CREATIVE LEARNING CLASSES, KARKALA KCET KEY ANSWERS WITH DETAILED SOLUTION  2022 DEPARTMENT OF PHYSICS VERSION CODE β A3 1. The kinetic energy of the photoelectrons increases by 0.52 eV when the wavelength of incident light is changed from 500 nm to another wavelength which is approximately (A) 1250 nm (B) 1000 nm (C) 700 nm (D) 400 nm Ans: (D) Here, change in kinetic energy, π₯πΎ = 0.52ππ, π = 500ππ, π2 =? βπ We know, πΎ1 = π β π 1 βπ πΎ2 = βπ π2 1 1 β΄ πΎ1 β πΎ2 = βπ ( β ) π1 π2 1 1 βπ₯πΎ = βπ (π β π ) 1 2 1 1 β0.52ππ = (1242ππππ) (500ππ β π ) 2 β0.52 1 1 = 500 β π 1242 2 1 1 0.52 = 500 + 1242 π2 π2 = 413ππ β 400ππ 2. The de Broglie wavelength of a particle of kinetic energy βKβ is π; the wavelength of the particle, if its πΎ kinetic energy is is 4 π (A) 2 (B) 4π (C) π (D) 2π Ans: (D) h π= π h π= β¦β¦β¦β¦...(1) β2ππΎ h πβ² = β¦β¦β¦β¦...(2) πΎ β2π 4 Divide equation (2) by (1), h πβ² β2ππΎ 4 = h π β2ππΎ πβ² = β4 π πβ² = 2π HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 0 3. The radius of hydrogen atom in the ground state is 0.53 π΄. After collision with an electron, it is found 0 to have a radius of 2.12 π΄, the principal quantum number βnβ of the final state of the atom is (A) n = 3 (B) n = 4 (C) n = 1 (D) n = 2 Ans: (D) π2 r = π0 π Given π0 = 0.53β« πππ π = 2.12β« For Hatom, Z= 1 π2 2.12 = 0.53 1 π=2 4. In accordance with the Bohrβs model, the quantum number that characterises the Earthβs revolution around the Sun in an orbit of radius 1.5Γ1011 m with orbital speed 3Γ104 ms1 is [given mass of Earth = 6Γ1024 kg] (A) 8.57Γ1064 (B) 2.57Γ1074 (C) 5.98Γ1086 (D) 2.57Γ1038 Ans: (B) According to Bohrβs model nh L = mvr = 2π 2πππ£π n= β 2Γ3.14Γ6Γ1024 Γ3Γ104 Γ1.5Γ1011 n= = 2.57 Γ 1074 6.625Γ10β34 0 5. If an electron is revolving in its Bohr orbit having Bohr radius of 0.529π΄, then the radius of third orbit is 0 0 (A) 4.761 π΄ (B) 5125 nm (C) 4234 nm (D) 4496 π΄ Ans: (A) rn = π0 π2 π3 = 0.529 ο΄ 10β10 Γ 32 π3 = 4.761 β« 6. Binding energy of a Nitrogen nucleus [ 147π], given in [ 147π] = 14.00307u (A) 206.5 MeV (B) 78 MeV (C) 104.7 MeV (D) 85 MeV Ans: (C) Mass defect Ξ m = [Z mp+ (AZ) mn] β M Ξ m = [7 x1.00783+ (147) 1.00867] β 14.00307=0.112434 Binding energy = 0.112434x931.6MeV= 104.67MeV HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 7. In a photo electric experiment, if both the intensity and frequency of the incident light are doubled, then the saturation photo electric current (A) is doubled (B) becomes four times (C) remain constant (D) is halved Ans: (A) Since the intensity of incident radiation is gets doubled, the number of electrons ejected also doubled. 8. Which of the following radiations is deflected by electric field? (A) πΎ  rays (B) πΌ  particles (C) X β rays (D) Neutrons Ans: (B) Only charged particle (like alpha particle) deflects in the electric field. 9. The resistivity of a semiconductor at room temperature is in between (A) 106 to 108 πΊ cm (B) 1010 to 1012 πΊcm (C) 102 to 105 πΊ cm (D) 103 to 106 πΊ cm Ans: (D) The resistivity of the semiconductor at room temperature is in between 103 to 10+6 Ξ©cm 10. The forbidden energy gap for βGeβ crystal at β0β K is (A) 2.57 eV (B) 6.57 eV (C) 0.071 eV (D) 0.71 eV Ans: (D) The forbidden energy gap for βGeβ crystal at β0β K is 0.72eV 11. Which logic gate is represented by the following combination of logic gate? (A) AND (B) NOR (C) OR (D) NAND Ans: (A) ____ ____ ____ ____ ____ ____ A B π¨ π© π¨+π© Μ Μ Μ Μ Μ Μ Μ Μ Μ Μ Y= π¨ + π© 0 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 1 0 1 1 0 0 0 1 The truth table of given combination is of AND gate 12. A metallic rod of mass per unit length 0.5 kg m1 is lying horizontally on a smooth inclined plane which makes an angle of 300 with the horizontal. A magnetic field of strength 0.25 T is acting on it in the vertical direction. When a current βIβ is flowing through it, the rod is not allowed to slide down. The quantity of current required to keep the rod stationary is HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI (A) 14.76 A (B) 11.32 A (C) 7.14 A (D) 5.98 A Ans: (B) IlBcosΞΈ = mgsinΞΈ mgsinΞΈ I= ππ΅πππ π π π π‘πππ I=(π) π΅ 9.8 tan300 = 0.5 Γ 0.25 = 11.32A 13. A nuclear reactor delivers a power of 109 W, the amount of fuel consumed by the reactor in one hour is (A) 0.72 g (B) 0.96 g (C) 0.04 g (D) 0.08g Ans: (C) πΈ π= π‘ πΈ = ππ‘ mc2 = Pt ππ‘ m= π2 109 Γ3600 m= (3Γ108 )2 = 0.04gm 14. The displacement βxβ (in metre) of a particle of mass βmβ (in kg) moving in the one dimension under the action of a force, is related to time βtβ (in sec) by π‘ = βπ₯ + 3. The displacement of the particle when its velocity is zero, will be (A) 6m (B) 2m (C) 4m (D) 0m Ans: (D) Given, t = βπ₯ + 3 βπ₯ = π‘ β 3 Square on both side x = (t3)2 ππ₯ v= = 2(t3) ππ‘ when velocity is zero, HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 2(t3) = 0 t=3 So, x = (t3)2 = (33)2 = 0 15. Two objects are projected at an angle π 0 and (90 β π 0 ), to the horizontal with the same speed. The ratio of their maximum vertical heights is (A) 1: tan π (B) tan2 π : 1 (C) 1 : 1 (D) tan π : 1 Ans: (B) π£0 2 sin2 ΞΈ π»1 = β¦β¦β¦β¦ (1) 2g π£0 2 (sin(90βπ))2 π£0 2 cos2 π π»2 = = = β¦β¦β¦β¦β¦ (2) 2g 2g π£0 2 sin2 ΞΈ π»1 2g = π£0 2 cos2 π π»2 2g π»1 π‘ππ2 π = π»2 1 π»1 : π»2 = π‘ππ2 π βΆ 1 16. A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 ms 1. A bob is suspended from the roof of the car by a light wire of length 1.0 m. The angle made by the wire with the vertical is (in radian) π π π (A) 0 (B) 3 (C) 6 (D) 4 Ans: (D) π cos π = ππ β¦β¦β¦β¦β¦.(1) π£2 π sin π = π β¦β¦β¦β¦β¦(2) π π£2 tan π = π π 100 = 10 Γ 10 π π = tanβ1 1 = 45Β° = 4 17. Two masses of 5 kg and 3 kg are suspended with the help of massless inextensible strings as shown in fig. when whole system is going upwards with acceleration 2 m/s2, the value of T1 is (use g = 9.8 ms2) (A) 23.6 N (B) 59 N (C) 94.4 N (D) 35.4 N HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI Ans: (C) For 3 kg, π2 β 3π = 3π π2 = 3(π + π) For 5 kg, π1 β π2 β 5π = 5π π1 = 8(π + π) π1 = 94.4 π 18. The Vernier scale of a travelling microscope has 50 divisions which coincides with 49 scale divisions. If each scale division is 0.5 mm, then the least count of the microscope is (A) 0.01 mm (B) 0.5 m (C) 0.01 cm (D) 0.5 mm Ans: (A) 50 πππ· = 1 πππ· 1 1 1 πππ· = πππ· = (0.5 ππ) 50 50 1πππ· = 0.01 ππ 19. The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 15 seconds. The angular acceleration of the motor wheel is ππππ (A) 6 (B) 8ππππ/π 2 (C) 2ππππ/π 2 (D) 4ππππ/π 2 π 2 Ans: (D) ππ = 1200 πππ ππ = 3120 πππ π‘ = 16 π ππ = ππ + πΌπ‘ ππ β ππ 3120 β 1200 π πΌ= = Γ π‘ 16 30 πΌ = 4π ππππ β2 20. The centre of mass of an extended body on the surface of the earth and its centre of gravity (A) Can never be at the same point. (B) Centre of mass coincides with the centre of gravity of a body if the size of the body is negligible as compared to the size (or radius) of the earth. (C) Are always at the same point for any size of the body. (D) Are always at the same point only for spherical bodies. Ans: (B) Centre of mass and Centre of gravity coincide if size of body is small compared to size of earth. 