the same brightness in both cases., Statement 2: The peak value of voltage for an A.C. source is √2 times the root mean, square voltage., A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: E, Q20:, Statement 1: Current is passed through a metallic wire, heating it red. When cold water is, poured on half of its portion, then the rest of the half portion becomes hotter., Statement 2: Resistances decrease due to a decrease in temperature and so current, through wire increases.

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Physics / XII (2020-21), A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: A, Q21:, Statement 1: Through the same current flows through the line wires and the filament, of the bulb but the heat produced in the filament is much higher than that in line wires., Statement 2: The filament of bulbs is made of a material of high resistance and a, high melting point., A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: A, Q22:, Statement 1: In practical application, the power rating of resistance is not important., Statement 2: Property of resistance remain the same even at high temperature., A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: D, Q23:, Statement 1: Leclanche cell is used, when a constant supply of electric current is not, required., Statement 2: The e.m.f. of a Leclanche cell falls, if it is used continuously.

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Physics / XII (2020-21), A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: A, Q24:, Statement 1: In the given circuit if lamp B or C fuses then the light emitted by lamp A, decreases., Statement 2: Voltage on A decreases., , A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: A, Q25:, Statement 1: If three identical bulbs are connected in series as shown in figure then on, closing the switches. Bulb C short-circuited and hence illumination of bulbs A and B, decreases., Statement 2: Voltage on A and B decreases, , A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: D

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Physics / XII (2020-21), Q26:, Statement 1: Heat is generated continuously is an electric heater but its temperature, becomes constant after some time., Statement 2: At the stage when heat produced in the heater is equal to the heat dissipated, to its surrounding the temperature of the heater becomes constant., A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: A, Q27:, Statement 1: Electric appliances with a metallic body; e.g. heaters, presses, etc, have, three-pin connections, whereas an electric bulb has a two pin connection., Statement 2: Three-pin connections reduce the heating of connecting cables., A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: C, Q28:, Statement 1: A domestic electrical appliance, working on a three-pin will continue working, even if the top pin is removed., Statement 2: The third pin is used only as a safety device., A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: A

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Physics / XII (2020-21), Q29:, Statement 1: The presence of water molecules makes separation of ions easier in an, electrolyte., Statement 2: The presence of water molecules in electrolyte decreases the resistance of, electrolyte., A) If both Statement 1 and Statement 2 are true and the statement 2 is the correct, explanation of the statement 1., B) If both Statement 1 and Statement 2 are true but statement 2 is not the correct, explanation of the statement 1., C) If Statement 1 is true but Statement 2 is false., D) If the Statement 1 and Statement 2 both are false., E) If Statement 1 is false but Statement 2 is true., Answer: B, , CASE BASED QUESTIONS, Q1:

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Physics / XII (2020-21), A: Efficiency of kettle A is, 1. 63.34%, 2. 83.34%, 3. 93.34%, 4. 73.34%, B: Efficiency of kettle B is, 1. 82.5%, 2. 72.5%, 3. 92.5%, 4. 62.5%, C: Ratio of efficiency consumed charges for one time boiling of tea in kettle A to that in, kettle B, 1. 3:5, 2. 2:3, 3. 3:4, 4. 1:1, D: If the resistance of the coil in kettle A and B is Ra and Rb then we can say, 1. Ra>Rb, 2. Ra=Rb, 3. Ra<Rb, 4. Data insufficient, E: If both the kettles are joined with the same source in series one after another then, boiling starts in kettle A and kettle B after, 1. 4 times of their original time, 2. Equal to their original time, 3. 2 times of their original time, 4. Data insufficient, Answers:, A. 2, B. 4, C. 3, D. 2, E. 1, Q2: Consider an evacuated cylindrical chamber of height h having rigid conducting plates, at the ends and an insulating curved surface as shown in the figure. A number of spherical, balls made of a light weight and soft material and coated with a conducting material are, placed on the bottom plate. The balls have a radius r«h. Now a high voltage source (HV) is, connected across the conducting plates such that the bottom plate is at + V, and the top, plate at - Vo. Due to their conducting surface, the balls will get charged, will become, equipotential with the plate and are repelled by it. The balls will eventually collide with the, top plate, where the coefficient of restitution can be taken to be zero due to the soft nature, of the material of the balls. The electric field in the chamber can be considered to be that of

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Physics / XII (2020-21), a parallel plate capacitor, Assume that there are no collisions between the balls and the, interaction between them is negligible. (Ignore gravity), , A: Which one of the following statements is correct?, 1. The balls will stick to the top plate and remain there, 2. The balls will bounce back to the bottom plate carrying the same charge they went up, with, 3. The balls will bounce back to the bottom plate carrying the opposite charge they went up, with, 4. The balls will execute simple harmonic motion between the two plates, B: The average current in the steady state registered by the ammeter in the, circuit will be, 1. zero, 2. proportional to the potential V, 3. proportional to the √𝑉, 4. Proportional to 𝑉2, Answers:, A. 3, B. 4, Q3:, , A: Equivalent resistance of the speakers is, 1. 12 ohm, 2. 7/3 ohm, 3. 8/3 ohm, 4. 0.375 ohm

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Physics / XII (2020-21), B: The total current supplied by music system, 1. 2.25 A, 2. 1 A, 3. 16 A, 4. 1.5 A, C: The power dissipated in the 4 ohm resistance is, 1. 9 W, 2. 4.5 W, 3. 13.5 W, 4. 0, Answers:, A. 3, B. 1, C. 1, Q4, , A: The value of EMF 𝐸1is, 1. 8 V, 2. 6 V, 3. 4 V, 4. 2 V

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Physics / XII (2020-21), B: The resistance 𝑅1 has value, 1. 10 ohm, 2. 20 ohm, 3. 30 ohm, 4. 40 ohm, C: The resistance 𝑅2 has value, 1. 10 ohm, 2. 20 ohm, 3. 30 ohm, 4. 40 ohm, Answers:, A. 3, B. 1, C. 1, Q5, , A: The relation between 𝑅𝐴and the actual value of R is, 1. R > 𝑅𝐴, 2. R < 𝑅𝐴, 3. R = 𝑅𝐴, 4. Dependent on E and r, B: The relation between 𝑅𝐵 and the actual value of R is, 1. R < 𝑅𝐵, 2. R > 𝑅𝐵, 3. R = 𝑅𝐵, 4. Dependent upon E and r, C: If the resistance of the voltmeter is 𝑅𝑉= 1 kilo ohm and that of ammeter is 𝑅𝐺 =, 1 ohm , the magnitude of percentage error in the measurement of R (the value of

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Physics / XII (2020-21), R is nearly 10 ohm ) is, 1. Zero in both cases, 2. Non-zero but equal in both cases, 3. More in circuit A, 4. More in circuit B, Answers:, A. 2, B. 2, C. 4, Q6, , A: Just after closing the switch, 1. 2 mA, 2. 3 mA, 3. 0 mA, 4. None of the above, B: Long time after the switch is closed, 1. 2 mA, 2. 3 mA, 3. 6 mA, 4. None of the above, C: Just after reopening the switch, 1. 2 mA, 2. 3 mA, 3. 6 mA, 4. None of the above

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Physics / XII (2020-21), Answers:, A. 3, B. 1, C. 1, Q7.

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Physics / XII (2020-21), , A: Two fuse wires of same potential material are having length ratio 1:2 and ratio, 4:1 Then respective ratio of their current rating will be, 1. 8:1, 2. 2:1, 3. 1:8, 4. 4:1, B: The maximum power rating of a 20.0 ohm fuse wire is 2.0 kW,then this fuse, wire can be connected safely to a DC source (negligible internal resistance) of, 1. 300 volt, 2. 190 volt, 3. 250 volt, 4. 220 volt, C: Efficiency of a battery when delivering maximum power is, 1. 100 %, 2. 50 %, 3. 90 %, 4. 40 %, Answers:, A. 1, B. 2, C. 4

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Physics / XII (2020-21), A: The ratio of, , 𝑅1, , is, , 𝑅2, , 1., , 2., , 3., , 4., B: The ratio of, , 𝑅2, , is, , 𝑅3, , 1., , 2., , 3., , 4., C: The current that passes through the resistance R2 nearest to the V0 is, , 1., , 2., , 3.

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Physics / XII (2020-21), , 4., Answers:, A. 4, B. 3, C. 4, Q10 An ammeter and a voltmeter are connected in series to a battery with an emf, of 10V. When a certain resistance is connected in parallel with the voltmeter, the, reading of the voltmeter decreases three times, whereas the reading of the, ammeter increases two times., A: Find the voltmeter reading after the connection of the resistance., 1. 1 V, 2. 2 V, 3. 3 V, 4. 4 V, B: If the resistance of the ammeter is 2 ohm, then the resistance of the voltmeter, is:1. 1 ohm, 2. 2 ohm, 3. 3 ohm, 4. 4 ohm, C: If the resistance of ammeter is 2 ohm ,then resistance of the resistor which is, added in parallel to the voltmeter is, 1. ⅗ ohm, 2. 2/7 ohm, 3. 3/7 ohm, 4. None of the above, Answers:, A. 2, B. 3, C. 1

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Physics / XII (2020-21), , UNIT-III MAGNETIC EFFECT OF ELECTRIC CURRENT AND, MAGNETISM, Assertion (A) & Reason(R), Two statements are given-one labelled Assertion (A) and the other labelled, Reason (R). Select the correct answer to these questions from the codes (a), (b),, (c) and (d) as given below., (a) Both A and R are true and R is the correct explanation of A, (b) Both A and R are true but R is not the correct explanation of A, (c) A is true but R is false, (d) A is false and R is also false, 1. Assertion(A):, The centripetal force on the test charge qo is qo vB, where v is the velocity of a, particle and B is the magnetic field., Reason (R):, When a charged particle is fired at right angles to the magnetic field, the radius of its, circular path is directly proportional to the kinetic energy of the particle., 2. Assertion (A):, Magnetic field due to an infinite straight conductor varies inversely as the distance, from it., Reason (R):, The magnetic field due to a straight conductor is in the form of concentric circles., 3. Assertion (A):, A rectangular current loop is in an arbitrary orientation in an external uniform, magnetic field. No work is required to rotate the loop about an axis perpendicular to, the plane of loop, Reason (R):, All positions represent the same level of energy., 4. Assertion (A):, The magnitude of magnetic field in a region is equal to the number of magnetic field, lines per unit area where area should be normal to the field., Reason (R):, Magnetic field is tangential to a magnetic field line., 5. Assertion (A):, If a proton and an α-particle enter a uniform magnetic field perpendicularly with the, same speed, the time period of revolution of α-particle is double than that of proton.

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Physics / XII (2020-21), Reason (R):, In a magnetic field, the period of revolution of a charged particle is directly, proportional to the mass of the particle and inversely proportional to the charge of, the particle., 6. Assertion (A):, A charged particle is moving in a circular path under the action of a uniform, magnetic field. During the motion, kinetic energy of the charged particle is constant., Reason (R):, During the motion, magnetic force acting on the particle is perpendicular to, instantaneous velocity., 7. Assertion (A):, When radius of a circular loop carrying current is doubled, its magnetic moment, becomes four times., Reason (R):, Magnetic moment depends on the area of the loop., 8. Assertion (A):, The magnetic field at the ends of a very long current carrying solenoid is half of that, at the centre., Reason (R):, If the solenoid is sufficiently long, the field within it is uniform., 9. Assertion (A):, If an electron and proton enter a magnetic field with equal momentum, then the, paths of both of them will be equally curved., Reason (R):, The magnitude of charge on an electron is same as that on a proton., 10. Assertion (A):, The coils of a spring come close to each other, when current is passed through it., Reason (R):, It is because, the coils of a spring carry current in the same direction and hence, attract each other., 11. Assertion (A):, The range of a voltmeter can be both increased or decreased., Reason (R):, The required resistance (to be connected in series) can be calculated by using the, relation,

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Physics / XII (2020-21), R=, , 𝑉, , –G, , 𝐼𝑔, , 12. Assertion (A):, Both A m2 and J T-1 are the units of magnetic dipole moment., Reason (R):, Both the units are equivalent to each other., 13. Assertion (A):, The true geographic north direction is found by using a compass needle., Reason (R):, The magnetic meridian of the earth is along the axis of rotation of the earth., 14. Assertion (A):, If a compass needle is kept at magnetic north pole of the earth, the compass needle, may stay in any direction., Reason (R):, Dip needle will stay vertical at the north pole of the earth., 15. Assertion (A):, The magnetic field at the centre of the current carrying circular coil shown in the fig., is zero., , Reason (R):, The magnitudes of magnetic fields are equal and the directions of magnetic fields, due to both the semicircles are opposite., 16. Assertion (A):, The voltage sensitivity may not necessarily increase on increasing the current, sensitivity., Reason (R):, Current sensitivity increases on increasing the number of turns of the coil., 17. Assertion (A):, The angle of dip is maximum at the poles of the earth., Reason (R):, The magnetic field lines are parallel to the surface of the earth at the poles., 18. Assertion (A):, An electron projected parallel to the direction of magnetic force will experience, maximum force.

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Physics / XII (2020-21), Reason (R):, Magnetic force on a charge particle is given by F = (IL x B)., 19. Assertion (A):, The torque acting on square and circular current carrying coils having equal areas,, placed in uniform magnetic field, will be same., Reason (R):, Torque acting on a current carrying coil placed in uniform magnetic field does not, depend on the shape of the coil, if the areas of the coils are same., 20. Assertion (A):, A phosphor bronze strip is used in a moving coil galvanometer., Reason (A):, Phosphor bronze strip has the maximum value of torsional constant k., ANSWER KEY:, 1. c, 2. b, 3. a, 4. b, 5. a, 6. a, 7. a, 8. b, 9. a, 10. a, , 11. a, 12. a, 13. d, 14. b, 15. a, 16. b, 17. c, 18. d, 19. a, 20. c, , CASE BASED QUESTIONS, Case 1. FORCE ON A CHARGE IN ELECTRIC AND MAGNETIC FIELD, , A point charge q (moving with a velocity v and located at r at a given time t) in the, presence of both the electric field E and magnetic field B. The force on an electric, charge q due to both of them can be written as, F = q [ E + v x B ] = Fel + Fmag, It is called the ‘Lorentz force’.

