Karnataka II PUC Supplementary Physics Exam May/June 2023 PDF

GOVERNMENT OF KARNATAKA KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD II PUC SUPPLEMENTARY EXAMINATION MAY/JUNE– 2023 Subject: PHYSICS SCHEME OF EVALUATION Subject code: 33 PART – A I. Pick the correct option among the four given option...

GOVERNMENT OF KARNATAKA KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD II PUC SUPPLEMENTARY EXAMINATION MAY/JUNE– 2023 Subject: PHYSICS SCHEME OF EVALUATION Subject code: 33 PART – A I. Pick the correct option among the four given options for ALL of the following questions: 15 × 1 = 15 1. The Sl unit of electric field is (a) NC (b) NC-2 (c) NC-1 (d) Vm -1 Ans (c) NC 1 2. The electric potential due to an electric dipole falls off, at large distances (along axis) as 1 1 1 (a) (b) (c) (d) r2 r r2 r3 Ans 1 1 (b) r2 3. The capacitance of a parallel plate capacitor is independent of (a) area of plates (b) distances between the plates (c) dielectric medium present between the plates (d) potential difference between the plates Ans (d) potential difference between the plates 1 4. Potential difference can be measured accurately using (a) galvanometer (b)ammeter (c) potentiometer (d) voltmeter Ans (c) potentiometer 1 5. The cyclotron frequency is given by the equation qB qm mB q (a) νC = (b) νC = (c) νC = (d) ν C = 2π m 2π B 2π q 2π mB Ans qB 1 (a) νC = 2π m 6. The magnetic susceptibility of a diamagnetic material is (a) small and positive (b) small and negative (c) large and positive (d) large and negative Ans (b) small and negative 1 7. "The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit". This is the statement of (a) Lenz's law (b) Faraday's law of electromagnetic induction (c) Ampere's circuital law (d) Gauss' law of magnetism Ans (b) Faraday's law of electromagnetic induction 1 8. The frequency of alternating current in an AC generator is decided by (a) area of the coil (b) number of turns of the coil (c) frequency of revolution of the coil (d) strength of magnetic fleld Ans (c) frequency of revolution of the coil 1 9. In case of a pure capacitor connected to an AC source, the phase difference between voltage and current through the circuit is (a) 180° (b) 90° (c) 0° (d) 45° Ans (b) 90° 1 10. Electromagnetic waves are produced by (a) accelerated charges (b) stationary charges (c) charges in uniform motion (d) a conductor carrying steady current Ans (a) accelerated charges 1 11. A concave mirror produces virtual image when the object is placed (a) at its centre of curvature (b) beyond its centre of curvature (c) between its principal focus and centre of curvature (d) within its principal focus Ans (d) within its principal focus 1 12. The bending of light around the corners of a small opaque object is called (a) polarisation (b) diffraction (c) interference (d) refraction Ans (b) diffraction 1 13. In photoelectric experiment, increase in the intensity of light (ν >ν0) (a) increases kinetic energy of photoelectrons (b) increases photoelectric current (c) decreases kinetic energy of photoelectrons (d) photoelectric current remains constant Ans (b) increases photoelectric current 1 14. The nuclides 13 H and 23 He are (a) isotopes (b) radioactive (c) isotones (d) isobars Ans (d) isobars 1 15. The universal logic gate among the following is (a) NOT gate (b) AND gate (c) NAND gate (d) OR gate Ans (c) NAND gate 1 II. Fill in the blanks by choosing appropriate answer given in the brackets for ALL the following questions: 5×1=5 (wavefront, zero, vacuum, hysteresis, beta decay) 16. The electrostatic force between two charges is maximum in ____________. Ans vacuum 1 17. Ferromagnetic materials exhibit the phenomenon of ___________. Ans hysteresis 1 18. ___________ is defined as a surface of constant phase. Ans wavefront 1 19. In____________a nucleus spontaneously emits an electron or a positron. Ans beta decay 1 