PHYSICS CET SOLVED.pdf - CREATIVE P.U COLLEGE

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CREATIVE LEARNING CLASSES, KARKALA K-CET KEY ANSWERS WITH DETAILED SOLUTION - 2022 DEPARTMENT OF PHYSICS VERSION CODE – 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, 𝛥𝐾 = 0.52𝑒𝑉, 𝜆 = 500𝑛𝑚, 𝜆2 =? ℎ𝑐 We know, 𝐾1 = 𝜆 − 𝜑 1 ℎ𝑐 𝐾2 = −𝜑 𝜆2 1 1 ∴ 𝐾1 − 𝐾2 = ℎ𝑐 ( − ) 𝜆1 𝜆2 1 1 −𝛥𝐾 = ℎ𝑐 (𝜆 − 𝜆 ) 1 2 1 1 −0.52𝑒𝑉 = (1242𝑒𝑉𝑛𝑚) (500𝑛𝑚 − 𝜆 ) 2 −0.52 1 1 = 500 − 𝜆 1242 2 1 1 0.52 = 500 + 1242 𝜆2 𝜆2 = 413𝑛𝑚 ≈ 400𝑛𝑚 2. The de- Broglie wavelength of a particle of kinetic energy ‘K’ is 𝜆; the wavelength of the particle, if its 𝐾 kinetic energy is is 4 𝜆 (A) 2 (B) 4𝜆 (C) 𝜆 (D) 2𝜆 Ans: (D) h 𝜆= 𝑝 h 𝜆= …………...(1) √2𝑚𝐾 h 𝜆′ = …………...(2) 𝐾 √2𝑚 4 Divide equation (2) by (1), h 𝜆′ √2𝑚𝐾 4 = h 𝜆 √2𝑚𝐾 𝜆′ = √4 𝜆 𝜆′ = 2𝜆 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 𝐴. After collision with an electron, it is found 0 to have a radius of 2.12 𝐴, the principal quantum number ‘n’ of the final state of the atom is (A) n = 3 (B) n = 4 (C) n = 1 (D) n = 2 Ans: (D) 𝑛2 r = 𝑟0 𝑍 Given 𝑟0 = 0.53Å 𝑎𝑛𝑑 𝑟 = 2.12Å For H-atom, Z= 1 𝑛2 2.12 = 0.53 1 𝑛=2 4. In accordance with the Bohr’s model, the quantum number that characterises the Earth’s revolution around the Sun in an orbit of radius 1.5×1011 m with orbital speed 3×104 ms-1 is [given mass of Earth = 6×1024 kg] (A) 8.57×1064 (B) 2.57×1074 (C) 5.98×1086 (D) 2.57×1038 Ans: (B) According to Bohr’s model nh L = mvr = 2𝜋 2𝜋𝑚𝑣𝑟 n= ℎ 2×3.14×6×1024 ×3×104 ×1.5×1011 n= = 2.57 × 1074 6.625×10−34 0 5. If an electron is revolving in its Bohr orbit having Bohr radius of 0.529𝐴, then the radius of third orbit is 0 0 (A) 4.761 𝐴 (B) 5125 nm (C) 4234 nm (D) 4496 𝐴 Ans: (A) rn = 𝑟0 𝑛2 𝑟3 = 0.529  10−10 × 32 𝑟3 = 4.761 Å 6. Binding energy of a Nitrogen nucleus [ 147𝑁], given in [ 147𝑁] = 14.00307u (A) 206.5 MeV (B) 78 MeV (C) 104.7 MeV (D) 85 MeV Ans: (C) Mass defect Δ m = [Z mp+ (A-Z) mn] – M Δ m = [7 x1.00783+ (14-7) 1.00867] – 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) 𝛾 - rays (B) 𝛼 - particles (C) X – 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 𝛺 cm (B) 1010 to 1012 𝛺cm (C) 10-2 to 10-5 𝛺 cm (D) 10-3 to 106 𝛺 cm Ans: (D) The resistivity of the semiconductor at room temperature is in between 10-3 to 10+6 Ωcm 10. The forbidden energy gap for ‘Ge’ crystal at ‘0’ 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 ‘Ge’ crystal at ‘0’ 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 𝑨 𝑩 𝑨+𝑩 ̅̅̅̅̅̅̅̅̅̅ Y= 𝑨 + 𝑩 