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Physics / XII (2020-21), 1. If the charge q is moving under a field, the force acting on the charge depends on, the magnitude of field as well as the velocity of the charge particle, what kind of field, is the charge moving in?, (a) Electric field, (b) Magnetic field, (c) Both electric and magnetic field perpendicular to each other, (d) None of these, 2. The magnetic force acting on the charge ‘q’ placed in a magnetic field will vanish, if, (a) if v is small, (b) If v is perpendicular to B, (c) If v is parallel to B, (d) None of these, 3. If an electron of charge -e is moving along + X direction and magnetic field is, along + Z direction, then the magnetic force acting on the electron will be along, (a), + X axis, (b), - X axis, (c), - Y axis, (d), + Y axis, 4. The vectors which are perpendicular to each other in the relation for magnetic, force acting on a charge particle are, (a) F and v, (b) F and B, (c) v and B, (d) All of these, 5. A particle moves in a region having a uniform magnetic field and a parallel,, uniform electric field. At some instant, the velocity of the particle is perpendicular to, the field direction. The path of the particle will be, (a) A straight line, (b) A circle, (c) A helix with uniform pitch, (d) A helix with non-uniform pitch

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Physics / XII (2020-21), CASE 2: HELICAL MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELD, , If velocity has a component along B, this component remains unchanged as the, motion along the magnetic field will not be affected by the magnetic field. The, motion in a plane perpendicular to magnetic field is a circular one, thereby, producing a helical motion., 1. The radius of the charge particle, (when v is perpendicular to B) placed in a, uniform magnetic field is given by, (a) R = mv/qB, (b) R = qB/mv, (c) R = Bqm/v, (d) R = vq/mB, 2. An electron, proton, He+ and Li++ are projected with the same velocity, perpendicular to a uniform magnetic field. Which one will experience maximum, magnetic force?, (a) Electron, (b) Proton, (c) He+, (d) Li++, 3. The work done by the magnetic field on the charge particle moving perpendicular, to a uniform magnetic field is, (a) Zero, (b) q (v x B). S, (c) Maximum, (d) qBS/v, 4. The distance moved by a charged particle along the magnetic field in one, rotation, when v has a component parallel to B is, , (a), (b), (c), , 2v cos, , qBm, 2mvcos, qB, qBm, 2v cos

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Physics / XII (2020-21), , , (d), , Bq, 2m, , CASE 3: AURORA BOREALIS, , During a solar flare, a large number of electrons and protons are ejected from the, sun. Some of them get trapped in the earth’s magnetic field and move in helical, paths along the field lines. The field lines come closer to each other near the, magnetic poles, hence the density of charges increases near the poles. The, particles collide with atoms and molecules of the atmosphere. Excited oxygen atoms, emit green light and excited nitrogen atoms emit pink light. This phenomenon is, called ‘Aurora Borealis’., 1. When will the path of the particle be helix, when it is moving in external magnetic, field?, (a) When v has a component parallel to B, (b) When v has a component perpendicular to B, (c) When v is parallel to B, (d) None of these, 2. When the charged particle travelling in a helical path enters a region where the, magnetic field is non-uniform, the pitch of helix of the charge particle will be, (a) Same as in uniform magnetic field, (b) Increases as the charge moves inside the magnetic field, (c) Decreases as the charge moves inside the magnetic field, (d) First increases then decreases as the charge moves inside the magnetic field, 3. The colour of Aurora Borealis is due to, (a) Excited ozone, chromium atoms, (b) Excited Oxygen and Nitrogen atoms, (c) Due to presence of water vapours in the atmosphere, (d) Excited electrons and protons in the atmosphere, 4. The density of magnetic field lines is greater, (a) At the poles, (b) Near the equator, (c) Uniform everywhere on the surface, (d) None of these, , on the earth

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Physics / XII (2020-21), CASE 4: VELOCITY SELECTOR, , A charge q moving with a velocity v in presence of both electric and magnetic fields, experience a force F = q [ E + v x B ]. If electric and magnetic fields are, perpendicular to each other and also perpendicular to the velocity of the particle, the, electric and magnetic forces are in opposite directions. If we adjust the value of, electric and magnetic field such that magnitude of the two forces are equal. The, total force on the charge is zero and the charge will move in the fields undeflected., 1.What will be the value of velocity of the charge particle, when it moves undeflected, in a region where the electric field is perpendicular to the magnetic field and the, charge particle enters at right angles to the fields., (a) v = E/B, (b) v = B/E, (c) v = EB, (d) v = EB/q, 2. Proton, neutron, alpha particle and electron enter a region of uniform magnetic, field with same velocities. The magnetic field is perpendicular to the velocity. Which, particle will experience maximum force?, (a) proton, (b) electron, (c) alpha particle, (d) neutron, 3. A charge particle moving with a constant velocity passing through a space without, any change in the velocity. Which can be true about the region?, (a) E = 0, B = 0, (b) E ≠ 0, B ≠ 0, (c) E = 0, B ≠ 0, (d) All of these, 4. Proton, electron and deuteron enter a region of uniform magnetic field with same, electric potential-difference at right angles to the field. Which one has a more curved, trajectory?

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Physics / XII (2020-21), (a) electron, (b) proton, (c) deuteron, (d) all will have same radius of circular path, CASE 5: MOTION OF A CHARGED PARTICLE IN A UNIFORM MAGNETIC FIELD, , A charged particle of mass m and charge q moves with a constant velocity along the, positive X direction v = ai. It enters a region of magnetic field which is directed, towards positive Z direction from x = a which is given by B = bk, 1. The initial acceleration of the particle is, (a) a = -𝑞𝑎𝑏 i, 𝑚, , (b) a = (c) a = -, , 𝑞𝑎𝑚, , 𝑏, 𝑞𝑎, , 𝑚𝑏, , j, , j, , (d) none of these, 2. The radius of the circular path which the particle moves is, , (a) 𝑚𝑏, 𝑞𝑎, , (b), (c), , 𝑚𝑎, , 𝑞𝑏, 𝑚𝑎𝑏, 𝑞, , (d) None of these, 3. Which of the following is true about the motion of the particle in uniform magnetic, , field, where the charged particle enters at right angles to the field?, (a) Force will always be perpendicular to the velocity., (b) Kinetic energy of the particle remains constant., (c) Velocity vector and magnetic field vector remains perpendicular to each other, during the motion., (d) All of these., 4. The frequency of the rotation, , (a) depends on the value of a, (b) depends on the value of b, (c) depends on the value of a and b both, (d) does not depend on a and b

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Physics / XII (2020-21), CASE 6: MOVING COIL GALVANOMETER, , The galvanometer is a device used to detect the current flowing in a circuit or a, small potential difference applied to it. It consists of a coil with many turns, free to, rotate about a fixed axis, in a uniform radial magnetic field formed by using concave, pole pieces of a magnet. When a current flows through the coil, a torque acts on it., 1.What is the principle of moving coil galvanometer?, (a) Torque acting on a current carrying coil placed in a uniform magnetic field., (b) Torque acting on a current carrying coil placed in a non-uniform magnetic field., (c) Potential difference developed in the current carrying coil., (d) None of these., 2. If the field is radial, then the angle between magnetic moment of galvanometer, coil and the magnetic field will be, (a) 0°, (b) 30°, (c) 60°, (d)90°, 3. Why pole pieces are made concave in the moving coil galvanometer?, (a) to make the magnetic field radial., (b) to make the magnetic field uniform., (c) to make the magnetic field non-uniform., (d) none of these., 4. What is the function of radial field in the moving coil galvanometer?, (a) to make the torque acting on the coil maximum., (b) to make the magnetic field strong., (c) to make the current scale linear., (d) all the above., 5. If the rectangular coil used in the moving coil galvanometer is made circular, then, what will be the effect on the maximum torque acting on the coil in magnetic field for, the same area of the coil?, (a) remains the same, (b) becomes less in circular coil, (c) becomes greater in circular coil, (d) depends on the orientation of the coil

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Physics / XII (2020-21), 6. What is the torque and force in the two cases as shown in the fig.?, , (a) τa < τb, Fa ≠ 0, Fb ≠ 0, (b) τa> τb, Fa = Fb = 0, (c)τa = τb = 0, Fa = Fb = 0, (d)τa = τb, Fa = Fb = 0, CASE 7: CONVERSION OF MOVING COIL GALVANOMETER INTO AN, AMMETER, , The galvanometer cannot be used as an ammeter to measure the value of the, current directly as it is a very sensitive device. It gives a full-scale deflection for, current of the order of µA. For measuring currents, the galvanometer has to be, connected in series, and as it has a large resistance, this will change the value of, current in the circuit., 1.How is a moving coil galvanometer converted into an ammeter of desired range?, (a)Connecting a shunt resistance in series., (b)Connecting a shunt resistance in parallel., (c)Connecting a large resistance in series., (d)Connecting a large resistance in parallel., 2.A moving coil galvanometer of resistance G gives a full-scale deflection for a, current Ig. It is converted into an ammeter of range 0- I ampere. What should be the, value of shunt resistance to convert it into an ammeter of desired range?, (a)S = 𝐼 G, (b)S =, , 𝐼− 𝐼 𝑔, 𝐼− 𝐼𝑔, 𝐼, , G

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Physics / XII (2020-21), (c)S = 𝐼 G, (d)S =, , 𝐼𝑔, 𝐼𝑔, 𝐼, , G, , 3. Which one will have the greatest resistance – a micro-ammeter, a milli-ammeter,, , an ammeter?, (a)Micro-ammeter, (b)Milli-ammeter, (c)Ammeter, (d)All will have the same resistance, 4. The resistance of the ammeter will be, , (a) 1 =, , 1, , 𝑅𝐴, , 𝐺, , +1, , 𝑆, , (b)RA = G + S, (c)RA = 𝐺+𝑆, 𝐺𝑆, , (d)None of these, CASE 8. MAGNETIC MOMENT OF ELECTRON, , In the Bohr model of the Hydrogen atom, the electron revolves around a positively, charged nucleus such as a planet revolves around the sun. The force which binds, the electron-proton system is the electrostatic force. There will be a magnetic, moment associated with this circulating current given by M = I A., 1. What will be the magnetic moment of the electron in the first orbit of H-atom?, 𝑒𝑣𝑟, (a), 2, (b)𝑒𝑣, 2𝑟, , 𝑒𝑣, (c) 2𝑟𝑚, , 𝑒𝑣𝑟, , (d)2𝑚, , 2. The relation between magnetic moment and angular momentum for an electron, , revolving in the first orbit of H-atom is, (a) M = 𝑒 L, (b) L =, (c) M =, (d) L =, , 2𝑚, , 𝑒, , M, , 2𝑚, , 𝑒𝐵, , 2𝑚, 𝑒𝐵, , 2𝑚, , L, M

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Physics / XII (2020-21), 3. The angle between magnetic moment vector and angular momentum vector is, , (a) 0°, (b)45°, (c)90°, (d)180°, 4. The value of gyroscopic ratio M/L, , (a) depends on the value of charge, (b) is a constant quantity, (c) depends on mass of the particle, (d) depends on the axis of rotation., CASE 9: EARTH’S MAGNETISM, , The magnetic field lines of the earth resemble that of a hypothetical magnetic dipole, located at the centre of the earth. The axis of the dipole does not coincide with the, axis of rotation of the earth but is presently tilted by approximately 11.3° with respect, to the later. If the magnetic needle is perfectly balanced about a horizontal axis so, that it can swing in a plane of the magnetic meridian, the needle would make an, angle with the horizontal. This is known as the angle of dip (also known as, inclination)., 1. What is the angle of dip at the equator?, (a) 0°, (b) 45°, (c) 60°, (d) 90°, 2. At the poles, the dip needle will, , (a) stay horizontal, (b) stay vertical, (c) stays at 45° angle with the horizontal, (d) does not remain steady in any fixed position

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Physics / XII (2020-21), 3. The angle of dip where the vertical component of the earth’s magnetic field is, , equal to the horizontal component of the earth’s magnetic field will be, (a)0°, (b)45°, (c)60°, (d)90°, 4. Which of the following independent quantities is not used to specify the earth’s, , magnetic field?, (a) Magnetic declination (θ), (b) Angle of dip (δ), (c) Horizontal component of earth’s magnetic field (BH ), (d) Vertical component of earth’s magnetic field (BV), CASE 10: FORCE BETWEEN TWO INFINITELY LONG PARALLEL CURRENTCARRYING WIRES, , Two current-carrying conductors placed near each other will exert magnetic forces, on each other. Ampere studied the nature of this magnetic force and its dependence, on the product of magnitude of currents in both the conductors, on the shape and, size of conductors as well as the distances between the conductors. Using, Fleming’s left hand rule, it is observed that currents flowing in the same direction, attract each other and currents flowing in the opposite directions repel each other., Thus, force per unit length acting on a conductor of infinite length is given by, F = µ0 𝐼1𝐼 2, 2𝜋 𝑑, 1.A vertical wire carries a current in upward direction. An electron beam sent, horizontally towards the wire will be deflected, (a)towards right, (b)towards left, (c)upwards, (d)downwards, 2. A current carrying, straight wire is kept along axis of a circular loop carrying a, current. The straight wire, (a)will exert an inward force on the circular loop., (b)will exert an outward force on the circular loop., (c)will not exert any force on the circular loop., (d)will exert a force on the circular loop parallel to itself.

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Physics / XII (2020-21), 3. A proton beam is going from north to south and electron beam is going from, south to north. Neglecting the earth’s magnetic field, the electron beam will be, deflected, (a)towards the proton beam, (b)away from the proton beam, (c)upwards, (d)downwards, 4. Consider the situation shown in fig. The straight wire is fixed but the loop can, move under magnetic force. The loop will, , (a)remain stationary, (b)move towards the wire, (c)move away from the wire, (d) rotate about the wire., CASE 11: TOROID, The toroid is a hollow circular ring on which a large number of turns of wire are, closely wound. It can be viewed as a solenoid which has been bent into a circular, shape to close on itself. The magnetic field vanishes in the open space inside and, outside the toroid. The magnetic field inside the toroid is constant in magnitude and, is given by B = µ0 n I, where n is the number of turns per unit length and I is the, current flowing in the toriod, µ0 is the absolute permeability of the free space.