20. Energy gap (Eg ) in case of conductors is ____________. Ans Zero 1 2  PART – B III. Answer any FIVE of the following questions: 5 × 2 = 10 21. Mention any two properties of equipotential surfaces. Ans i) For any charge configuration, equipotential surface through a point is normal to the electric field at that point. ii) Work done to move a charge on an equipotential surface is zero. 2 iii) The equipotential surfaces corresponding to different potentials will be (a) very close in case of strong field and (b) far apart in case of weak field. iv) Two equipotential surfaces do not intersect Any 2 properties 22. Write the expression for the magnitude of force experienced by a charged particle moving in a magnetic field and explain the terms. Ans 1 The magnitude of force on a charged particle moving in a magnetic field is F= q v B sinθ where, q is the charge on the particle, v is the speed of the particle, B is the magnetic field, → → θ is the angle subtended by velocity vector ( v ) with magnetic field vector ( B ) ( Explanation of any 2 terms) 1 → → → NOTE: Even if F = q( v× B) = q v B sinθ n̂ is written, mark must be awarded. 23. State and explain Gauss' law in magnetism. Ans Statement: The net magnetic flux through any closed surface placed in magnetic field is zero. 1 Explanation: The magnetic field lines always form a closed loop. Therefore the total magnetic flux → → 1 through the closed surface Φ =  B ⋅ ∆S = 0 OR The isolated magnetic poles do not exist. Therefore the total magnetic flux through the closed surface → → Φ =  B ⋅ ∆S = 0 24. A coil of self inductance 2H is carrying a steady current of 1 A. Calculate the energy stored in the coil. Ans 1 2 1 Energy stored in the coil, U = LI 2 1 1 = x 2 x 12 = 1 J 2 25. What is a transformer? Mention its principle of working. Ans Transformer is a device used to increase or decrease alternating voltage (i.e. to vary ac) 1 It works on the principle of mutual induction. 1 26. What are displacement currents and conduction currents? Ans The current due to time varying electric field/flux is called displacement current. 1 OR A time varying electric field between the plates of a capacitor produces a current. It is called as displacement current. The current due to flowing charges is called conduction current. 1 27. Give any two uses of Polaroids. Ans Polaroids are used i) to produce or analyse plane polarised light, 3  ii) to control the intensity of light in sunglasses, windowpanes, etc. and 2 iii) in photographic cameras and 3D movie cameras. Any 2 uses 28. Name the spectral series of hydrogen atom which lies in (i) visible region and (ii) ultraviolet region. Ans (i) Visible region : Balmer series 1 (ii) Ultraviolet region : Lyman series 1 29. Draw the schematic diagram of a nuclear reactor based on thermal nuclear fission and label the parts. Ans Diagram 1 Labelling any 2 parts 1 PART – C IV. Answer any FIVE of the following questions: 5 × 3 = 15 30. State and explain Coulomb's law of electrostatics. Ans Statement: The electrostatic force of attraction or repulsion between two stationary point charges 1 is directly proportional to the product of the magnitude of the two charges and inversely proportional to the square of the distance between charges. Explanation: If q1 and q2 are the two point charges at rest separated by a distance ‘r’, then by Coulomb’s law. q1q 2 1 Fα r2 q1q 2  F=K r2 1 Where, K is proportionality constant and K = for air/vacuum in SI system 1 4πε 0 1 q1q 2 OR F= 4πε 0 r2 31. Derive the expression J= oE. Ans By Ohm’s law, V = RI. 1 ρ L L therefore V =  I Because, R = ρ  .  A  A I OR V = ρL J (Q J = is the current density) A V V 1 OR = ρ J => E = ρ J (because = E ) L L 4  E 1 1 Therefore J = = σ E where, σ = is the conductivity of the material of the conductor. ρ ρ 32. Explain the conversion of galvanometer into an ammeter with a circuit diagram. Ans A galvanometer can be converted into an ammeter by connecting a low resistance in parallel with it. 1 Circuit diagram 1 RG – resistance of galvanometer G. rs – shunt resistance in parallel with the galvanometer. 