21. A metallic rod breaks when strain produced is 0.2%. The Youngβs modulus of the material of the rod is7 Γ 109 N/π2 . The area of cross section to support a load of 104 π is (A) 7.1Γ 10β4 π2 (B) 7.1 Γ 10β2 π2 (C) 7.1Γ 10β8 π2 (D) 7.1Γ 10β6 π2 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI Ans: (A) βπ ππ‘ππππ = = 0.2% = 0.0 02, π = 7 Γ 109 ππβ2 , πΉ = 104 π πΉπ π= π΄βπ πΉπ 104 β΄π΄= = = 7.1 Γ 10β4 π2 πβπ 7Γ109 Γ0.002 22. A tiny spherical oil drop carrying a net charge q is balanced in still air, with a vertical uniform electric 81 field of strength π Γ 105 π/π. When the field is switched off, the drop is observed to fall with terminal 7 velocity 2Γ 10β3 ππ β1 . Here g = 9.8 m/s2, Viscosity of air is 1.8Γ 10β5 ππ /π2 and the density of oil is 900 kg πβ3 . The magnitude of βqβ is (A) 1.6Γ 10β19 πΆ (B) 3.2Γ 10β19 πΆ (C) 0.8Γ 10β19 πΆ (D) 8Γ 10β19 πΆ Ans: (D) 81 πΊππ£ππ πΈ = π Γ 105 π/π π£ = 2 Γ 10β3 ππ β1 π = 1.8 Γ 10β5 ππ /π2 7 For equilibrium, qE = mg β¦β¦β¦β¦.(1) ππ = 6ππππ£ 4 ππ 3 ππ = 6ππππ£ β¦β¦β¦β¦.(2) 3 Solving (1) and (2) we get π = 8 Γ 10β19 πΆ 23. βHeat cannot be itself flow from a body at lower temperature to a body at higher temperatureβ. This statement corresponds to (A) Conservation of mass (B) First law of thermodynamics (C) Second law of Thermodynamics (D) Conservation of momentum Ans: (C) The statement represents second law of thermodynamics 24. A smooth chain of length 2 m is kept on a table such that its length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. The work done in pulling the entire chain on the table is, (Take g=10 m/π 2 ) (A) 3.6 J (B) 2.0 J (C) 12.9 J (D) 6.3 J Ans: (A) Mass of chain = 4 kg 4 6 Mass of 60 cm chain = 200 Γ 60 = 5 ππ Work done to pull chain = change in potential energy = ππ ( βπ β βπ ) 6 = (10)(30 Γ 10β2 β 0) = 3.6 π½ 5 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 25. Electrical as well as gravitational affects can be thought to be caused by fields. Which of the following is true for an electrical or gravitational field? (A) Fields are useful for understanding forces acting through a distance. (B) There is no way to verify the existence of a force field since it is just a concept. (C) The field concept is often used to describe contact forces. (D) Gravitational or Electric field does not exist in the space around an object. Ans: (A) Fields are useful for understanding forces acting through a distance. 26. Four charges +q, +2q, +q and 2q are placed at the corners of a square ABCD respectively. The force on a unit positive charge kept at the centre βOβ is (A) along the diagonal AC (B) perpendicular to AD (C) zero (D) along the diagonal BD Ans: (D) 27. An electric dipole with dipole moment 4Γ 10β9 πΆ m is aligned at 30Β° with the direction of a uniform electric field of magnitude 5Γ 104 ππΆ β1 , the magnitude of the torque acting on the dipole is (A) 10β5 π π (B) 10Γ 10β3 π π (C) 10β4 π π (D) β3 Γ 10β4 π π Ans: (C) π = ππΈ sin π = (4 Γ 10β9 )(5 Γ 104 ) sin 30Β° = 10β4 ππ 28. A charged particle of mass βmβ and charge βqβ is released from rest in an uniform electric field πΈβ . Neglecting the effect of gravity, the kinetic energy of the charged particle after βtβ second is πΈππ πΈ2 π2π‘ 2 (A) (B) π‘ 2π 2πΈ 2 π‘ 2 πΈπ 2 π (C) (D) ππ 2π‘ 2 Ans: (B) πΉ = ππΈ πΉ ππΈ π= π = π ππΈ Velocity after time t, π£ = π’ + ππ‘ = 0 + π‘ π HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 1 1 ππΈπ‘ 2 π2πΈ2 π‘ 2 Kinetic energy after time t, πΎ = ππ£ 2 = π( π ) = 2 2 2π 29. The electric field and the potential of an electric dipole vary with distance r as 1 1 1 1 (A) π 2 πππ (B) π 3 πππ π3 π2 1 1 1 1 (C) π πππ (D) π 2 πππ π2 π Ans: (B) 1 Electric field due to dipole β π3 1 Electric potential due to dipole β π2 π 30. The displacement of a particle executing SHM is given by X= 3 sin [2ππ‘ + 4 ]where βxβ is in metres and βtβ is in seconds. The amplitude and maximum speed of the particle is (A) 3 m, 6π ππ β1 (B) 3 m, 8π ππ β1 (C) 3 m, 2π ππ β1 (D) 3 m, 4π ππ β1 Ans: (A) π π₯ = 3 sin( 2ππ‘ + ) 4 πΉπππ πππ’ππ‘πππ π΄ = 3π, π = 2π ππππ β1 πππ₯πππ’π π ππππ = π΄π = 6π ππ β1 31. A parallel plate capacitor is charged by connecting a 2 V battery across it. It is then disconnected from the battery and a glass slab is introduced between plates. Which of the following pairs of quantities decrease? (A) Energy stored and capacitance (B) Capacitance and charge (C) Charge and potential difference (D) Potential difference and energy stored. Ans: (D) When the capacitor is determined from 2v battery, Q remains constant π πΆ= π π βπ= πΆ 1 βπβ πΆ When glass slab is introduced, C increases β΄V decreases π2 Energy stored = 2πΆ So, when C increases, energy stored decreases HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 32. A charged particle is moving in an electric field of 3 x 1010 V m1 with mobility 2.5 Γ106 m2/v/s, its drift velocity is (A) 2.5 104 m/s (B) 1.2 x 104 m/s (C) 7.5 x 104 m/s (D) 8.33 x 104 m/s Ans: (C) π£π π= πΈ π£π 2.5 Γ 106 = 3Γ10β10 π£π = 7.5 Γ 10β4m/s 33. Wire bound resistors are made by (A) winding the wires of an alloy of Ge, Au, Ga (B) winding the wires of an alloy of manganin, constantan, nichrome (C) Winding the wires of an alloy of Cu, Al, β¦ (D) winding the wires of an alloy of Si, Tu, Fe Ans: (B) Wire bound resistors are made by winding the wire of an alloys i.e manganin, constantan, nichrome etc. Their resistivity is relatively less sensitive to temperature. 34. Ten identical cells each of potential 'E' and internal resistance 'r', are connected in series to form a closed circuit. An ideal voltmeter connected across three cells, will read (A) 13 E (B) 7E (C) 10 E (D) 3 E Ans: (D) Voltage across each cell is E So, voltage across 3 cells is 3E. 35. In an atom electron revolve around the nucleus along a path of radius 0.72 A making 9.4 x 1018 revolutions per second. The equivalent current is  [given e = 1.6 Γ 1019 C] (A) 1.4 A (B) 1.8 A (C) 1.2 A (D) 1.5 A Ans: (D) 0 π = 0.72π΄ π = 9.4 Γ 1018 π πΌ=π = ef = 1.6 Γ 10β19 Γ 9.4 Γ 1018 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI = 15.05 Γ 10β1 = 1.5A 36. When a metal conductor connected to left gap of a meter bridge is heated, the balancing point (A) remains unchanged (B) shifts to the center (C) shifts towards right (D) shifts towards left Ans: (C) π Unknown resistance, π = π₯ 100βπ (1) Where l is the balancing length from the left. When conductor placed on left side is heated resistance of conductor increases. By equation (1) balancing length also increases. Therefore, balancing point shift towards right. 