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Physics / XII (2020-21), 1. The magnetic field inside a toroid of radius R is B. If the current through it is doubled and, , its radius is also doubled keeping the number of turns per unit length the same, magnetic, field produced by it will be, (a) B/2, (b) B/4, (c)B, (d)2B, 2. What is the magnetic field in the empty space enclosed by the toroid of radius R?, , (a), , µ0 2𝐼, 4𝜋 𝑅, , (b) Infinity, (c) Zero, (d), , µ0 𝜋𝐼, 4𝜋 𝑅, , 3. A toroid of 300 turns/m and radius 2 cm is carrying a current of 5 A. What is the, , magnitude of magnetic field intensity in the interior of the toriod?, (a) 1.9 T, (b) 1.9 x 10-6 T, (c) 1.9 x 10-3 T, (d) 1.9 x 10-7 T, 4. Magnetic field due to a current carrying toroid is independent of, , (a)Its number of turns, (b) Current, (c) Radius, (d) None of these, 5. How can you increase the magnetic field inside a toroid?, , (a)by increasing the radius, (a) by decreasing the current, (b) by introducing a soft iron core inside a toroid, (d)by decreasing the total number of turns, ANSWER KEY OF CASE-BASED QUESTIONS, CASE 1, 1 (b), 2 (c), CASE 2, 1 (a), 2 (d), CASE 3, 1 (a), 2, CASE 4, 1 (a), 2 (c), , 3 (d), 3 (a), 3 (b), 3 (d), , 4 (d), 4 (b), 4 (a), 4 (a), , 5 (d), 5, 5, 5, , CASE 5, CASE 6, CASE 7, CASE 8, CASE 9, CASE 10, CASE 11, , 3 (d), 3 (a), 3 (a), 3 (d), 3 (b), 3 (a), 3 (c), , 4 (b), 4 (d), 4 (a), 4 (b), 4 (d), 4 (b), 4 (b), , 5, 5 (a) 6 (b), 5, 5, 5, 5, 5, , 1 (a), 1 (a), 1 (b), 1 (a), 1 (a), 1 (c), 1 (d), , 2 (b), 2 (d), 2 (a), 2 (a), 2 (b), 2 (c), 2 (c)

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Physics / XII (2020-21), , UNIT-IV, ELECTROMAGNETIC INDUCTION AND ALTERNATING, CURRENT, ASSERTION (A) & REASONING (R) QUESTIONS, Of the following statements, mark the correct Answers asA - if both Assertion and Reason -- are true and Reason -- is correct explanation of the Assertion., B - if both Assertion and Reason -- are true but Reason -- is not correct explanation of Assertion., C - if Assertion is true but Reason -- is false., D - if both Assertion and Reason -- are false., E - if Assertion is false but Reason -- is true, , 1. Assertion-- The mutual induction of two coils is doubled, if the self-inductance of the, primary or secondary coil is doubled, Reason -- Mutual induction is proportional to self-inductance of primary and secondary, coils, Answer - C, 2. Assertion- Making and breaking of current in a coil produce no momentary current in the, neighboring coil of another circuit, Reason -- Momentary current in the neighboring coil of another circuit is an eddy current, Answer - D, 3. Assertion- If primary coil is connected by voltmeter and secondary coil by ac source. If, large copper sheet is placed between two coils, induced emf in primary coil is reduced, Reason -- Copper sheet between coils has no effect on induced emf in primary coil, Answer – A, 4. Assertion- An electric motor will have maximum efficiency when back emf becomes, equal to half of applied emf, Reason -- Efficiency of electric motor depends only on magnitude of back emf, Answer – C, 5. Assertion- Armature current in DC motor is maximum when the motor has just started, Reason -- Armature current is given by I=E-e/R where e is back emf, R is resistance of, armature, Answer – B, 6. Assertion- Eddy current is produced in any metallic conductor when magnetic flux is, changed around it, Reason -- Electric potential determine the flow of charge, Answer - B, 7. Assertion -- The quantity L/R possesses dimensions of time, Reason -- to reduce the rate of increase of current through a solenoid should increase the, time constant L/R, Answer - B

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Physics / XII (2020-21), 8. Assertion- Faraday laws are consequence of conservation of energy, Reason -- In a purely resistive AC circuit, the current lags behind the emf in phase, Answer - C, 9. Assertion- Only a change in magnetic flux through a coil maintain a current in the coil if, the current is continues, Reason -- The presence of large magnetic flux through a coil maintain a current in the coil, if the current is continues, Answer - C, 10. Assertion- magnetic flux can produce induced emf, Reason -- Faraday established induced emf experimentally, Answer - E, 11. Assertion- Inductance coil are made of copper, Reason -- Induced current is more in wire having less resistance, Answer - A, 12. Assertion- When two coils are wound on each other, the mutual induction between coil, is maximum, Reason -- Mutual induction doesn’t depends on the orientation of the coil s, Answer – C, 13. Assertion- an aircraft flies along the meridian, the potential at the ends of its wings will, be the same., Reason -- Whenever there is change in magnetic flux emf induce, Answer – E, 14. Assertion- A spark occur between the poles of a switch when the switch is opened, Reason -- Current flowing in the conductor produce magnetic field, Answer – B, 15. In the phenomenon of mutual induction self-induction of each of coils persists, Reason -- Self-induction arises when strength of current in same coil change in the mutual, induction, current is changing in both the individual, Answer – B, 16. An induced emf is generated when magnet is withdrawn from the solenoid, Reason -- The relative motion between the magnet and solenoid induced emf, Answer - A, 17. A transformer can’t work on DC supply, Reason -- DC changes neither in magnitude nor in direction, Answer - A, 18. Soft iron is used as a core of transformer, Reason -- Area of hysteresis is loop for soft iron is small, Answer - A

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Physics / XII (2020-21), 19. An AC generator is based on the phenomenon of self-induction, Reason -- in single coil we consider, self-induction only, Answer - E, 20. An electric motor will maximum efficient, when back emf is equal to applied emf, Reason -- Efficiency of electric motor is depends only on magnitude of back emf, Answer - D, 21. An AC doesn’t show any magnetic effect, Reason -- AC doesn’t vary with time, Answer - D, 22. Assertion- A variable capacitor is connected in series with a bulb through AC source if, the capacitance of variable capacitor is decrease the brightness of bulb is reduced, Reason -- The reactance of capacitor increase if capacitance is reduced, Answer - A, 23. A capacitor of suitable capacitance can be used in AC circuit in the place of choke coil, Reason -- A capacitor blocks DC and allow only AC, Answer - B, 24. An AC doesn’t show any magnetic effect, Reason -- AC varies with time, Answer - B, 25. The division are equally marked on the scale of AC ammeter, Reason -- heat produced is directly proportion to current, Answer - D, 26. Average value of AC over a complete cycle is always zero, Reason -- Average value of AC is always defined over half cycle, Answer – B, 27. Eddy current is produced in any metallic conductor when magnetic flux is changed, around it, Reason -- electric potential determine the flow of charge, Answer – B, 28. In LCR circuit resonance can take place, Reason -- resonance can take place if inductance and capacitive reactance are equal and, opposite, Answer - A, 29. When capacitive reactance is smaller than the inductive reactance in LCR circuit, emf, leads the current, Reason -- The phase angle is angle between alternating emf and alternating current of the, circuit, Answer – B, 30. The DC and AC both can be measured by a hot wire instrument, Reason -- The hot wire instrument is based on the principle of magnetic effect of current, Answer - C

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Physics / XII (2020-21), , CASE STUDY QUESTIONS, TOPIC: ELECTROMAGNETIC INDUCTION AND AC, Question 1: An inductor is simply a coil or a solenoid that has a fixed inductance. It is, referred to as a choke. The usual circuit notation for an inductor is as shown., , Let a current i flows through the inductor from A to B. Whenever electric current changes, through it, a back emf is generated. If the resistance of inductor is assumed to be zero, (ideal inductor) then induced emf in it is given by, , e=VB-VA = - L di / dt, Thus, potential drops across an inductor as we movein the direction of current. But, potential also drops across a pure resistor when we move in the direction of the current., The main difference between a resistor and an inductor is that while a resistor opposes the, current through it, an inductor opposes the change in current through it., Now answer the following questions., (1) How does inductor behave when, (a) a steady current flow through it?, (b) a steadily increasing, current flows through it?, (c) a steadily decreasing current flows through it?, (d) Name the phenomenon in which change in current in a coil induces EMF in coil itself?, ANS: (i) (a) As electric current is steady therefore, , di / dt = 0;, :: induced emf = e = 0 and the inductor behaves as short circuit., (b) in the expression, , e= - L di / dt, as di / dt is positive EMF is negative. that is VB < VA., That is back EMF is genreted that opposses the increase in current., (c ) di / dt is negative, therefore EMF is positive. that is V B > VA. Forward EMF is, generated that opposses fall in current., (d) Self induction., Question 2:, (a) A closed loop is held stationary in the magnetic field between the north and south, poles of two permanent magnets held fixed. Can we hope to generate current in the loop, by using very strong magnets?, (b) A closed loop moves normal to the constant electric field between the plates of a large, , capacitor. Is a current induced in the loop, (i) when it is wholly inside the region between the capacitor plates, (ii) when it is partially outside the plates of the capacitor? The electric field is normal to the, plane of the loop.

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Physics / XII (2020-21), (c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region, , (Figure) to a field-free region with a constant velocity v. In which loop do you expect the, induced emſ to be constant during the passage out of the field region? The field is normal, to the loops., , (d) Predict the polarity of the capacitor in the situation described by the figure, , Solution:, (a) No. However strong the magnet may be current can be induced only by changing the, magnetic flux through the loop., (b) No current is induced in either case. Current can not be induced by changing the, electric flux., (c) The induced emf is expected to be constant only in the case of the rectangular loop. In, the case of circular loop, the rate of change of area of the loop during its passage out of, the field region is not constant, hence induced emf will vary accordingly,, (d) The polarity of plate 'A' will be positive with respect to plate 'B' in the capacitor., Question 3:, Given figure shows a metal rod PQ resting on the smooth rails AB and positioned between, the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three, mutual perpendicular directions. A galvanometer G connects the rails through a switch K., Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod =, 9.0 mΩ. Assume the field to be uniform., (a) Suppose K is open and the rod is moved with a speed of 12 cm s-1 in the direction, shown. Give the polarity and magnitude of the induced emf.

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Physics / XII (2020-21), (b) Is there an excess charge built up at the ends of the rods when K is open? What if K is, closed?, (c) With K open and the rod moving uniformly, there is no net force on the electrons in the, rod PQ even though they do experience magnetic force due to the motion of the rod., Explain., (d) What is the retarding force on the rod when K is closed?, (e) How much power is required (by an external agent) to keep the rod moving at the same, speed (=12 cm/ sec) when K is closed? How much power is required when K is open?, (f) How much power is dissipated as heat in the closed circuit? What is the source of this, power?, (g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails, instead of being perpendicular?, Answers:, (a) EMF = vBL = 0.12 0.50 x 0.15 = 9.0 mV;, P positive end and Q negative end., (b) Yes. When K is closed, the excess charge is maintained by the continuous flow of, current., (c) Magnetic force is cancelled by the electric force set-up due to the excess charge of, opposite signs at the ends of the rod., (d) Retarding force = IBL, 9 mV / 9 mΩ x 0.5 T x 0.15 m, = 75 x 10-3 N, e) Power expended by an external agent against the above retarding, force to keep the rod moving uniformly at 12 cm s', = 75 x 10-3 x 12 x 10-2 = 9.0 x 10-3 W, When K is open, no power is expended., (f) I2 R = 1x1x 9 x 10-3 = 9.0 x 10-3 W, The source of this power is the power provided by the external agent as calculated above., g) Zero: motion of the rod does not cut across the field lines. [Note: length of Pg has been, considered above to be equal to the spacing between the rails.], Question 3:, A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away, from an electric plant generating power at 440 V. The resistance of the two wire line, carrying power is 0.5Ω per km. The town gets power from the line through a 4000-220 V, step-down transformer at a sub-station in the town., (a) Estimate the line power loss in the form of heat., (b) How much power must the plant supply, assuming there is negligible power loss due to, leakage?, (c) Characterise the step up transformer at the plant., Answers:, Line resistance = 30 X 0.5 = 15Ω, rms current in the line . 800 x 1000 W / 4000 V = 200 A

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Physics / XII (2020-21), (a) Line power loss = (200 A) 2 x 15 Ω = 600 kW., (b) Power supply by the plant = 800 kW + 600 kW = 1400 kW., (c) Voltage drop on the line = 200 A 15Ω = 3000 V., The step-up transformer at the plant is 440 V - 7000 V., Question 4. Electromagnetic induction is defined as the production of an electromotive, force across an electric conductor in the changing magnetic field. The discovery of, induction was done by Michael Faraday in the year 1831. Electromagnetic induction finds, many applications such as in electrical components which includes transformers, inductors,, and other devices such as electric motors and generators., Alternating current is defined as an electric current which reverses in direction periodically., In most of the electric power circuits, the waveform of alternating current is the sine wave., 1. How to increase the energy stored in an inductor by four times?, (a) By doubling the current, (b) This is not possible, (c) By doubling the inductance, (d) By making current 2–√ times, Answer: (a) By doubling the current, 2. Consider an inductor whose linear dimensions are tripled and the total number of, turns per unit length is kept constant, what happens to the self-inductance?, (a) 9 times, (b) 3 times, (c) 27 times, (d) 13 times, Answer: (b) 3 times, 3. Lenz law is based on which of the following conservation>, (a) Charge, (b) Mass, (c) Momentum, (d) Energy, Answer: (d) Energy, 4. What will be the acceleration of the falling bar magnet which passes through the ring, such that the ring is held horizontally and the bar magnet is dropped along the axis, of the ring?, (a) It depends on the diameter of the ring and the length of the magnet, (b) It is equal due to gravity, (c) It is less than due to gravity, (d) It is more than due to gravity, Answer: (c) It is less than due to gravity

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Physics / XII (2020-21), , UNIT-V, ELECTROMAGNETIC WAVES, ASSERTION (A) AND REASONING (R) QUESTIONS, A. Both assertion and reason are True, and reason is the correct explaination ., B. Both assertion and reason are True, but reason is not the correct explaination ., C. Assertion is True , but reason is False ., D. Both assertion and reason are False ., Assertion:, Reason:, Answer: C, , Electromagnetic waves do not require medium for their propagation., They can’t travel in a medium., , Assertion:, Reason:, Answer: B, , A changing electric field produces a magnetic field., A changing magnetic field produces an electric field., , Assertion:, Reason:, Answer: A, , X-rays travel with the speed of light., X-rays are electromagnetic rays., , Assertion:, Reason:, Answer: D, , Environmental damage has increased amount of Ozone in atmosphere., Increase of ozone increases amount of ultraviolet radiation on earth, , Assertion:, Reason:, Answer: B, , Electromagnetic radiation exert pressure., Electromagnetic waves carry both - Momentum & Energy., , Assertion:, Reason:, Answer: A, , During discharging, there is magnetic field between plates of capacitor., Time varying electric field produces magnetic field., , Assertion:, , In electromagnetic waves, electric and magnetic Field are perpendicular to, each other., E and B, are self-sustaining., , Reason:, Answer: B, Assertion:, Reason:, Answer: A, , The earth without its atmosphere would be inhospitably Cold., All heat would escape in the absence of atmosphere.

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Physics / XII (2020-21), Assertion:, Reason:, Answer: C, , The EM waves of shorter wavelength can travel longer distances on earth’s, surface than those of longer wavelengths., Shorter the wavelength, the larger is the Velocity of propagation., , Assertion:, Reason:, Answer: A, , EM waves follow Superposition principle., Differential expression of EM wave is linear., , Assertion:, Reason:, Answer: A, Assertion:, Reason:, Answer: B, , Sound waves cannot travel in vacuum, but light waves can., Light is an electromagnetic wave - but sound is a Mechanical wave., , Assertion:, Reason:, Answer: B, , Transverse waves are not produced in liquids and gases., Shorter the wavelength, the larger is the Velocity of propagation., , Assertion:, , The energy contained in a small volume through which an em wave is, passing, oscillates with the frequency of the wave., , Reason:, Answer: D, , Energy density of the wave is given by : ½, , Assertion :, Reason:, Answer: B, , Like Light radiation, thermal radiations are also e.m. radiations ., Thermal radiations require no medium for propagation ., , Assertion :, Reason:, Answer: A, , X-rays cannot be deflected by electric or magnetic fields ., These are electromagnetic waves ., , Assertion :, Reason :, Answer: C, , EM waves are transverse in nature ., Waves of wavelength 10mm are radiowave and microwave ., , Assertion :, Reason:, Answer: A, , Dipole oscillations produce em waves., Accelerated charge produces em waves., , The Microwaves are better carriers of signals than radio waves., The electromagnetic waves do not require any medium to propagate., , e0 E2 .