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 m-1 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 “I’ 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) IlBcosθ = mgsinθ mgsinθ I= 𝑙𝐵𝑐𝑜𝑠𝜃 𝑚 𝑔 𝑡𝑎𝑛𝜃 I=(𝑙) 𝐵 9.8 tan300 = 0.5 × 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) 𝐸 𝑃= 𝑡 𝐸 = 𝑃𝑡 mc2 = Pt 𝑃𝑡 m= 𝑐2 109 ×3600 m= (3×108 )2 = 0.04gm 14. The displacement ‘x’ (in metre) of a particle of mass ‘m’ (in kg) moving in the one dimension under the action of a force, is related to time ‘t’ (in sec) by 𝑡 = √𝑥 + 3. The displacement of the particle when its velocity is zero, will be (A) 6m (B) 2m (C) 4m (D) 0m Ans: (D) Given, t = √𝑥 + 3 √𝑥 = 𝑡 − 3 Square on both side x = (t-3)2 𝑑𝑥 v= = 2(t-3) 𝑑𝑡 when velocity is zero, HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 2(t-3) = 0 t=3 So, x = (t-3)2 = (3-3)2 = 0 15. Two objects are projected at an angle 𝜃 0 and (90 − 𝜃 0 ), to the horizontal with the same speed. The ratio of their maximum vertical heights is (A) 1: tan 𝜃 (B) tan2 𝜃 : 1 (C) 1 : 1 (D) tan 𝜃 : 1 Ans: (B) 𝑣0 2 sin2 θ 𝐻1 = ………… (1) 2g 𝑣0 2 (sin(90−𝜃))2 𝑣0 2 cos2 𝜃 𝐻2 = = = …………… (2) 2g 2g 𝑣0 2 sin2 θ 𝐻1 2g = 𝑣0 2 cos2 𝜃 𝐻2 2g 𝐻1 𝑡𝑎𝑛2 𝜃 = 𝐻2 1 𝐻1 : 𝐻2 = 𝑡𝑎𝑛2 𝜃 ∶ 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) 𝑇 cos 𝜃 = 𝑚𝑔 …………….(1) 𝑣2 𝑇 sin 𝜃 = 𝑚 ……………(2) 𝑅 𝑣2 tan 𝜃 = 𝑅𝑔 100 = 10 × 10 𝜋 𝜃 = tan−1 1 = 45° = 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, 𝑇2 − 3𝑔 = 3𝑎 𝑇2 = 3(𝑔 + 𝑎) For 5 kg, 𝑇1 − 𝑇2 − 5𝑔 = 5𝑎 𝑇1 = 8(𝑔 + 𝑎) 𝑇1 = 94.4 𝑁 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 𝑉𝑆𝐷 = 1 𝑀𝑆𝐷 1 1 1 𝑉𝑆𝐷 = 𝑀𝑆𝐷 = (0.5 𝑚𝑚) 50 50 1𝑉𝑆𝐷 = 0.01 𝑚𝑚 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 𝜋𝑟𝑎𝑑 (A) 6 (B) 8𝜋𝑟𝑎𝑑/𝑠 2 (C) 2𝜋𝑟𝑎𝑑/𝑠 2 (D) 4𝜋𝑟𝑎𝑑/𝑠 2 𝑠2 Ans: (D) 𝜔𝑖 = 1200 𝑟𝑝𝑚 𝜔𝑓 = 3120 𝑟𝑝𝑚 𝑡 = 16 𝑠 𝜔𝑓 = 𝜔𝑖 + 𝛼𝑡 𝜔𝑓 − 𝜔𝑖 3120 − 1200 𝜋 𝛼= = × 𝑡 16 30 𝛼 = 4𝜋 𝑟𝑎𝑑𝑠 −2 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 Young’s modulus of the material of the rod is7 × 109 N/𝑚2 . The area of cross section to support a load of 104 𝑁 is (A) 7.1× 10−4 𝑚2 (B) 7.1 × 10−2 𝑚2 (C) 7.1× 10−8 𝑚2 (D) 7.1× 10−6 𝑚2 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI Ans: (A) ∆𝑙 𝑆𝑡𝑟𝑎𝑖𝑛 = = 0.2% = 0.0 02, 𝑌 = 7 × 109 𝑁𝑚−2 , 𝐹 = 104 𝑙 𝐹𝑙 𝑌= 𝐴∆𝑙 𝐹𝑙 104 ∴𝐴= = = 7.1 × 10−4 𝑚2 𝑌∆𝑙 7×109 ×0.002 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 × 105 𝑉/𝑚. When the field is switched off, the drop is observed to fall with terminal 7 velocity 2× 10−3 𝑚𝑠 −1 . Here g = 9.8 m/s2, Viscosity of air is 1.8× 10−5 𝑁𝑠/𝑚2 and the density of oil is 900 kg 𝑚−3 . The magnitude of ‘q’ is (A) 1.6× 10−19 𝐶 (B) 3.2× 10−19 𝐶 (C) 0.8× 10−19 𝐶 (D) 8× 10−19 𝐶 Ans: (D) 81 𝐺𝑖𝑣𝑒𝑛 𝐸 = 𝜋 × 105 𝑉/𝑚 𝑣 = 2 × 10−3 𝑚𝑠 −1 𝜂 = 1.8 × 10−5 𝑁𝑠/𝑚2 7 For equilibrium, qE = mg ………….