37. Two tiny spheres carrying charges 1.8 ΞΌC and 2.8 Β΅C are located at 40 cm apart. The potential at the midpoint of the line joining the two charges is (A) 4.3 x 104 V (B) 3.6 x 105 V (C) 3.8 x 104 V (D) 2.1 x 105 V Ans: (D) π½ = π½π + π½π π πΈπ π πΈπ π½= + ππ πΊπ ππ ππ πΊπ ππ πΎ π= (1.8 + 2.8) Γ 10β6 20Γ10β2 9Γ109 = 20Γ10β2 (4.6) Γ 10β6 = 20.7 Γ 104 = 21 Γ 104 V π = 2.1 x 105 V 38. A wire of a certain material is stretched slowly by 10%. Its new resistance and specific resistance becomes respectively (A) 1.21 times, same (B) both remains the same (C) 1.1 times, 1.1 times (D) 1.2 times, 1.1 times Ans: (A) Since length is stretched specific resistance remains same. So new length is π β² = 1.1π Then the new area of cross section is given by Volume = constant So π΄π = π΄β² π β² π΄π = π΄β² (1.1π) π΄π π΄β² = 1.1π = 0.91π΄ ππβ² π(1.1π) New resistance π β² = = π΄β² 0.91π΄ HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI π β² = 1.21π New resistance exchanges by 1.21 times but specific resistance remains same. 39. A proton moves with a velocity of 5 x 106 ms1 through the uniform electric field,πΈβ = 4 x 106[2πΜ + 0.2πΜ + 0.1πΜ]Vm1 and the uniform magnetic field π΅ β = 0.2[πΜ + 0.2πΜ + πΜ]π. The approximate net force acting on the proton is (A) 2.2 x 1013 N (B) 20 x 1013 N (C) 5 x 1013 N (D) 25 x 1013 N Ans: (B) πΉ = π(πΈβ + π Γ π΅ β) = π[4 Γ 106 (2πΜ + 0.2πΜ + 0.1πΜ + 106 (βπΜ) + πΜ)] πΉ = π Γ 106 [9πΜ + 0.8πΜ β 0.6πΜ] πΉ = 1.6 Γ 10β13 (β92 + 0. 82 + 0. 62 ) πΉ = 1.6 Γ 10β12 Γ 9 = 14.4 Γ 10β13 π β 20 Γ 10β13 π 40. A solenoid of length 50 cm having 100 turns carries a current of 2.5A. The magnetic field at one end of the solenoid is (A) 1.57 x 104 T (B) 9.42 x 104 T (C) 3.14 x 104 T (D) 6.28 x 104 T Ans: (C) π0 ππΌ π΅= 2 4πΓ10β7 Γ200Γ2.5 π 100 π΅= (Since, n = = = 200) 2 πΌ 0.5 π΅ = 3.14 Γ 10β4 T 41. A galvanometer of resistance 50 πΊ is connected to a battery of 3 V along with a resistance 2950 πΊ in series. A fullscale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be (A) 5050 πΊ (B) 4450 πΊ (C) 6050 πΊ (D) 5550 πΊ Ans: (B) π 3 Current flowing in galvanometer πΌ = π = 50+2950 = 10β3 π΄ π +π If current for 30 division is 103A 10β3 Γ20 2 Then current for 20 division is = 3 Γ 10β3 π΄ 30 Let the series resistance = R 2 3 Γ 10β3 = 3 50 + π Then π = 4950πΊ 42. A circular coil of wire of radius βrβ has βnβ turns and carries a current βIβ. The magnetic induction βBβ at a point on the axis of the coil at a distance β3r from its centre is π0 ππΌ π0 ππΌ π0 ππΌ π0 ππΌ (A) (B) (C) (D) 16π 4π 32π 8π HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI Ans: (A) ππ0 ππ2 π΅= 2(π2 +π₯ 2 )3/2 ππ πΌπ 2 0 = 2(π 2+3π 2 )3/2 ππ πΌπ 2 0 = 2Γ8Γπ 3 π0 ππΌ π΅= 16π 43. If voltage across a bulb rated 220 V, 100 W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is (A) 5% (B) 10% (C) 20% (D) 2.5% Ans: (A) π2 Power π = π As the resistance bulb is constant π₯π 2π₯π β΄ = π π π₯π % decrease in power = Γ 100 π 2π₯π£ = Γ 100 π = 2 Γ 2.5% = 5% 44. A long solenoid has 500 turns, when a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 Γ 10β3 Wb, then selfinduction of the