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Physics / XII (2020-21), Assertion :, Reason:, Answer: A, , In an electromagnetic wave, magnitude of magnetic field vector B is much, smaller than the magnitude of vector E ., This is because in an electromagnetic wave E/B = c = 3x108., , Assertion :, Reason:, Answer: A, , The gyrating electron can be a source of EM wave ., The electron in circular motion is accelerated motion ., , Assertion :, Reason:, Answer: C, , EM waves interacts with matter and set up oscillations ., Interaction is independent of em wave’s Wavelength ., , Assertion :, , When an em wave going through vacuum is described as :, E = E0 . sin (kx-wt) , then w/k is independent of wavelength., w / k is speed of the wave., , Reason:, Answer: A, Assertion :, Reason:, Answer: A, , Ozone layer is essential for sustaining life on earth ., Ozone layer absorbs UV radiation, hence preventing it to reach on earth ., , Assertion :, Reason:, , Microwaves are considered suitable for radar ,used in navigation, Microwaves have wavelength of few millimeters. Due to this reason, they ], suffer very small diffraction ., , Answer: A, Assertion : Ratio of speed of uv rays & infrared waves (in vacuum) is 1., Reason: Both; infrared and uv rays are electromagnetic waves ., Answer: A, Assertion : Welders wear face mask,goggles during welding - on eyes., Reason: ‘Gamma’ rays are produced by welding, is harmful for eyes., Answer: C, Assertion :, Reason:, Answer: A, , Infrared radiation are referred as Heat wave., they get readily absorb ny molecules in most material ., , Assertion :, , Ratio of frequencies of ultraviolet waves to infrared waves is greater than 1., Frequency of u.v. rays is more than infrared rays ., , Reason:, Answer: A

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Physics / XII (2020-21), Assertion :, Reason :, , Gamma rays are more energetic than X-rays., Gamma rays are of nuclear origin- but X-rays are produced to sudden, deceleration of high energy electrons while falling on a metal of high atomic, number ., , Answer: B, Assertion :, Reason:, Answer: B, , The velocity of em wave depends on Electric and Magnetic properties of, medium ., Velocity of em waves in free space is constant., , CASE STUDY QUESTIONS, TOPIC: ELECTROMAGNETIC WAVES, Q1)Microwave in aircraft navigation, Microwave are used in aircraft navigation. A radar guns out short bursts of microwave and, it reflect back from oncoming aircraft and are detected by receiver in gun. The frequency of, reflected wave used to compute speed of aircraft, , 1 Q) How are microwave produced?, a) klystron and magnetron valve, b) sudden deceleration of electron in xray tube, c)accelerated motion of charge in conducting wire, d)hot bodies and molecules

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Physics / XII (2020-21), 2 Q) why microwave use for aircraft navigation?, A) due to high wavelength, B) due to low wavelength, c) due to low frequency, d) due to their frequency modulation power, 3 Q) which is use of microwave?, a) in treatment of cancer, b) to observe changing blood flow, c) used to kill microbes, d) studying details of atoms and molecule, 4 Q) where do microwave fall in electromagnetic spectrum?, a) between u.v region and infrared, b) between gamma and u.v, c) between infrared and radio wave, d) between gamma and infrared, ANSWER KEY, 1) a, 3)d, 2) b, 4)c, Q 2) GAMMA RAYS IN TREATMENT OF CANCER, Gamma rays are used in radiotherapy to Treat cancer. They are used to spot, they kill the living cells and damage malignant tumor., , tumors.

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Physics / XII (2020-21), 1 Q) what is the source of gamma rays?, a) radioactive decay of nucleus, b) accelerated motion of charges in conducting wire, c) hot bodies and molecule, d) klystron valve, 2 Q) how is wavelength of gamma rays, a) low, b) high, c) infinite, d) zero, 3 Q)choose the one with correct radiation order?, a) alpha>beta>gamma, b) beta>alpha>gamma, c) gamma>beta>alpha, d) gamma>alpha>beta, 4 Q) what is other use of gamma rays?, a) used to change white topaz to blue topaz, b) used in aircraft navigation, c) used in kill microbes, d) checking fractures of bone, ANSWER KEY, 1) a, 3)c, 2) b, 4)a, Q3) X- Rays, X-rays are a form of electromagnetic radiation, similar to visible light. Unlike light, however,, x-rays have higher energy and can pass through most objects, including the body. Medical, x-rays are used to generate images of tissues and structures inside the body.

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Physics / XII (2020-21), Q1. What is the most common method of preparation of X rays ?, a) magnetron valve, b) vibration of atoms and molecules, c) bombardment of metal by high energy electrons, d) radioactive decay of nucleus, Q2) which of the following set of instrument /equipment can detect X- rays, a) Photocells ,photographic film, b) Thermopiles ,bolometer, c) Photographic film ,Geiger tube, d) Geiger tube ,human eye, Q3) where do X rays fall on the electromagnetic spectrum?, a) Between UV region and infrared region, b) Between gamma rays and UV region, c) Between infrared and microwaves, d) Between microwaves and radio waves, Q4) what is the use of rays lying beyond X ray region in electromagnetic spectrum, a) used to kill microbes, b) used to detect heat loss in insulated systems, c) used in standard broadcast radio and television, d) used In oncology, to kill cancerous cells., ANSWER KEY, Q1 c, Q3 b, , Q2 c, Q4 d, , Q4). Green house effect, The greenhouse effect is a natural process that warms the Earth's surface. When the Sun's, energy reaches the Earth's atmosphere, some of it is reflected back to space and the rest, is absorbed and re-radiated by greenhouse gases. The absorbed energy warms the, atmosphere and the surface of the Earth

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Physics / XII (2020-21), Q1 The one which is not considered as naturally occurring greenhouse gas is, (a) methane, (b) CFCs, (c) carbon dioxide, (d) nitrous oxide, Q2) Which of the following is not a use of infrared waves, a) Used in treatment for certain forms of cancer, b) in military and civilian applications include target acquisition, surveillance, night vision,, homing, and tracking., c) to observe changing blood flow in the skin, d) In imaging cameras, used to detect heat loss in insulated systems, Q3) which of the following is the best method for production of infrared waves, a) bombardment of metal by high energy electrons, b) radioactive decay of nucleus, c) magnetron valve, d) vibration of atoms and molecules, Q4) Wavelength of infrared radiations is, (a) shorter (b) longer (c) infinite (d) zero, (ANSWER KEY), Q1 b, Q2 a, Q3 d, Q4 b, , , , Q5 ) ELECTROMAGNETIC (EM) SPECTRUM, The electromagnetic (EM) spectrum is the range of all types of EM radiation. Radiation is, energy that travels and spreads out as it goes – the visible light that comes from a lamp in, your house and the radio waves that come from a radio station are two types of, electromagnetic radiation. The other types of EM radiation that make up the electromagnetic, spectrum are microwaves, infrared light, ultraviolet rays, X- rays and gamma rays., Q1. The classification is roughly based on?, I) Wavelength and frequency of waves., II) Production and detection of waves., III) The way of travelling of waves., IV) Year discovered., Q2. Which of the following is NOT an example of EM RAYS., I) Radiotherapy(medicine)., II) Checking fractures., III) Sterilisation., IV) Explosives., Q3. Identify the pair having highest frequency and highest wavelength EM WAVES.

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Physics / XII (2020-21), I), II), III), IV), , UV rays and X- rays, Gamma rays and Microwaves., Gamma rays and Radio waves., Radio waves and UV rays., , Q4. What physical quantity is the same for X rays of wavelength 10-10m, red light of, wavelength 6800 Ao and radiowaves of wavelength 500m?, I) Speed in vacuum (c), II) frequency (f), III) Scattering, IV) Energy (e), ANSWER KEY, 1. II) PRODUCTION AND DETECTION OF WAYS, 2. IV) EXPLOSIVES, 3. III) GAMMA RAYS AND RADIO WAVES, 4. I) SPEED IN VACUUM

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Physics / XII (2020-21), , UNIT-VI, OPTICS, Instructions:, Two statements are given-one labelled Assertion (A) and the other labelled Reason, (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d), as given below., a) Both A and R are true and R is the correct explanation of A, b) Both A and R are true but R is NOT the correct explanation of A, c) A is true but R is false, d) A is false and R is also false, 1) Assertion : The stars twinkle while the planets do not., Reason : The stars are much bigger in size than the planets., Correct Answer: B, Solution : The stars twinkle while the planets do not. It is due to variation in density of, atmospheric layer. As the stars are very far and giving light continuously to us. So, the light, coming from stars is found to change their intensity continuously. Hence they are seen, twinkling. Also stars are much bigger in size than planets but it has nothing to deal with, twinkling phenomenon., 2) Assertion : The air bubble shines in water., Reason : Air bubble in water shines due to refraction of light, Correct Answer: C, Solution : Shining of air bubble in water is on account of total internal reflection., 3) Assertion : A double convex lens ( μm = 1.5) has focal length 10 cm. When the lens is, immersed in water ( μl = 4/3) its focal length becomes 40 cm., Reason : 1/f = [( μl−μm)/μm](1/R1−1/R 2), Correct Answer: A, Solution : Focal length of lens immersed in water is four times the focal length of, lens in air. It means, fw = 4fa = 4×10 = 40 cm, 4) Assertion : The colour of the green flower seen through red glass appears to be dark., Reason : Red glass transmits only red light., Correct Answer: A, Solution : The red glass absorbs the radiations emitted by green flowers; so flower, appears black., 5) Assertion : The mirrors used in search lights are parabolic and not concave spherical., Reason : In a concave spherical mirror the image formed is always virtual., Correct Answer: C

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Physics / XII (2020-21), Solution : In search lights, we need an intense parallel beam of light. If a source is placed, at the focus of a concave spherical mirror, only paraxial rays are rendered parallel. Due to, large aperture of mirror, marginal rays give a divergent beam. But in case of parabolic, mirror, when source is at the focus, beam of light produced over the entire cross-section of, the mirror is a parallel beam., 6) Assertion : The size of the mirror affect the nature of the image., Reason : Small mirrors always forms a virtual image., Correct Answer: D, Solution : The size of the mirror does not affect the nature of the image except that a, bigger mirror forms a brighter image., 7) Assertion : Within a glass slab, a double convex air bubble is formed. This air bubble, behaves like a converging lens., Reason : Refractive index of air is more than the refractive index of glass., Correct Answer: D, Solution : The air bubble would behave as a diverging lens, because refractive index of air, is less than refractive index of glass. However, the geometrical shape of the air bubble, shall resemble a double convex lens., 8) Assertion : The focal length of lens does not change when red light is replaced by blue, light., Reason : The focal length of lens does not depends on colour of light used., Correct Answer: D, Solution : Focal length of the lens depends upon it's refractive index as 1/f 𝖺 (μ−1). Since, μb > μr so fb < fr . Therefore, the focal length of a lens decreases when red light is replaced, by blue light., 9) Assertion : There is no dispersion of light refracted through a rectangular glass slab., Reason : Dispersion of light is the phenomenon of splitting of a beam of white light into its, constituent colours., Correct Answer: B, Solution : After refraction at two parallel faces of a glass slab, a ray of light emerges in a, direction parallel to the direction of incidence of white light on the slab. As rays of all, colours emerge in the same direction (of incidence of white light), hence there is no, dispersion, but only lateral displacement., 10) Assertion : A beam of white light gives a spectrum on passing through a hollow prism., Reason : Speed of light outside the prism is different from the speed of light inside the, prism., Correct Answer: D

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Physics / XII (2020-21), Solution : Dispersion of light cannot occur on passing through air contained in a hollow, prism. Dispersion take place because the refractive index of medium for different colour is, different. Therefore when white light travels from air to air, refractive index remains same, and no dispersion occurs., 11) Assertion : If objective and eye lenses of a microscope are interchanged then it can, work as telescope. Reason : The objective of telescope has small focal length., Correct Answer: D, Solution :, We cannot interchange the objective and eye lens of a microscope to make a, telescope. The reason is that the focal length of lenses in microscope are very small, of the, order of mm or a few cm and the difference (fo & fe) is very small, while the telescope, objective have a very large focal length as compared to eye lens of microscope., 12) Assertion : Although the surfaces of a goggle lens are curved, it does not have any, power., Reason : In case of goggles, both the curved surfaces have equal radii of curvature., Correct Answer: A, Solution : The focal length of a lens is given by 1/f=(μ−1)(1/R 1−1/R2) For, goggle, R1 = R 2, 1/ f= (μ−1)(1/R 1−1/R2) = 0. Therefore, P = 1/f = 0., 13) Assertion : If the angles of the base of the prism are equal, then in the position of, minimum deviation, the refracted ray will pass parallel to the base of prism., Reason : In the case of minimum deviation, the angle of incidence is equal to the angle of, emergence., Correct Answer: A, Solution : In case of minimum deviation of a prism ∠i=∠e. so, ∠r1=∠r2, 14) Assertion : An empty test tube dipped into water in a beaker appears silver, when, viewed from a suitable direction., Reason : Due to refraction of light, the substance in water appears silvery., Correct Answer: C, Solution : The ray of light incident on the water air interface suffers total internal, reflections, in that case the angle of incidence is greater than the critical angle. Therefore,, if the tube is viewed from suitable direction (so that the angle of incidence is greater than, the critical angle), the rays of light incident on the tube undergoes total internal reflection., As a result, the test tube appears as highly polished i.e. silvery., 15) Assertion : Spherical aberration occur in lenses of larger aperture., Reason : The two rays, paraxial and marginal rays focus at different points., Correct Answer: A, Solution :, In wide beam of light, the light rays of light which travel close to the principal, axis are called paraxial rays, while the rays which travel quite away from the principal axis

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Physics / XII (2020-21), is called marginal rays. In case of lens having large aperture, the behaviour of the paraxial, and marginal rays are markedly different from each other. The two types of rays come to, focus at different points on the principal axis of the lens, thus the spherical aberration, occur. However in case of a lens with small aperture, the two types of rays come to focus, quite close to each other., 16) Assertion : The frequencies of incident, reflected and refracted beam of monochromatic, light incident from one medium to another are same, Reason : The incident, reflected and refracted rays are coplanar., Correct Answer: B, Solution : If both assertion and reason are true but reason is not the correct explanation, of the assertion., 17) Assertion : By roughening the surface of a glass sheet its transparency can be, reduced., Reason : Glass sheet with rough surface absorbs more light., Correct Answer: C, Solution : When glass surface is made rough then the light falling on it is scattered in, different direction due to which its transparency decreases., 18) Assertion : Diamond glitters brilliantly., Reason : Diamond does not absorb sunlight., Correct Answer: B, Solution : Diamond glitters brilliantly because light enters in diamond suffers total internal, reflection. All the light entering in it comes out of diamond after number of reflections and, no light is absorb by it., 19) Assertion : The cloud in sky generally appear to be whitish., Reason : Diffraction due to cloud is efficient in equal measure at all wavelengths., Correct Answer: C, Solution : The clouds consist of dust particles and water droplets. Their size is very large, as compared to the wavelength of the incident light from the sun. So there is very little, scattering of light. Hence the light which we receive through the clouds has all the colours, of light. As a result of this, we receive almost white light. Therefore, the cloud are generally, white., Case based Questions (Ray optics), 1) Total internal reflection.