1 R G rs OR OR The resistance of the arrangement = R G + rs IG R G Ig G OR Shunt resistance: rs = OR S= I − IG I − Ig 33. Mention any three properties of magnetic field lines. Ans i) The magnetic field lines form closed loops. ii) The tangent to the field line at a given point represents the direction of the net magnetic field at that point. iii) The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field. 3 iv) The magnetic field lines do not intersect. Any 3 properties 34. Derive an expression for motional emf induced in a straight conductor moving perpendicular to a uniform magnetic field. Ans Labelled diagram (current not necessary in figure) 1 Magnetic flux enclosed by the loop PQRS is φB =BAcos0 = Blx 1 dφB Induced emf ε = − dt d dx ε=− ( B l x ) = −B l = B l v dt dt 1 (because –dx/dt = v) 35. What is meant by total internal reflection? Mention two uses of optical fibres. Ans When light travelling from an optically denser medium to a rarer medium incident on the interface 1 at an angle greater than a particular angle (i.e. critical angle) is completely reflected back into the same medium. This phenomenon is called the total internal reflection. Uses of optical fibres 1) Optical fibres are used in communication for the transmission of signals. 2) Optical fibres are used in endoscopy. 2 3) Plastic optical fibres are used in decorative lamps. Any 2 uses 36. e2 Show that the total energy of an electron revolving in hydrogen atom is given by, E= − 8π ε 0 r Ans +Ze is the charge of the nucleus of hydrogen atom, r is the radius of the circular orbit of an electron, v is the orbital velocity of the electron. For stable orbit, centripetal force = electrostatic force mv 2 1 e× e =. (For H atom Z = 1) r 4πε0 r 2 Where, m is the mass of electron, e - charge of an electron e2 e2 mv2 = or v2 = 4πε0 r 4πmε0 r 1 2 1 e2 e2 1 ∴ KE = mv = m = --- (1) 2 2 4πmε0 r 8πε0 r 1 1 e ( −e ) e2 P.E = =− --- (2) 4πε0 r 4πε0 r The total energy of an electron is the sum of potential energy and kinetic energy. 1 e2 e2 e2  1  e2 ∴ Total energy, TE = KE + PE = − = −1 = −. 8πε0 r 4πε0 r 4πε0 r  2  8πε0 r 37. The half-life of a radioactive sample is 4.5 x 105 years. Calculate (i) the radioactive decay constant and (ii) mean life of the sample. Ans T = 4.5 x 105 yrs = 4.5 x 105 x 365 x 24 x 60 x 60 = 1.42 x 1013 s 0.693 1 Radioactive decay constant, λ = T1 / 2 0.693 0.693 1 λ= = 4.88 x 10 −14 s −1 OR λ= = 1.54 x 10 − 6 year −1 1.42x1013 4.5 x 10 5 1 1 1 1 1 Mean life, τ = = −14 = 2.05 x 1013 s OR τ = = = 6.49x 105 year λ 4.88 x 10 λ 1.54 x 10− 6 38. Mention any three differences between intrinsic and extrinsic semiconductors. Ans Intrinsic semiconductors Extrinsic semiconductors i) It is a pure semiconductor i) It is an impure semiconductor ii) Number of holes and electrons will be ii) Number of holes and electrons will be equal unequal iii) Conductivity is zero at very low iii) Conductivity is not zero even at low 3 temperatures. temperatures. iv) Conductivity depends only on temperature iv) Conductivity depends on temperature and doping concentration. v) Conductivity is relatively less v) Conductivity is relatively more. Any three differences 6  PART – D V. Answer any THREE of the following questions: 3 × 5 = 15 39. Derive an expression for the electric field at a point due to an infinitely long thin charged straight wire using Gauss's law. Ans Diagram with direction of electric field or field lines 1 The electric field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero. Let l → length of the cylinder and λ → linear charge density The surface area of the curved part is 2πrl. Magnitude of E is same through the curved surface of the cylinder. The