solenoid is (A) 2.0 henry (B) 1.0 henry (C) 4.0 henry (D) 2.5 henry Ans: (B) π = πΏπ π(ππππβπ‘π’ππ ) = πΏπ 500(4 Γ 10β3 ) = πΏ(2) 2000 Γ 10β3 πΏ= = 1π» 2 45. A fully charged capacitor βCβ with initial charge β²π0 β²is connected to a coil of self inductance βLβ at t = 0. The time at which the energy is stored equally between the electric and the magnetic field is π (A) πβπΏπΆ (B) 4 βπΏπΆ (C) 2πβπΏπΆ (D) βπΏπΆ Ans: (B) Change on the capacitor at any time t π = π0 πππ π π‘ β¦β¦β¦β¦β¦(1) At time energy stored equally in Electric and magnetic field. 1 Energy of capacitor = Total energy 2 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI π2 1 π 2 0 = 2 ( 2πΆ ) 2πΆ π0 π= β2 From equation (1) π0 = π0 πππ π π‘ β2 1 πππ π π‘ = β2 1 ππ‘ = πππ β1 ( ) β2 π 1 ππ‘ = (πππππ π = ) 4 βπΏπΆ π π‘ = βπΏπΆ 4 46. A magnetic field of flux density 1.0 Wb m2 acts normal to a 80 turn coil of 0.01 m2 area. If this coil is removed from the field in 0.2 second, the emf induced in it is (A) 0.8 V (B) 5 V (C) 4V (D) 8 V Ans: (C) B = 1 W bm2 N = 80 A = 0.01 m2 dt = 0.2 βππ π= ππ‘ ππ΄π΅ =β ππ‘ 80Γ0.01Γ1 π=β =4V 0.2 47. An alternating current is given by π = π1 π ππ π π‘ + π2 πππ π π‘. The r.m.s current is given by π 2 +π22 π 2 +π22 π1 +π2 π1 βπ2 (A) β 1 2 (B) β 1 (C) (D) β2 β2 β2 Ans: ( π = π1 π ππ π π‘ + π2 πππ π π‘ = (βπ12 + π22 ) (π ππ(ππ‘ + π)) βπ12 +π22 = β2 π1 π2 π = βπ12 + π22 ( π ππ π π‘ πππ π π‘) βπ12 +π22 βπ12 +π22 i = βπ12 + π22 (sin(Οt + Ο)) where tan Ο = i2 1 βπ12 +π22 ππππ = β2 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 48. Which of the following statements proves the Earth has a magnetic field? (A) Earth is surrounded by ionosphere (B) a large quantity of ironore is found in the Earth (C) The intensity of cosmic rays stream of charged particles is more at the poles than at the equator. (D) Earth is a planet rotating about the North South axis. Ans: (C) The intensity of cosmic rays stream of charged particles is more at the poles than at the equator. 49. In a series LCR circuit R=300 πΊ, L = 0.9 H, C= 2.0 ππΉ and w = 1000 rad/sec., then impedance of the circuit is (A) 500 πΊ (B) 400 πΊ (C) 1300 πΊ (D) 900 πΊ Ans: (A) π = 300πΊ, πΏ = 0.9π», πΆ = 2ππΉ, π = 1000πππ/π π = βπ 2 + (ππΆ β ππΆ )2 1 1 ππΆ = ππΆ = 103 Γ2Γ10β6 = 0.5 Γ 103 = 500πΊ ππΏ = ππΏ = 103 Γ 0. = 0.9 Γ 103 = 900πΊ β΄ π = β3002 + (900 β 500)2 = β9 Γ 104 + 42 Γ 104 = β25 Γ 102 = 500πΊ 50. Which of the following radiation of electromagnetic waves has the highest wavelength? (A) IR β rays (B) Microwaves (C) Xrays (D) UVrays Ans: (B) Microwaves 51. The power of a equiconcave lens is β 4.5 D and is made of a material of R.I. 1.6, the radii of curvature of the lens is (A) β 2.66 cm (B) 115.44 cm (C) β 26.6 cm (D) +36.6 cm Ans: (C) P = 4.5 D, n = 1.6, R=? 1 1 1 = (π β 1) (π + π ) π 1 1 π = (π β 1) ( + ) π π 2 β4.5 = 0.6 (π ) 1.2 π = β4.5 = β26.6cm 52. A ray of light passes through an equilateral glass prism in such a manner that the angle of incidence is 3 equal to the angle of emergence and each of these angles is equal to 4 of the angle of prism. The angle of deviation is HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI (A) 200 (B) 300 (C) 450 (D) 390 Ans: (B) A = 600 3 π = 4π΄ Angle of deviation πΏ = 2π β π΄ 3 = 2 (4 π΄) β π΄ 3 π΄ = 2π΄ β π΄ = = 300 2 53. A convex lens of focal length βf βis placed somewhere in between an object and a