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Physics / XII (2020-21), (i) What is refractive index of a medium(in terms of speed of light), a) Speed of light in medium/speed of light in vacuum, b) Speed of light in vacuum/speed of light in medium, c) Speed of light in medium  speed of light in vacuum, d) None of the above., (ii) In the above diagram, calculate the speed of light in the liquid of unknown refractive, index., a) 1.2 × 108 m/𝑠, b) 1.4 × 108 m/𝑠, c) 1.6 × 108 m/𝑠, d) 1.8 × 108 m/𝑠, (iii) What is refractive index of a medium(in terms of real and apparent depth)., a) Real depth/ App depth, b) App/ Real depth, c) App  Real depth, d) Real + App depth, (iv) What is the relation between refractive index and critical angle for a medium., a) n = 1/sin ic, b) n = sin ic, c) 1 = n/ sin ic, d) None of the above, Answer:, i) (b), ii) (d), iii) (a), iv) (a), 2) Advance sunrise and delayed sunset, , (i) What is the principal behind Advance sunrise and delayed sunset., (a) Reflection., (b) Refraction., c) Dispersion, d) Total internal reflection., (ii) For how much time the sun is visible apparently after sunset., (a) Approx. 5 minutes, b) Approx. 10 minutes

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Physics / XII (2020-21), c) Approx. 2 minutes, d) None of the above, (iii) The Sun looks reddish while sunset or sunrise, because., a) red colour is highly scattered., b) red colour is least scattered., c) of refraction of light., d) of dispersion of light., iv) The above given phenomenon cannot be observed on the moon because, a) total internal reflection could not take place on the moon, b) there is no atmosphere, c) Sun is not visible from the moon, d) None of the above, Answers, i) (b), ii) (c), iii) (b), iv) (b), 3) Optical fibres: Now-a-days optical fibres are extensively used for transmitting audio and, video signals through long distances. Optical fibres too make use of the phenomenon of, total internal reflection. Optical fibres are fabricated with high quality composite, glass/quartz fibres. Each fibre consists of a core and cladding. The refractive index of the, material of the core is higher than that of the cladding. When a signal in the form of light is, directed at one end of the fibre at a suitable angle, it undergoes repeated total internal, reflections along the length of the fibre and, finally comes out at the other end. Since light undergoes total internal reflection at each, stage, there is no appreciable loss in the intensity of the light signal. Optical fibres are, fabricated such that light reflected at one side of inner surface strikes the other at an angle, larger than the critical angle. Even if the fibre is bent, light can easily travel along its length., Thus, an optical fibre can be used to act as an optical pipe., , i) Which of the following statement is not true., a) Optical fibres is based on the principle of total internal reflection., b) The refractive index of the material of the core is less than that of the cladding., c) an optical fibre can be used to act as an optical pipe., d) there is no appreciable loss in the intensity of the light signal while propagating through, an optical fibre.

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Physics / XII (2020-21), ii) What is the condition for total internal reflection to occur?, a) angle of incidence must be equal to the critical angle., b) angle of incidence must be less than the critical angle., c) angle of incidence must be greater than the critical angle., d) None of the above., iii) Which of the following is not an application of total internal reflection?, a) Mirage, b) Sparkling of diamond, c) Splitting of white light through a prism., d) Totally reflecting prism., iv) Optical fibres are used extensively to transmit, a) Optical Signal, b) current, c) Sound waves, d) None of the above, Answers, i) (b), ii) (c), iii) (c), iv) (a), PART - B, Wave Optics (Assertion and Reasoning Based Questions), 1) Assertion : When a light wave travels from a rarer to a denser medium, it loses speed., The reduction in speed imply a reduction in energy carried by the light wave., Reason : The energy of a wave is proportional to velocity of wave., Correct Answer: D, Solution : When a light wave travel from a rarer to a denser medium it loses speed, but, energy carried by the wave does not depend on its speed. Instead, it depends on the, amplitude of wave., 2) Assertion : No interference pattern is detected when two coherent sources are infinitely, close to each other., Reason : The fringe width is inversely proportional to the distance between the two slits., Correct Answer: A, Solution :, When d is negligibly small, fringe width β which is proportional to 1/d may, become too large. Even a single fringe may occupy the whole screen. Hence the pattern, cannot be detected.

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Physics / XII (2020-21), 3) Assertion : For best contrast between maxima and minima in the interference pattern of, Young’s double slit experiment, the intensity of light emerging out of the two slits should be, equal., Reason : The intensity of interference pattern is proportional to square of amplitude., Correct Answer: B, Solution : When intensity of light emerging from two slits is equal, the intensity at minima,, Imin = (√ Ia - √Ib) 2=0, or absolute dark. It provides a better contrast., 4) Assertion: In Young’s experiment, the fringe width for dark fringes is different from that, for white fringes., Reason : In Young’s double slit experiment the fringes are performed with a source of, white light, then only black and bright fringes are observed., Correct Answer: D, Solution : In Young’s experiments fringe width for dark and white fringes are same while in, Young’s double slit experiment when a white light as a source is used, the central fringe is, white around which few coloured fringes are observed on either side., 5) Assertion : When a tiny circular obstacle is placed in the path of light from some, distance, a bright spot is seen at the centre of shadow of the obstacle., Reason : Destructive interference occurs at the centre of the shadow., Correct Answer: C, Solution : As the waves diffracted from the edges of circular obstacle, placed in the path, of light interfere constructively at the centre of the shadow resulting in the formation of a, bright spot., 6) Assertion : Interference pattern is made by using blue light instead of red light, the, fringes becomes narrower., Reason : In Young?s double slit experiment, fringe width is given by relation β = λD/d., Correct Answer: A, Solution :, β = λD/d., 7) Assertion: Diffraction is common in sound but not common in light waves., Reason : Wavelength of light is more than the wavelength of sound., Answer (c), Solution: If assertion is true but reason is false, 8) Assertion : In Young's double slit experiment if wavelength of incident monochromatic, light is just doubled, number of bright fringe on the screen will increase., Reason : Maximum number of bright fringe on the screen is directly proportional to the, wavelength of light used., Answer: (d), Solution: Wavelength is inversely proportional to the number of fringes, hence by doubling, the wavelength the number of fringes decreases. Hence Assertion and reason are false.

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Physics / XII (2020-21), 9) Assertion : In interference and diffraction, light energy is redistributed., Reason :There is no gain or loss of energy, which is consistent with the principle of, conservation of energy., Answer: (a), Solution: In interference and diffraction, light energy is redistributed. If it reduces in one, region, producing a dark fringe, it increases in another region, producing a bright fringe., There is no gain or loss of energy, which is consistent with the principle of conservation of, energy., 10) Assertion : If complete YDSE (Young’s Double Slit Experiment) is dipped in the liquid, from the air, then fringe width decreases., Reason : Wavelength of light decreases, when we move from air to liquid., Answer: (a), 11) Assertion: No sustained interference pattern is obtained when two electric bulbs of the, same power are taken., Reason: Phase difference between waves coming out of electric bulbs is not constant., Answer: (a), 12) Assertion: The maximum intensity in YDSE (Young’s Double Slit Experiment) is four, times the intensity due to each slit when they are identical., Reason: The phase difference between the interfering waves is 2nπ at the position of, maxima where n = 0, 1, 2, ......, Answer: (a), , CASE BASED QUESTIONS (WAVE OPTICS), 1) Refraction of a plane wave, , i) What is the angle made by the ray of light on the wavefront?, a) 90˚, b) 0˚, c) 45˚, d) None of the above

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Physics / XII (2020-21), ii) Which parameter remains unchanged while a ray of light propagates from one medium, to another?, a) velocity, b) Wave length, c) frequency, d) None of the above, iii) According to the above given fig., identify the correct expression for Snell’s law., a) n1 sin i = n2 sin r, b) n2 sin i = n1 sin r, c) n21 = sin r/ sin i, d) None of the above, iv) When a ray of light travels from a denser to a rarer medium, it, a) it bends towards the normal, b) it travels in a straight line irrespective of angle of incidence., c) it bends away from the normal, d) None of the above, Answers:, i) (a), ii) (c), iii) (a), iv) (c), 2) Interference (Young’s Double slit experiment)

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Physics / XII (2020-21), i) What is the path difference between the two light waves coming from coherent sources,, which produces 3rd maxima., a) , b) 2, c) 3, d0, ii) What is the correct expression for fringe width()., a) d/D, (b) dD, (c) d/D, (d) D/d, iii) what is the phase diff. between two interfering waves producing 1st dark fringe., a) π, b) 2π, c) 3π, d) 4π, iv) The ratio of the widths of two slits in Young’s double slit experiment is 4 : 1. Evaluate, the ratio of intensities at maxima and minima in the interference pattern., a) 1:1, b) 1:4, c) 3:1, d) 9:1, v) In a Young’s double slit experiment, the separation between the slits is 0.1 mm, the, wavelength of light used is 600 nm and the interference pattern is observed on a screen, 1m away. Find the separation between bright fringes., (a), 6.6 mm, (b), 6.0 mm, (c), 6m, (d), 60cm, Answers:, i) (c), ii) (d), iii) (a), iv) (d), v) (b)

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Physics / XII (2020-21), 3) Diffraction at a single slit, , (i) In the phenomena of Diffraction of light when the violet light is used in the experiment is, used instead of red light then,, (a), Fringe width increases, (b), No change in fridge width, (c), Fringe width decreases, (d), Colour pattern is formed, (ii) Diffraction aspect is easier to notice in case of the sound waves then in case of the light, waves because sound waves, (a), Have longer wavelength, (b), Shorter wavelength, (c), Longitudinal wave, (d), Transverse waves, (iii) Diffraction effects show that light does not travel in straight lines. Under what condition, the concepts of ray optics are valid. ( D = distance of screen from the slit)., (a), D < Zf, (b), D = Zf, (c), D > Zf, (d), D << Zf, (iv) when 2nd secondary maxima is obtained in case of single slit diffraction pattern, the, angular position is given by, (a) , (b) /2, (c) 3/2, (d) 5/2, Answers:, (i) (c), (ii) (a), (iii) (d), (iv) (d)

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Physics / XII (2020-21), , UNIT-VII, DUAL NATURE OF RADIATION AND MATTER, Directions: In each of the following questions, a statement of Assertion (A) is given followed, by a corresponding statement of Reason (R) just below it. Of the statements, mark the, correct answer as:, (A)If both assertion and reason are true and reason is the correct explanation of assertion, (B)If both assertion and reason are true but reason is not the correct explanation of, assertion, (C)If assertion is true and reason is false, (D)If both assertion and reason are false, , 1. Assertion: A photon has no rest mass , yet it carries definite momentum., Reason: Momentum of photon is due to its energy and hence its equivalent mass., (a)A, (b)B, (c)C, (d)D, 2. Assertion: Mass of moving photon varies inversely as the wavelength., Reason: Energy of the particle = mass x (speed of light)2, (a)A, (b)B, (c)C, (d)D, 3. Assertion: In photoelectron emission, the velocity of electron ejected from near the, surface is larger, than that coming from interior of metal., Reason. The velocity of ejected electron will be zero., (a)A, (b)B, (c)C, (d)D, 4. Assertion: A photocell is called an electric eye., Reason. When light is incident on some semiconductor, its electrical resistance is, reduced ., (a)A, (b)B, (c)C, (d)D, 5. Assertion: The de Broglie equation has significance for any microscopic or submicroscopic particle., Reason: The de Broglie wavelength is inversely proportional to the mass of the object if, velocity is constant., (a)A, (b)B, (c)C, (d)D, 6. Assertion : A particle of mass M at rest decay into particles of masses m 1 and m2,having, non-zero, velocities will have ratio of de-Broglie wavelengths unity., Reason. Here we cannot apply conservation of linear momentum., (a)A, (b)B, (c)C, (d)D, 7. Assertion: Photoelectric effect demonstrates the wave nature of light., Reason. The number of photoelectrons is proportional to the frequency of light., (a)A, (b)B, (c)C, (d)D, 8. Assertion:When acertain wavelength of light falls on a metal surface it ejects electron., Reason. Light has wave nature., (a)A, (b)B, (c)C, (d)D

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Physics / XII (2020-21), 9. Assertion: As work function of a material increases by some mechanism, it requires, greater energy to excite the electrons from its surface., Reason. A plot of stopping potential (V2) versus frequency (v) for different materials,has, greater slope for metals with greater work functions., (a)A, (b)B, (c)C, (d)D, 10. Assertion : Light of frequency 1.5 times the threshold frequency is incident on photosensitive material.If the frequency is halved and intensity is doubled the photo current, remains unchanged., Reason.The photo electric current varies directly with the intensity of light and frequency of, light., (a) A, (b)B, (c)C, (d) D, 11. Assertion. The de-Broglie wavelength of a neutron when its kinetic energy is k is λ. Its, wavelength is 2 λ when its kinetic energy is 4k., Reason. The de - Broglie wavelength λ is proportional to square root of the kinetic energy., (a)A, (b)B, (c)C, (d)D, 12. Assertion. The de – Broglie wavelength of a molecule varies inversely as the square, root of temperature., Reason. The root mean square velocity of the molecule depends on the temperature., (a)A, (b)B, (c)C, (d)D, Answers, Q1. (a), Q6. (a), Q11. (d), , Q2. (a), Q7. (d), Q12. (a), , Q3. (c), Q8. (b), Q13. (b), , Q4. (c), Q9. (c), , Q5. (a), Q10. (d), , CASE BASED QUESTIONS, DUAL NATURE OF RADIATION AND MATTER, 1. The photoelectric emission is possible only if the incident light is in the form of packets of, energy, each having a definite value, more than the work function of the metal. This shows, that light is not of wave nature but of particle nature. It is due to this reason that, photoelectric emission was accounted by quantum theory of light., Q1. Packet of energy are called, (a)electron, (b) quanta, (c)frequency, (d)neutron