electric flux φ through the Gaussian surface is φ = Electric field × area = E × 2πrl ……………. (1) 1 q 1 From Gauss’s law, electric flux: φ = ε0 λl 1 The enclosed by Gaussian surface: q = λl then φ = …… (2) ε0 λl From equn(1) and (2) , E × 2πrl = ε0 λ 1 Thus, the electric field: E = 2πε 0 r 40. Derive an expression for the effective emf and effective internal resistance of two cells connected in parallel. Ans Let us consider the parallel combination of two cells with positive terminals connected to B1 and negative terminals connected B2. E1, E 2 are the emf’s of the two cells and r1, r2 their internal 1 resistances, respectively. I1 and I2 are the currents leaving the positive electrodes of the cells. Since as much charge flows in as out, we have I = I1 + I2 Let V (B1) and V (B 2) be the potentials at B1 and B2, respectively. Then, considering the first cell, the potential difference across its terminals is E1 − V V=V (B1) – V (B2) = E1 – I1 r1  I1= r1 1 Similarly for the second cell, the potential difference across its terminals is E2 − V V=V (B1) – V (B2) = E2 – I2 r2  I2= r2 E1 − V E2 − V E1 E2  1 1 Therefore, I = + = + - V  +  r1 r2 r1 r2  r1 r2  1  E1 r2 + E 2 r1   r1r2  V=   − I   ……(1)  r1 + r2   r1 + r2   If the combination of cells is replaced by a single cell of emf Eeq and 1 internal resistance req, then V = Eeq – I req …………(2) E1 r2 + E2 r1 r1r2 From equations (1) and (2), Eeq = and req = 1 r1 + r2 r1 + r2 41. Derive an expression for magnetic dipole moment of an electron revolving in hydrogen atom. Ans According to Bohr model of hydrogen and hydrogen like atoms, the negatively charged electron revolves round the nucleus of charge +Ze. Let ‘r’ be the radius of the orbit, ‘v’ be the constant speed with which electron is revolving and ‘T’ be the period of revolution of the electron. e The current associated with revolving electron, I = --- (1) 1 T 2π r 1 The period of revolution of the electron is given by T = ---(2) v ev 1 ∴I = ---(3) 2π r The magnetic moment associated with the orbital motion which is equivalent to a current loop is 1 given by μl = I A ev evr = × π r2 = --- (4) 2π r 2 1 (The derivation of expression for magnetic moment in terms of angular momentum considering following steps not compulsory) The magnitude of angular momentum of revolving electron is given by l l = me v r or vr = --- (5) Where, me  mass of the electron me el ∴μ l = --- (6) 2me 42. Derive Lens Maker's formula. Ans O be a point object placed on the principal axis of a thin convex lens of focal length ‘f ’. n1 be the RI of the medium in which object is present and n2 be the RI of 1 the material of the lens. (i) For the refraction at the surface QP1R of radius of curvature R1 At I1 a real image is formed in the medium of RI n2. n1 n 2 (n 2 − n 1 ) For this refraction + =..............(1) 1 - u v1 R1 (ii) For the refraction at the surface Q P2 R of radius of curvature R2 The final image is formed at I and I1 acts as virtual object. n2 n (n − n 1 ) 1 For this refraction 1 + 1 = 2 …………….(2) -v v - R2 8  n1 n (n − n 1 ) (n 2 − n 1 ) 1 Equation (1) + (2) gives, + 1 = 2 - -u v R1 R2  1 1  1 1   1 1 n  1 1   n1  +  = (n 2 − n 1 )  -    +  = ( 2 − 1)  -  -u v  R1 R2   - u v  n1  R1 R2   1 1 n  1 1  If u = ∞ then v = f. Therefore  +  = ( 2 − 1)  -   - ∞ f  n1  R1 R 2  1  n2  1 1  1  1 1  =  − 1  -  OR = (n 21 − 1)  -  This is lens maker’s formula. 