screen. The distance between the object and the screen is βxβ. If the numerical value of the magnification produced by the lens is βmβ, then the focal length of the lens is (m+1)2 π₯ (πβ1)2 π₯ mx mx (A) (B) (C) (m+1)2 (D) (m1)2 π π Ans: (C) π π π + (βπ’) = π₯ (Since π = π’ = βπ’ β΄ π = βππ’) βππ’ β π’ = π₯ β΄ βπ’(π + 1) = π₯ βπ₯ β΄ π’ = π+1 ππ₯ π = π+1 1 1 1 π+1 π+1 π+1 Now, =πβπ’= + = (π + 1) [ ππ₯ ] π ππ₯ π₯ ππ₯ β΄π= (π+1)2 54. A series resonant ac circuit contains a capacitance 106 F and an inductor of 104 H. The frequency of electrical oscillations will be 105 10 (A) 2π Hz (B) 2π Hz (C) 105 Hz (D) 10 Hz Ans: (A) 1 π = 2πβπΏπΆ 1 1 1 π = 2π [ ] = 2π β1010 π»π§ β(10β4 )(10β6 ) HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 105 π= π»π§ 2π 55. Focal length of a convex lens will be maximum for (A) Green light (B) Red light (C) Blue light (D) Yellow light Ans: (B) 1 1 1 Focal length, f given by lens makers formula. = (π β 1) (π β π 2 ), πis the refractive. Since red light π has least π, its focal length will be maximum. 56. For light diverging from a finite point source (A) The wave front is parabolic (B) The intensity at the wave front does not depend on the distance. (C) The wave front is cylindrical (D) The intensity decreases in proportion to the distance squared Ans: (D) 1 Point source has spherical wavefronts, hence intensity β π 2 where x is the distance from the point source. 57. The fringe width for red colour as compared to that for violet colour is approximately (A) 4 times (B) 8 times (C) 3 times (D) Double Ans: (D) π·π Fringe width, π½ = π β΄ratio of fringe width for red and violet π½πππ ππππ = ππ£πππππ‘ π½π£πππππ‘ π½πππ = 2π½π£πππππ‘ 58. In case of Fraunhofer diffraction at a single slit the diffraction pattern on the screen is correct for which of the following statements? (A) Central dark band having uniform brightness on either side. (B) Central bright band having dark bands on either side. (C) Central dark band having alternate dark and bright bands of decreasing intensity on either side. (D) Central bright band having alternate dark and bright bands of decreasing intensity on either side. Ans: (D) Central bright band having alternate dark and bright bands of decreasing intensity on either side 59. When a Compact Disc (CD) is illuminated by small source of white light coloured bands are observed. This is due to (A) Interference (B) Reflection (C) Scattering (D) Diffraction Ans: (A) Interference HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 60. Consider a glass slab which is silvered at one side and the other side is transparent. Given the refractive index of the glass slab to be 1.5. If a ray of light is incident at an angle of 450 on the transparent side, the deviation of the ray of light from its initial path, when it comes out of the slab is (A) 1200 (B) 450 (C) 900 (D) 1800 Ans: (C) n = 1.5 i = 450 Angle of deviation = 900 DEPARTMENT OF PHYSICS β Dr. Gananath Shetty B Mr. Sharath Rai A N β Mr. Abhishek Kumar Mr. Amith Guha β Mr. Thirumala Reddy Mr. Karthik M B β Mrs. Nidhi B Shetty Mr. Mohammed Fanu β Mr. Lakshman Bajila Mr. Joel M F β Mr. Suhas Gore P Mr. Ranjith CREATIVE EDUCATION FOUNDATION MOODBIDRI (R) Website : www.creativeedu.in Phone No. : 9019844492 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI
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