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Physics / XII (2020-21), Q2. One quantum of radiation is called, (a)meter, (b) meson, (c) photon, (d)quark, Q3. Energy associated with each photon, (a)hc, (b) mc, (c) hv, (d)hk, Q4. Which of the following waves can produce photo electric effect, (a). UV radiation, (b). Infrared radiation, (c). Radio waves, (d) .Microwaves, Q5. Work function of alkali metals is, (a)less than zero, (b) just equal to other metals, (c) greater than other metals, (d) quite less than other metals, Answer, Q1.(b), , Q2.(c), , Q3.(c), , Q4.(a), , Q5.(d), , Q2. According to de-Broglie a moving material particle sometimes acts as a wave and, sometimes as a particle or a wave is associated with moving material particle which, controls the particle in every respect. The wave associated with moving material particle is, called matter wave or de-Broglie wave whose wavelength called de-Broglie wavelength, is, given by λ = h/mv, 1. The dual nature of light is exhibited by, (a) diffraction and photo electric effect, (b) photoelectric effect, (c) refraction and interference, (d)diffraction and reflection., 2. If the momentum of a particle is doubled , then its de-Broglie wavelength will, (a)remain unchanged (b)become four times, (c) become two times (d)become half

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Physics / XII (2020-21), 3. If an electron and proton are propagating in the form of waves having the same, , λ , it implies that they have the same, (a)energy, (b)momentum, (c)velocity, (d)angular momentum, 4. Velocity of a body of mass m, having de-Broglie wavelength λ , is given by relation, (a) v = λ h/m, (b) v = λm/h, (c) v = λ/hm, (d) v = h/ λm, 5. Moving with the same velocity , which of the following has the longest de Broglie, wavelength?, (a)ᵦ -particle, (b) α -particle, (c) proton, (d) neutron., Answer, Q1.(a), , Q2.(d), , Q3.(b), , Q4.(d), , Q5.(a)

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Physics / XII (2020-21), , UNIT-VIII, ATOM AND NUCLEUS, Instructions:, A) If both assertion and reason are true and the reason is the correct explanation of, the assertion., B) If both assertion and reason are true but reason is not the correct explanation of, the assertion., C) If assertion is true but reason is false., D) If the assertion and reason both are false., E) If assertion is false but reason is true., 1., Assertion: It is not possible to use 35Cl as the fuel for fusion energy., Reason: The binding energy of 35Cl is too small., Correct Answer: C, Solution : In fusion, lighter nuclei are used so, fusion is not possible with, binding energy of 35Cl is not too small., , 35Cl., , Also, , 2., Assertion : 90Sr from the radioactive fall out from a nuclear bomb ends up in the, bones of human beings through the milk consumed by them. It causes impairment of, the production of red blood cells., Reason : The energetics b-particles emitted in the decay of 90Sr damage the bone, marrow., Correct Answer: A, Solution : 90Sr38 decays to 90Y39 by the emission of β − rays. Sr gets absorbed in bones, along with calcium. Reason is also true., 3., Assertion : Neutrons penetrate matter more readily as compared to protons., Reason, : Neutrons are slightly more massive than protons., Correct Answer: B, Solution : Neutron is about 0.1 more massive than proton. But the unique thing about, the neutron is that while it is heavy, it has no charge (it is neutral). This lack of, charge gives it the ability to penetrate matter without interacting as quickly as the, beta particles or alpha particles.

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Physics / XII (2020-21), 4., Assertion : Neutrons penetrate matter more readily as compared to protons., Reason, : Neutrons are slightly more massive than protons., Correct Answer: B, Solution : Neutron is about 0.1 more massive than proton. But the unique thing about, the neutron is that while it is heavy, it has no charge (it is neutral). This lack of, charge gives it the ability to penetrate matter without interacting as quickly as the, beta particles or alpha particles., 5., Assertion :, Reason, :, , Radioactive nuclei emit β−1 particles., Electrons exist inside the nucleus., , Correct Answer: C, Solution : Nuclear stability depends upon the ratio of neutron to proton. If the n/p ratio, is more than the critical value, then a neutron gets converted into a proton forming, a β− particle in the process. So electrons do not exist in the nucleus but they result in, some nuclear transformation., 6., Assertion : ZXA undergoes 2α decays 2β− decays and 2Ƴ decays and the daughter, product is Z−2YA−8., Reason : In a-decay the mass number decreases by 4 and atomic, number, decreases by 2. In b- decay the mass number remains unchanged, but atomic, number increases by 1 only., Correct Answer: A, Solution :, , ZXA, , →, , 2(2He4) + 2(−1e0) + 2γ + z−2XA−8, , 7., Assertion : Density of all the nuclei is same., Reason : Radius of nucleus is directly proportional to the cube root of mass number., Correct Answer: A, Solution : Experimentally it is found that the average radius of a nucleus is given by R, = R0A1/3 where R0 = 1.1×10−15m = 1.1 fm and A = mass number.

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Physics / XII (2020-21), 8., Assertion : Isobars are the element having same mass number but different atomic, number., Reason : Neutrons and protons are present inside nucleus., Correct Answer: B, 9., Assertion : The force of repulsion between atomic nucleus and a-particle varies with, distance according to inverse square law., Reason : Rutherford did a-particle scattering experiment., Correct Answer: B, Solution : Rutherford confirmed the repulsive force on a-particle due to nucleus, varies with distance according to inverse square law and that the positive charges, are concentrated at the centre and not distributed throughout the atom., , 10., Assertion : The positively charged nucleus of an atom has a radius of almost 10−15m., Reason : In a-particle scattering experiment, the distance of closest approach for aparticles is ≃ 10−15m., Correct Answer: A, Solution : In a-particle scattering experiment, Rutherford found a small number of aparticles which were scattered back through an angle approaching to 180∘. This is, possible only if the positive charges are concentrated at the centre or nucleus of the, atom., 11., Assertion : According to classical theory, the proposed path of an electron in, Rutherford atom model will be parabolic., Reason : According to electromagnetic theory an accelerated particle continuously, emits radiation., Correct Answer: E, Solution : According to classical electromagnetic theory, an accelerated charge, continuously emits radiation. As electrons revolving in circular paths are constantly, experiencing centripetal acceleration, hence they will be losing their energy

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Physics / XII (2020-21), continuously and the orbital radius will go on decreasing and form spiral and finally, the electron will fall on the nucleus., 12., Assertion : Electrons in the atom are held due to coulomb forces., Reason : The atom is stable only because the centripetal force due to Coulomb?s, law is balanced by the centrifugal force., Correct Answer: C, Solution : According to postulates of Bohr?s atom model, the electron revolve round, the nucleus in fixed orbit of definite radii. As long as the electron is in a certain orbits, it does not radiate any energy., 13., Assertion : The electron in the hydrogen atom passes from energy level n=4 to the, n=1 level. The maximum and minimum number of photon that can be emitted are six, and one respectively., Reason : The photons are emitted when electron make a transition from the higher, energy state to the lower energy state., Correct Answer: B, Solution : Maximum number of photon is given by all the transitions possible =4C2 =, 6. Minimum number of transition = 1, that is directly jump from 4 to 1., 14., Assertion : Hydrogen atom consists of only one electron but its emission spectrum, has many lines., Reason : Only Lyman series is found in the absorption spectrum of hydrogen atom, whereas in the emission spectrum, all the series are found., Correct Answer: B, Solution : When the atom gets appropriate, rises to some higher energy level. Now it, energy level or come to the lowest energy, energy lends, hence all possible transitions, are seen in the spectrum., , energy from outside, then this electron, can return either directly to the lower, level after passing through other lower, take place in the source and many lines

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Physics / XII (2020-21), 15., Assertion : It is essential that all the lines available in the emission spectrum will, also be available in the absorption spectrum., Reason : The spectrum of hydrogen atom is only absorption spectrum., Correct Answer: D, Solution : Emission transitions can take place between any higher energy level and, any energy level below it while absorption transitions start from the lowest energy, level only and may end at any higher energy level. Hence number of absorptions, transitions between two given energy levels is always less than the number of, emission transitions between same two levels., 16., Assertion : For the scattering of a-particles at a large angles, only the nucleus of the, atom is responsible., Reason : Nucleus is very heavy in comparison to α particle., Correct Answer: A, Solution : We know that an electron is very light particle as compared to an, particle. Hence electron cannot scatter the a-particle at large angles, according, law of conservation of momentum. On the other hand, mass of nucleus, comparable with the mass of a-particle, hence only the nucleus of atom, responsible for scattering of a-particles., , ato, is, is, , 17., Assertion : All the radioactive elements are ultimately converted in lead., Reason : All the elements above lead are unstable., Correct Answer: C, Solution : All those elements which are heavier than lead are radioactive. This is, because in the nuclei of heavy atoms, besides the nuclear attractive, forces,, repulsive forces between the protons are also effective and these forces reduce the, stability of the nucleus. Hence, the nuclei of heavier elements are being converted, into lighter and lighter elements by emission of radioactive radiation. When they are, converted into lead, the emission is stopped because the nucleus of lead is stable, (or lead is most stable elements in radioactive series).

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Physics / XII (2020-21), 18., Assertion : Amongst alpha, beta and gamma rays, a-particle, penetrating power., Reason : The alpha particle is heavier than beta and gamma rays., , has maximum, , Correct Answer: D, Solution : The penetrating power is maximum in case of gamma rays because, gamma rays are an electromagnetic radiation of very small wavelength., 19., Assertion : The ionising power of β-particle is less compared to a-particles but their, penetrating power is more., Reason : The mass of β-particle is less than the mass of a-particle., Correct Answer: B, Solution : β particles, being emitted with very high speed compared to α particles,, pass very little time near the atoms of the medium. So the probability of the atoms, being ionised is comparatively less. But due to this reason, their loss of energy is, very slow and they can penetrate the medium through a sufficient depth., 20., Assertion : The mass of b-particles when they are emitted is higher than the mass of, electrons obtained by other means., Reason : β-particle and electron, both are similar particles., Correct Answer: B, Solution : β-particles are emitted with very high velocity (up to 0.99 c)., So,, according to Einstein?s theory of relatively, the mass of a β -particle is much higher, compared to is` its rest mass (m0). The velocity of electrons obtained by other, means is very small compared to c (Velocity of light). So its mass remains nearly m0., But b-particle and electron both are similar particles., 21. Assertion : Radioactivity of 108 un-decayed radioactive nuclei of half-life of 50, days is equal to that of 1.2×108 number of un-decayed nuclei of some other material, with half-life of 60 days., Reason : Radioactivity is proportional to half-life., Correct Answer : C, Solution : Radioactivity –dN/dt = λN = 0.693/T1/2 Radioactivity is proportional to 1/T1/2, and not to T1/2.

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Physics / XII (2020-21), 22., Assertion : Fragments produced in the fission of U235 are radioactive., Reason : The fragments have abnormally high proton to neutron ratio., Correct Answer: C, Solution : Fragments produced in the fission of U235 are radioactive. When uranium, undergoes fission, barium and krypton are not the only products. Over 100 different, isotopes of more than 20 different elements have been detected among fission, products. All of these atoms are, however, in the middle of the periodic table, with, atomic numbers ranging from 34 to 58. Because the neutron-proton ratio needed for, stability in this range is much smaller than that of the original uranium nucleus, the, residual nuclei called fission fragments, always have too many neutrons for stability., A few free neutrons are liberated during fission and the fission fragments undergo a, series of beta decays (each of which increases Z by one and decreases N by one), until a stable nucleus is reached. During decay of the fission fragments, an average, of 15 MeV of additional energy is liberated., 23., Assertion : The mass of a nucleus can be either less than or more than the sum of, the masses of nucleons present in it., Reason : The whole mass of the atom is considered in the nucleus., Correct Answer: E, Solution : The whole mass of the atom is concentrated at nucleus and M(nucleus) <, (Sum of the masses of nucleus) because, when nucleous combines, some energy is, wasted., 24., Assertion : Only those nuclei which are heavier than, Reason : Nuclei of elements heavier than lead are unstable., , lead, , are, , radioactive., , Correct Answer : D, Solution : Some lighter nuclei are also radioactive., 25., Assertion : In one half-life of a radioactive substance more number of nuclei are decayed, than in one average life., Reason : Average life = Half -life/Ln (2), Correct Answer : D)

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Physics / XII (2020-21), Solution : Average life is more Hence more nuclei decay in one average life ., 26., Assertion : Nucleus of the atom does not contain electrons, yet it emits β−particles in the form of, electrons., Reason : In the nucleus, protons and neutrons exchange mesons frequently., Correct Answer D, Solution : Nucleus of the atom does not contain electrons, yet it emits β− particles in the form of, electrons. The nucleus contains only protons and neutrons but In the nucleus, protons and, neutrons do not exchange mesons frequently., , CASE BASED QUESTIONS, (ATOMS AND NUCLEI), Everything around us which has mass and, occupies space is matter. An atom is the basic unit of, matter. It cannot be broken down further using any, chemical means because it is the basic building block of, an element. Every state of matter solid, liquid, gas, and, plasma is composed of either atom either it is neutral (unionized), or ionized atoms. An atom is made up of three, particles known as protons, neutrons, and electrons. And, these particles are also made up from sub-particles., Among these three particles, protons have a positive, charge while electrons carry a negative charge and the, third particle neutrons have no electrical charge. And the, charge of atoms depends on the number of protons and, electrons, i.e an atom is electrically neutral if the number of protons and electrons are, equal. If an atom has more or fewer electrons than protons, then it has an overall negative, or positive charge, respectively. These atoms are extremely small or you can say their, typical sizes are around 100 picometers. So the dense region consisting of protons and, neutrons at the center of an atom is known as the atomic nucleus of an atom. Every atom, is composed of such nucleus and some elections will be surrounding it. Studying these, atoms and Nuclei will help us to have a thorough understanding of matter. Studying about, the nucleus and its reactions will help us to understand more about nuclear energy, which, is a very useful renewable energy. That's why it is very important to study about Atoms and, Nuclei., Q1. What is the basic unit of matter?, a) Atom, b) Electron, c) Proton, d) Neutron

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Physics / XII (2020-21), Q2. Which particle is responsible for the ionization of the atom?, a) Positron, b) Electron, c) Proton, d) Neutron, Q3. If number of protons in an atom is equal to (number of electrons + 2). Then the atom is, said to be, a) Single ionized positive ion, b) Single ionized positive atom, c) Double ionized positive ion, d) Double ionized positive atom, Q4. Which is the most dense part of an atom?, a) The exact central part of the atom., b) The region at the center of atom containing neutrons and protons., c) Outer edge of the atom, d) None of the above, Answer : Q1 – a; Q2 – b; Q3 – c; Q4 – b, Bohr’s Atomic Model, To study about atom various scientists perform various experiments and, suggest various models of an atom with some explanation. For example, Thomson gives, the "plum pudding" model in which he said the atom consists of a positive material known, as "pudding" with some negative materials ("plums") distributed throughout. Later, famous, scientist, Rutherford gives Rutherford's model of the atom after performing an Alpha, Particle scattering experiment., This model is a modification of the earlier Rutherford Model. According to, this model, an atom consists of a small, positively-charged nucleus and negatively-charged, electrons orbiting around it in an orbital. These orbital can have different sizes, energy, etc., And the energy of the orbit is also related to its size, I.e The lowest energy is found in the, smallest orbit. So if the electron is orbiting in nth orbit then we will study about its Velocity, in nth orbital, Radius of nth orbital, Energy of electron in nth orbit, etc. Energy is also, emitted due to the transition of electrons from one orbit to another orbit. This energy is, emitted in the form of photons with different wavelengths. This wavelength is given by the, Rydberg formula. When electrons make transitions between two energy levels in an atom, various spectral lines are obtained. The emission spectrum of the hydrogen atom has been, divided into various spectral series like Lyman series, Balmer series, Paschen series Etc., Q1. The formula which gives the wavelength of emitted photon when electron jumps from, higher nergy state to lower was given by, a) Balmer, b) Paschen, c) Lymen, d) Rydberg