1 f  n1   R1 R2  f  R1 R2  43. (i) Give Einstein's explanation of photoelectric effect. (3) (ii) Mention any two properties of photons. (2) Ans (i) Albert Einstein proposed a new picture of electromagnetic radiation to explain photoelectric effect. According to him, photoelectric emission does not take place by continuous absorption of 1 energy from radiation. Radiation energy consists of discrete units called quanta of energy of radiation. Each quantum of radiant energy or photon has energy hν, where h is Planck’s constant and 1 ν the frequency of light. In photoelectric effect, an electron absorbs a photon of energy (hν) of radiation. If the energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (i.e. 1 work function φ0), the electron is emitted with maximum kinetic energy (Kmax). (ii) Properties of photons (a) In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons. (b) Each photon has energy E = hν where ν is the frequency, momentum p = hν/c where c is the 2 speed of light. (c) All photons of light of a particular frequency (ν), or wavelength (λ), have the same energy and momentum. The photon energy is independent of intensity of radiation. (d) Photons are electrically neutral and are not deflected by electric and magnetic fields. (e) In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, the number of photons may not be conserved in a collision. Any 2 properties 44. (i) What is a rectifier? (1) (ii) Draw the circuit diagram and input-output waveforms of a half wave rectifier. (2) (iii) Explain the working of a half wave rectifier. (2) Ans The device which converts ac into dc is called a rectifier. 1 1 9  1 During +ve half cycle of AC input the diode (D) is forward biased and hence it conducts. 1 But during the –ve half cycle of AC input, the diode is reverse biased and hence it does not conduct. 1 So the current flows through the load only during one half cycle of AC input. Hence the name half wave rectifier. The current flowing through the RL is a rectified output in the form of pulsating D.C. VI. Answer any TWO of the following questions: 2 × 5 = 10 45. Two capacitors of capacitances 3μF and 7μF are connected in series and the combination is connected to a source of emf 10V. Calculate the effective capacitance of the combination. Also find the potential difference across each capacitor. Ans C1 C2 To find CS: Effective capacitance of series combination, CS = 1 C1 + C2 3 x10-6 x 7x10-6 = = 2.1 µ F 1 ( ) ( ) 3x10- 6 + 7 x10 -6 1 To find V1 and V2: Charge stored in the combination, Q = CS V = 2.1 x 10 -6 x 10 = 21 x 10 -6 C Q 21 x 10 −6 Potential difference across 1 st capacitor, V1 = = = 7V 1 C1 3 x 10-6 Q 21 x 10−6 Potential difference across 2 nd capacitor, V2 = = = 3V 1 C2 7 x 10- 6 Alternate method to find V1 and V2 In series combination V1 : V2 = C2 : C1 = 7 : 3 7 V1 = x 10 = 7 V 10 3 V2 = x 10 = 3 V 10 46. Calculate the current through the galvanometer in the following network. 10 Ans Applying junction rule and representing branch currents 1 From loop rule: 1 For the loop ABDA -100 I – 15 Ig + 60 (0.2 – I) = 0 ……..(1) 1 For the loop BCDB -10 (I – Ig) + 5 (0.2 – I + Ig) + 15 Ig = 0 ……(2) Simplifying (1) and (2) to get -160 I – 15 Ig = -12 ………(3) -15 I + 30 Ig = -1 ………(4) 1 Solving (3) and (4) to get Ig = 3.98 x 10-3 A 1 47. A series LCR circuit containing an inductor of 1.5 H, a capacitor of 35μF and a resistor of 50Ω is connected to ac source of 200V and 50Hz. Calculate (i) the impedance and (ii) power factor of the circuit. Ans a) XL = 2πν C = 2 x 3.14 x 50 x 1.5 = 471 Ω 1 1 1 XC = = = 90.99 Ω ≈ 91 Ω 2πν C 2 x3.14 x 50 x35 x10 − 6 1 Impedance: Z = R 2 + ( X L − X C )2 1 2 2 = 50 + ( 471 − 91) = 383.27 Ω ≈ 383 Ω 1 R 50 Power factor, cos φ = = = 0.13 Z 383.27 1 48. In a Young's double slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used. Also find the distance of fifth dark fringe from the central bright fringe. Ans nλD 1 i) Distance of nth bright fringe, (xn)B = d ( xn ) B d 1.2x10-2 x 0.28 x 10-3 1 => λ = = = 0.6x10− 5 m = 6000 x 10-10 m nD 4 x 1.4 (2n - 1)λD 1 ii) Distance of nth dark fringe, (xn)D = 2d (2 x 5 - 1) x 6000 x 10-10 x 1.4 1 Distance of 5th dark fringe, (x5)D = 2 x 0.28 x 10-3 9 x 6000 x 10-10 x 1.4 1 = = 13.5 × 10−3 m 2 x 0.28 x 10- 3 (2n + 1)λD OR If the formula (xn)D = is used, then for 5th dark fringe n=4 2d Note: Any other alternate correct method/answer should be considered. 11