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Physics / XII (2020-21), Q2. What is true about Bohr’s atomic Model, a) His model was unique totally different from other, b) His model is a modification of Rutherford atomic model., c) His model is a modification of Thomson atomic model., d) None of the above, Q3. Bohr’s atomic model is applicable for, a) All types of atoms, b) Only for hydrogen atom, c) For hydrogen like atoms, d) For H2 gas., Q4. The cause of rejection of Rutherford atomic model was, a) It was totally wrong, b) It could not justify its stability, c) Rutherford was unable to explain it, d) None of the above., Answer : Q1 – d; Q2 – b; Q3 – c; Q4 – b, SIZE OF THE NUCLEUS, Rutherford was the pioneer who postulated and established the, existence of the atomic nucleus. At Rutherford’s suggestion, Geiger and Marsden, performed their classic experiment: on the scattering of α-particles from thin gold foils., Their experiments revealed that the distance of closest approach to a gold nucleus of an αparticle of kinetic energy 5.5 MeV is about 4.0 × 10 –14 m. The scattering of α-particle by the, gold sheet could be understood by Rutherford by assuming that the coulomb repulsive, force was solely responsible for scattering. Since the positive charge is confined to the, nucleus, the actual size of the nucleus has to be less than 4.0 × 10 –14 m. If we use αparticles of higher energies than 5.5 MeV, the distance of closest approach to the gold, nucleus will be smaller and at some point the scattering will begin to be affected by the, short range nuclear forces, and differ from Rutherford’s calculations. Rutherford’s, calculations are based on pure coulomb repulsion between the positive charges of the α, particle and the gold nucleus. From the distance at which deviations set in, nuclear sizes, can be inferred. By performing scattering experiments in which fast electrons, instead of αparticles, are projectiles that bombard targets made up of various elements, the sizes of, nuclei of various elements have been accurately measured. It has been found that a, nucleus of mass number A has a radius R = R 0 A1/3, where R0 = 1.2 × 10–15 m. This means, the volume of the nucleus, which is proportional to R3 is proportional to A. Thus the density, of nucleus is a constant, independent of A, for all nuclei. Different nuclei are likes drop of, liquid of constant density. The density of nuclear matter is approximately 2.3 × 10 17 kgm–3., This density is very large compared to ordinary matter, say water, which is 10 3 kg m–3. This, is understandable, as we have already seen that most of the atom is empty. Ordinary, matter consisting of atoms has a large amount of empty space.

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Physics / XII (2020-21), Q1. Relative density of nucleus with respect to water is, a) 2.3 × 1017 kgm–3, b) 2.3 × 1014 kgm–3, c) 23 × 1017 kgm–3, d) .23 × 1017 kgm–3, Q2. From R = R 0 A1/3 how can we conclude that density of almost all the nucleus is same, a) Volume being proportional to square of R density becomes independent of mass, number A, b) Volume being proportional to cube of R density becomes independent of mass, number A, c) Volume being proportional to R density becomes independent of mass number A, d) Density has no relation with R, Q3. What is the kinetic energy of α-particles bombarded towards the gold nucleus in, Geiger and Marsden classic experiment?, a) 8.8x10-13 Joule, b) 8.8x10-15 Joule, c) 8.8x10-13 Joule, d), x 10-13 Joule, e) 8.5x10-13 Joule, Q4. What is the range of volume of hydrogen nucleus?, a) 10-45m., b) 10-30m., c) 10-15m., d) 10-60m., Answer : Q1 – b; Q2 – b; Q3 – a; Q4 – a, Graphical representation Of Scattering of α Particles By Gold Nucleus, A typical graph of the total number of αparticles scattered at different angles, in a, given interval of time, is shown in Fig. The, dots in this figure represent the data points, and the solid curve is the theoretical, prediction based on the assumption that the, target atom has a small, dense, positively, charged nucleus. Many of the α-particles, pass through the foil. It means that they do, not suffer any collisions. Only about 0.14%, of the incident α-particles scatter by more, than 1º; and about 1 in 8000 deflect by more, than 90º. Rutherford argued that, to deflect, the α-particle backwards, it must experience, a large repulsive force. This force could be provided if the greater part of the mass of the, atom and its positive charge were concentrated tightly at its centre. Then the incoming α-

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Physics / XII (2020-21), particle could get very close to the positive charge without penetrating it, and such a close, encounter would result in a large deflection. This agreement supported the hypothesis of, the nuclear atom. This is why Rutherford is credited with the discovery of the nucleus., In Rutherford’s nuclear model of the atom, the entire positive charge and most, of the mass of the atom are concentrated in the nucleus with the electrons some distance, away. The electrons would be moving in orbits about the nucleus just as the planets do, around the sun. Rutherford’s experiments suggested the size of the nucleus to be about 10–, 15 m to 10–14 m. From kinetic theory, the size of an atom was known to be 10–10 m,, about 10,000 to 100,000 times larger than the size of the nucleus. Thus, the electrons, would seem to be at a distance from the nucleus of about 10,000 to 100,000 times the size, of the nucleus itself. Thus, most of an atom is empty space. With the atom being largely, empty space, it is easy to see why most α -particles go right through a thin metal foil., However, when α-particle happens to come near a nucleus, the intense electric field there, scatters it through a large angle. The atomic electrons, being so light, do not appreciably, affect the α-particles. The scattering data shown in Fig. can be analysed by employing, Rutherford’s nuclear model of the atom. As the gold foil is very thin, it can be assumed that, α-particles will suffer not more than one scattering during their passage through it., Therefore, computation of the trajectory of an alpha-particle scattered by a single nucleus, is enough. Alpha particles are nuclei of helium atoms and, therefore, carry two units,2e, of, positive charge and have the mass of the helium atom. The charge of the gold nucleus is, Ze, where Z is the atomic number of the atom; for gold Z = 79. Since the nucleus of gold is, about 50 times heavier than α-particle, it is reasonable to assume that it remains stationary, throughout the scattering process. Under these assumptions, the trajectory of an alphaparticle can be computed employing Newton’s second law of motion and the Coulomb’s, law for electrostatic force of repulsion between the alpha-particle and the positively, charged nucleus., Q1. What percentage of α particle scattered at an angle more than 90º?, a) .0125%, b) .125%, c) 1.25%, d) 12.5%, Q2. Why the nucleus of gold is about remains stationary throughout the scattering, process?, a) Because its mass is 100 times the mass of proton., b) Because its mass is 50 times the mass of proton., c) Because its mass is 150 times the mass of proton., d) Because its mass is 200 times the mass of proton., Q3. Why electrons around the gold nucleus were unable to deflect α particles?, a) Size of α particle is much greater than that of electron., b) Number of electrons around gold nucleus is very small, c) α particles is much heavier than electron., d) Electrons are negatively charged.

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Physics / XII (2020-21), Q4. What is the ratio of charge on α particle and gold nucleus?, a) .025, b) .25, c) .2, d) .5, Answer : Q1 – a; Q2 – d; Q3 – c; Q4 – a, THE LINE SPECTRA OF THE HYDROGEN ATOM, According to the third postulate of Bohr’s, model, when an atom makes a transition from the, higher energy state with quantum number ni to the, lower energy state with quantum number nf (nf < ni),, the difference of energy is carried away by a photon, of frequency ν such that hν = Eni – Enf. Since both, nf and ni are integers, this immediately shows that in, transitions between different atomic levels, light is, radiated in various discrete frequencies. For, hydrogen, spectrum,, the, Balmer, formula, corresponds to nf = 2 and ni = 3, 4, 5 etc. The, results of the Bohr’s model suggested the presence, of other series spectra for hydrogen atom–those, corresponding to transitions resulting from nf = 1, and ni = 2, 3, etc; nf = 3 and ni = 4, 5, etc. and so, on. Such series were identified in the course of, spectroscopic investigations and are known as the, Lyman, Balmer, Paschen, Brackett, and Pfund, series. The electronic transitions corresponding to these series are shown in Fig. The, various lines in the atomic spectra are produced when electrons jump from higher energy, state to a lower energy state and photons are emitted. These spectral lines are called, emission lines. But when an atom absorbs a photon that has precisely the same energy, needed by the electron in a lower energy state to make transitions to a higher energy, state, the process is called absorption. Thus if photons with a continuous range of, frequencies pass through a rarefied gas and then are analysed with a spectrometer, a, series of dark spectral absorption lines appear in the continuous spectrum. The dark lines, indicate the frequencies that have been absorbed by the atoms of the gas. The explanation, of the hydrogen atom spectrum provided by Bohr’s model was a brilliant achievement,, which greatly stimulated progress towards the modern quantum theory., Q1. The series of spectrum when electron jumps from n = 5 to n = 3 is, a) Lymen, b) Balmer, c) Paschen, d) Bracket

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Physics / XII (2020-21), Q2. Balmer series is obtained when electron transits from, a) n = 1,2,3, … to n = 5, b) n = 3,4,5 … to n = 2, c) n = 1,2,3, … to n = 4, d) n = 1,2,3, … to n = 6, Q3. From Fig. shown predict which series has waves of maximum frequency, a) Lymen, b) Balmer, c) Paschen, d) Bracket, Q4. What is the maximum energy of photon in emission spectrum of hydrogen atom, a) 13.6 eV, b) 1.36 eV, c) 1.5 eV, d) 1eV, Answer : Q1 – c; Q2 – b; Q3 – a; Q4 – a

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Physics / XII (2020-21), , UNIT-IX, ELECTRONIC DEVICES, Two statements are given – One labeled assertion (A) and other labeled reason (R)., Select the correct answer to these questions from the codes (a), (b), (c) and (d) as, given below:, a) Both A and R are true and R is the correct explanation of A, b) Both A and R are true but R is not the correct explanation of A., c) A is true but R is false., d) A is false but R is true., 1. Assertion (A): A Pure semiconductor has negative temperature coefficient of, resistance., Reason (R): On raising the temperature, more charge carriers are released,, conductance increases and resistance decreases., 2. Assertion (A): At a fix temperature, silicon will have a minimum conductivity when it, has a smaller accepter doping., Reason (R): The conductivity of and intrinsic semiconductor is slightly higher than of, a lightly doped p-type., 3. Assertion (A): The electrons in the conduction band have higher energy than those in, the valance band of a semi-conductor., Reason (R): The conduction band lies above the energy gap and valance band lies, below the energy gap., 4. Assertion (A): The energy gap between the valance band and conduction band is, greater in silicon than a germanium., Reason (R): Thermal energy produces fewer minority carriers in silicon than in, germanium., 5. Assertion (A): p- n junction diode can be used even at ultra-high frequencies., Reason (R): Capacitative reactance p- n junction diode increases as frequency, increases., 6. Assertion (A): The colour of light emitted by LED depends on its forward biasing., Reason (R): The reverse biasing of p-n junction will lower the width of depletion, layer., 7. Assertion (A): Two p-n junction diodes placed back to back will work as n-p-n, transistor., Reason (R): The p- reasons of two p-n junction diodes placed back to back will, form the base of n-p-n transistor., 8. Assertion ( A) : The number of electrons in a p- type silicon semi-conductor is less, than the number of electrons in a pure silicon semiconductor at room temperature., Reason (R): It is due to law of mass action., 9. Assertion (A): Electron has higher mobility than hole in a semiconductor., Reason (R): Mass of electron is less than the mass of hole., 10. Assertion (A): An n type semiconductor has a large number of electrons but still it is, electrically neutral.

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Physics / XII (2020-21), Reason (R): A n type semiconductor is obtained by doping an intrinsic, semiconductor with a penta valent impurity., 11. Assertion (A): V-I characteristic of p-n junction diode is same as that of any other, conductor., Reason (R): p-n junction diode behave as conductor at room temperature., 12. Assertion (A): At 0K germanium is a super conductor., Reason (R): At 0K germanium offers zero resistance., 13. Assertion (A): Semiconductor do not obey’s Om’s Law., Reason (R): Current is determined by the rate of flow of charge carriers., 14. Assertion (A): Silicon is preferred over germanium for making semiconductors, device., Reason (R): The energy gap for germanium is more than the energy gap of silicon., , CASE STUDY BASED QUESTIONS, ELECTRONIC DEVICES, 1. SEMICONDUCTOR :, A pure semiconductor germanium or silicon, free of every, impurity is called intrinsic semiconductor. At room temperature, a pure, semiconductor has very small number of current carriers (electrons and holes), .Hence its conductivity is low., When the impurity atoms of valance five or three are doped in a pure, semiconductor, we get respectively n- type or p- type extrinsic semiconductor. In, case of doped semiconductor ne nh=ni2. Where ne and nh are the number density, of electron and hole charge carriers in a pure semiconductor. The conductivity of, extrinsic semiconductor is much higher than that of intrinsic semiconductor., Answer the following questions:, Q (1). Which of the following statements is not true?, a. The resistance of intrinsic semiconductor decreases with increase of, temperature., b. Doping pures Si with trivalent impurities gives p- type semiconductors., c. The majority charges in n- type semiconductors are holes., d. A p-n junction can act as semiconductor diode., Q (2). The impurity atoms with which pure Si should be doped to make a p- type, semiconductor is, a. Phosphorus, b. Boron, c. Arsenic, d. Antimony, Q (3). Holes are majority charge carriers in, a. Intrinsic semiconductors., b. Ionic Solids, c. p- type semiconductors, d. Metals

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Physics / XII (2020-21), Q (4). At absolute zero, Si acts as, a. Non- metal, b. Metal, c. Insulator, d. None of these, Answers, 1. (c) The majority Charge carriers in n-type semiconduch as holes, 2. (b) BORON, 3. (c) p-type semiconductors, 4. (c) Insulators, 2. p-n junction diode :, p-n junction is a semiconductor diode. It is obtained by bringing, p-type semiconductor in close contact with n- type semiconductor. A thin layer is, developed at the p- n junction which is devoid of any charge carrier but has, immobile ions. It is called depletion layer. At the junction a potential barrier, appears, which does not allow the movement of majority charge carriers across, the junction in the absence of any biasing of the junction. p-n junction offers low, resistance when forward biased and high resistance when reverse biased., Q (1). In the middle of depletion layer of reverse biased p- n junction, the, a. Electric field is zero, b. Potential is zero, c. Potential is maximum, d. Electric field is maximum, Q (2). The energy band gap is maximum in, a. Metals, b. Superconductors, c. Insulators, d. Semiconductors, Q (3). The number of majority carriers crossing the junction of diode depends primarily, on the, a. Concentration of doping impurities, b. Magnitude of potential barriers, c. Magnitude of the forward bias voltage, d. Rate of thermal generation of electron –hole pairs, Q (4). Hole is, a. Antiparticle of electron, b. A vacancy created when an electron leaves covalent bond, c. Absence of free electrons, d. An artificially created particle., Answers:, 1., (c) potential is maximum, 2., (c) Insulators, 3. (d) Rate of thermal Generation of eleeton-holepair

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Physics / XII (2020-21), 4. (b) A Vacancy created when an electon leaves covalent bond, 3. Rectifiers :, A semiconductor device is used as a rectifier that allows the voltage to flow in positive, direction and very small value in the reverse direction. Now a days, there is a problem, of supply of less voltage that damages the household appliances., Q (1). In the depletion region of a diode, a. There are no mobile charges, b. Equal number of holes and electrons exist, making the region neutral., c. Recombination of holes and electrons has taken place., d. Immobile charge ions exist., Q(2).When a p-n junction diode is reverse biased then, a. No Current flows, b. The depletion reason is increased, c. The depletion reason is reduced, d. Height of potential barrier is reduced, Q(3). Diode is used as, a. Oscillator, b. Amplifier, c. Rectifier, d. Modulator, Q(4).Which one statement is incorrect?, a. Diode is used as rectifier, b. Diode is used as half wave rectifier, c. Diode is used as Amplifier, d. Diode is used as full wave rectifier, Answers:, 1. (d), 2. (b), 3. (c), 4. (c), 4. Zener diode :, Zener diode is a specially designed p-n junction diode in which both p- side and, n- side of p-n junction are heavily dopped. The zener diode is designed specially to, operate in the reverse break down voltage region continuously without being, damaged. Zener diode is used to remove the fluctuations from given voltage and, thereby provides a voltage of constant magnitude (i.e. zener diode is used as, voltage regulator)., Q (1). Zener diode is mostly used as, a. Half wave rectifier, b. Full wave rectifier, c. Voltage regulator, d. LED

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Physics / XII (2020-21), Q (2). Zener diode is designed to specially work in which region without getting, damaged ?, a. Active region, b. Break down region, c. Forward Biased, d. Reverse biased, Q (3). The depletion region of the zener diode is, a. Thick, b. Normal, c. Very thin, d. Very thick, Q (4). What is the level of dopping in zener diode?, a. Lightly dopped, b. Heavily dopped, c. Moderately dopped, d. No dopping, Answers:, 1. (c), 2. (b), 3. (c), 4. (b)

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Class: XII Session: 2020-2021, Subject: Physics, Sample Question Paper (Theory), , Maximum Marks: 70 Marks, , Time Allowed: 3 hours, , General Instructions:, (1) All questions are compulsory. There are 33 questions in all., (2) This question paper has five sections: Section A, Section B, Section C, Section D and, Section E., (3) Section A contains ten very short answer questions and four assertion reasoning, MCQs of 1 mark each, Section B has two case based questions of 4 marks each,, Section C contains nine short answer questions of 2 marks each, Section D contains, five short answer questions of 3 marks each and Section E contains three long answer, questions of 5 marks each., (4) There is no overall choice. However internal choice is provided. You have to attempt, only one of the choices in such questions., , Sr., No., , Marks, , Section – A, All questions are compulsory. In case of internal choices, attempt, any one of them., 1, , Name the physical quantity having unit J/T., , 1, , 2, , Mention one use of part of electromagnetic spectrum to which a wavelength, of 21 cm (emitted by hydrogen in interstellar space) belongs., , 1, , OR, Give the ratio of velocity of the two light waves of wavelengths 4000Å and, 8000Å travelling in vacuum., 3, , An electron with charge -e and mass m travels at a speed v in a plane 1, perpendicular to a magnetic field of magnitude B. The electron follows a, circular path of radius R. In a time, t, the electron travels halfway around the, circle. What is the amount of work done by the magnetic field?, , Page 1 of 10

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4, , A solenoid with N loops of wire tightly wrapped around an iron-core is 1, carrying an electric current I. If the current through this solenoid is reduced, to half, then what change would you expect in inductance L of the solenoid., OR, An alternating current from a source is given by i=10sin314t. What is the, effective value of current and frequency of source?, , 5, , What is the value of angular momentum of electron in the second orbit of, Bohr’s model of hydrogen atom?, , 1, , 6, , In a photoelectric experiment, the potential required to stop the ejection of 1, electrons from cathode is 4V. What is the value of maximum kinetic energy, of emitted Photoelectrons?, , 7, , In decay of free neutron, name the elementary particle emitted along with, proton and electron in nuclear reaction., , 1, , OR, In the following nuclear reaction, Identify unknown labelled X., , 8, , How does the width of a depletion region of a pn junction vary if doping, concentration is increased?, , 1, , OR, In half wave rectification, what is the output frequency if input frequency is, 25 Hz., 9, , When a voltage drop across a pn junction diode is increased from 0.70 V to 1, 0.71V, the change in the diode current is 10 mA .What is the dynamic, resistance of diode?, , 10, , Which specially fabricated pn junction diode is used for detecting light, intensity?, , 1, , For question numbers 11, 12, 13 and 14, two statements are given-one, labelled Assertion (A) and the other labelled Reason (R). Select the, correct answer to these questions from the codes (a), (b), (c) and (d), as given below., a), b), c), d), , Both A and R are true and R is the correct explanation of A, Both A and R are true but R is NOT the correct explanation of A, A is true but R is false, A is false and R is also false, , Page 2 of 10

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11, , Assertion(A) :, In a nonuniform electric field, a dipole will have translatory as well as, rotatory motion., Reason(R):, In a nonuniform electric field, a dipole experiences a force as well as, torque., , 1, , 12, , Assertion(A):, Electric field is always normal to equipotential surfaces and along the, direction of decreasing order of potential, Reason(R):, Negative gradient of electric potential is electric field., , 1, , 13, , Assertion (A):, A convex mirror cannot form real images., Reason (R):, Convex mirror converges the parallel rays that are incident on it., , 1, , 14, , Assertion(A):, A convex lens of focal length 30 cm can’t be used as a simple microscope, in normal setting., Reason (R):, For normal setting, the angular magnification of simple microscope is, M=D/f, , 1, , Section – B, Questions 15 and 16 are Case Study based questions and are, compulsory. Attempt any 4 sub parts from each question. Each, question carries 1 mark., 15, , Faraday Cage:, , 4, , A Faraday cage or Faraday shield is an enclosure made of a conducting, material. The fields within a conductor cancel out with any external fields,, so the electric field within the enclosure is zero. These Faraday cages act, as big hollow conductors you can put things in to shield them from electrical, fields. Any electrical shocks the cage receives, pass harmlessly around the, outside of the cage., , Page 3 of 10

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1. Which of the following material can be used to make a Faraday, cage?, a) Plastic, b) Glass, c) Copper, d) Wood, 2., a), b), c), d), , Example of a real-world Faraday cage is, car, plastic box, lightning rod, metal rod, , 3. What is the electrical force inside a Faraday cage when it is struck, by lightning?, a) The same as the lightning, b) Half that of the lightning, c) Zero, d) A quarter of the lightning, 4. An isolated point charge +q is placed inside the Faraday cage. Its, surface must have charge equal toa) Zero, b) +q, c) –q, d) +2q, 5. A point charge of 2C is placed at centre of Faraday cage in the shape, of cube with surface of 9 cm edge. The number of electric field lines, passing through the cube normally will bea), b), c), d), 16, , 1.9105 Nm2/C entering the surface, 1.9105 Nm2/C leaving the surface, 2.0105 Nm2/C leaving the surface, 2.0105 Nm2/C entering the surface, , Sparking Brilliance of Diamond:, , 4, , Page 4 of 10

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The total internal reflection of the light is used in polishing diamonds to, create a sparking brilliance. By polishing the diamond with specific cuts, it, is adjusted the most of the light rays approaching the surface are incident, with an angle of incidence more than critical angle. Hence, they suffer, multiple reflections and ultimately come out of diamond from the top. This, gives the diamond a sparking brilliance., 1. Light cannot easily escape a diamond without multiple internal, reflections. This is because:, a) Its critical angle with reference to air is too large, b) Its critical angle with reference to air is too small, c) The diamond is transparent, d) Rays always enter at angle greater than critical angle, 2. The critical angle for a diamond is 24.4o. Then its refractive index isa), 2.42, b), 0.413, c), 1, d), 1.413, 3. The basic reason for the extraordinary sparkle of suitably cut, diamond is that, a) It has low refractive index, b) It has high transparency, c) It has high refractive index, d) It is very hard, 4. A diamond is immersed in a liquid with a refractive index greater, than water. Then the critical angle for total internal reflection will, a) will depend on the nature of the liquid, b) decrease, c) remains the same, d) increase, 5. The following diagram shows same diamond cut in two different, shapes., , Page 5 of 10

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The brilliance of diamond in the second diamond will be:, a), b), c), d), , less than the first, greater than first, same as first, will depend on the intensity of light, , Section – C, All questions are compulsory. In case of internal choices, attempt, anyone., 17, , Two straight infinitely long wires are fixed in space so that the current in the 2, left wire is 2 A and directed out of the plane of the page and the current in, the right wire is 3 A and directed into the plane of the page. In which, region(s) is/are there a point on the x-axis, at which the magnetic field is, equal to zero due to these currents carrying wires? Justify your answer., , 18, , Draw the graph showing intensity distribution of fringes with phase angle, due to diffraction through single slit., , 2, , OR, What should be the width of each slit to obtain n maxima of double slit, pattern within the central maxima of single slit pattern?, 19, , Deduce an expression for the potential energy of a system of two point, charges q1 and q2 located at positions r1 and r2 respectively in an external, field (⃗E→), , 2, , OR, Establish the relation between electric field and electric potential at a, point., Draw the equipotential surface for an electric field pointing in +Z direction, with its magnitude increasing at constant rate along –Z direction, 20, , Explain with help of circuit diagram, the action of a forward biased p-n 2, junction diode which emits spontaneous radiation. State the least band gap, energy of this diode to have emission in visible region., , Page 6 of 10

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21, , A coil of wire enclosing an area 100 cm 2 is placed with its plane making an 2, angle 600 with the magnetic field of strength 10-1T. What is the flux through, the coil? If magnetic field is reduced to zero in 10 -3 s, then find the induced, emf?, , 22, , Two waves from two coherent sources S and S’ superimpose at X as shown 2, in the figure. If X is a point on the second minima and SX – S’X is 4.5 cm., Calculate the wavelength of the waves., , 23, , Draw the energy band diagram when intrinsic semiconductor (Ge) is doped 2, with impurity atoms of Antimony (Sb). Name the extrinsic semiconductor so, obtained and majority charge carriers in it., , 24, , Define the terms magnetic inclination and horizontal component of earth’s 2, magnetic field at a place. Establish the relationship between the two with, help of a diagram., OR, Horizontal component of earth’s magnetic field at a place is √3 times the, vertical component. What is the value of inclination at that place?, , 25, , Write two characteristics of image formed when an object is placed between, the optical centre and focus of a thin convex lens. Draw the graph showing, variation of image distance v with object distance u in this case., , 2, , Section -D, All questions are compulsory. In case of internal choices, attempt, any one., 26, , A rectangular loop which was initially inside the region of uniform and time, - independent magnetic field, is pulled out with constant velocity 𝑣 as shown, in the figure., , 3, , Page 7 of 10

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a) Sketch the variation of magnetic flux, the induced current, and, power dissipated as Joule heat as function of time., b) If instead of rectangular loop, circular loop is pulled out; do you, expect the same value of induced current? Justify your answer., Sketch the variation of flux in this case with time., 27, , A variable resistor R is connected across a cell of emf E and internal, resistance r., a), b), c), , 3, , Draw the circuit diagram., Plot the graph showing variation of potential drop across R as, function of R., At what value of R current in circuit will be maximum., OR, , A storage battery is of emf 8V and internal resistance 0.5 ohm is being, charged by d.c supply of 120 V using a resistor of 15.5 ohm, , 28, , a), b), , Draw the circuit diagram., Calculate the potential difference across the battery., , c), , What is the purpose of having series resistance in this circuit?, , a) Explain de-Broglie argument to propose his hypothesis. Show that deBroglie wavelength of photon equals electromagnetic radiation., , 3, , b) If, deuterons and alpha particle are accelerated through same potential,, find the ratio of the associated de-Broglie wavelengths of two., OR, State the main implications of observations obtained from various, photoelectric experiments. Can these implications be explained by wave, nature of light? Justify your answer., , Page 8 of 10

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29, , Derive an expression for the frequency of radiation emitted when a 3, hydrogen atom de-excites from level n to level (n – 1). Also show that for, large values of n, this frequency equals to classical frequency of revolution, of an electron., , 30, , a) Give one point of difference between nuclear fission and nuclear, fusion., , 3, , b) Suppose we consider fission of a 5626Fe into two equal fragments of, 28, 13Al nucleus. Is the fission energetically possible? Justify your answer, by working out Q value of the process., Given (m)5626Fe = 55.93494 u and (m)2813Al = 27.98191, Section – E, All questions are compulsory. In case of internal choices, attempt, any one., 31, , a) State Gauss’s law in electrostatics. Show that with help of suitable figure 5, that outward flux due to a point charge Q, in vacuum within gaussian, surface, is independent of its size and shape., b) In the figure there are three infinite long thin sheets having surface, charge density +2σ, -2σ and +σ respectively. Give the magnitude and, direction of electric field at a point to the left of sheet of charge density, +2σ and to the right of sheet of charge density +σ., , Page 9 of 10

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OR, a) Define an ideal electric dipole. Give an example., b) Derive an expression for the torque experienced by an electric, dipole in a uniform electric field. What is net force acting on this, dipole., c) An electric dipole of length 2cm is placed with its axis making an, angle of 600 with respect to uniform electric field of 105N/C., If it experiences a torque of 8√3 Nm, calculate the (i) magnitude of, charge on the dipole, and its potential energy., 32, , a) Derive the expression for the current flowing in an ideal capacitor, , 5, , and its reactance when connected to an ac source of voltage, V=V0sinωt., b) Draw its phasor diagram., c) If resistance is added in series to capacitor what changes will, occur in the current flowing in the circuit and phase angle between, voltage and current., OR, a) State the principle of ac generator., b) Explain with the help of a well labelled diagram, its working and, obtain the expression for the emf generated in the coil., c) Is it possible to generate emf without rotating the coil? Explain, 33, , a) Define a wave front., , 5, , b) Draw the diagram to show the shape of plane wave front as they, pass through (i) a thin prism and (ii) a thin convex lens. State the, nature of refracted wave front., c) Verify Snell’s law of refraction using Huygens’s principle., OR, a), b), c), , State two main considerations taken into account while, choosing the objective of astronomical telescope., Draw a ray diagram of reflecting type telescope. State its, magnifying power., State the advantages of reflecting type telescope over the, refracting type?, , Page 10 of 10