(1) 𝑚𝑔 = 6𝜋𝜂𝑟𝑣 4 𝜋𝑟 3 𝜌𝑔 = 6𝜋𝜂𝑟𝑣 ………….(2) 3 Solving (1) and (2) we get 𝑞 = 8 × 10−19 𝐶 23. “Heat cannot be itself flow from a body at lower temperature to a body at higher temperature”. 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/𝑠 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 × 60 = 5 𝑘𝑔 Work done to pull chain = change in potential energy = 𝑚𝑔 ( ℎ𝑓 − ℎ𝑖 ) 6 = (10)(30 × 10−2 − 0) = 3.6 𝐽 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 ‘O’ 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 4× 10−9 𝐶 m is aligned at 30° with the direction of a uniform electric field of magnitude 5× 104 𝑁𝐶 −1 , the magnitude of the torque acting on the dipole is (A) 10−5 𝑁 𝑚 (B) 10× 10−3 𝑁 𝑚 (C) 10−4 𝑁 𝑚 (D) √3 × 10−4 𝑁 𝑚 Ans: (C) 𝜏 = 𝑝𝐸 sin 𝜃 = (4 × 10−9 )(5 × 104 ) sin 30° = 10−4 𝑁𝑚 28. A charged particle of mass ‘m’ and charge ‘q’ is released from rest in an uniform electric field 𝐸⃗ . Neglecting the effect of gravity, the kinetic energy of the charged particle after ‘t’ second is 𝐸𝑞𝑚 𝐸2 𝑞2𝑡 2 (A) (B) 𝑡 2𝑚 2𝐸 2 𝑡 2 𝐸𝑞 2 𝑚 (C) (D) 𝑚𝑞 2𝑡 2 Ans: (B) 𝐹 = 𝑞𝐸 𝐹 𝑞𝐸 𝑎= 𝑚 = 𝑚 𝑞𝐸 Velocity after time t, 𝑣 = 𝑢 + 𝑎𝑡 = 0 + 𝑡 𝑚 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 1 1 𝑞𝐸𝑡 2 𝑞2𝐸2 𝑡 2 Kinetic energy after time t, 𝐾 = 𝑚𝑣 2 = 𝑚( 𝑚 ) = 2 2 2𝑚 29. The electric field and the potential of an electric dipole vary with distance r as 1 1 1 1 (A) 𝑟 2 𝑎𝑛𝑑 (B) 𝑟 3 𝑎𝑛𝑑 𝑟3 𝑟2 1 1 1 1 (C) 𝑟 𝑎𝑛𝑑 (D) 𝑟 2 𝑎𝑛𝑑 𝑟2 𝑟 Ans: (B) 1 Electric field due to dipole ∝ 𝑟3 1 Electric potential due to dipole ∝ 𝑟2 𝜋 30. The displacement of a particle executing SHM is given by X= 3 sin [2𝜋𝑡 + 4 ]where ‘x’ is in metres and ‘t’ is in seconds. The amplitude and maximum speed of the particle is (A) 3 m, 6𝜋 𝑚𝑠 −1 (B) 3 m, 8𝜋 𝑚𝑠 −1 (C) 3 m, 2𝜋 𝑚𝑠 −1 (D) 3 m, 4𝜋 𝑚𝑠 −1 Ans: (A) 𝜋 𝑥 = 3 sin( 2𝜋𝑡 + ) 4 𝐹𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐴 = 3𝑚, 𝜔 = 2𝜋 𝑟𝑎𝑑𝑠 −1 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑝𝑒𝑒𝑑 = 𝐴𝜔 = 6𝜋 𝑚𝑠 −1 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 𝑄 𝐶= 𝑉 𝑄 ⇒𝑉= 𝐶 1 ⇒𝑉∝ 𝐶 When glass slab is introduced, C increases ∴V decreases 𝑄2 Energy stored = 2𝐶 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 10-10 V m-1 with mobility 2.5 ×106 m2/v/s, its drift velocity is (A) 2.5 104 m/s (B) 1.2 x 10-4 m/s (C) 7.5 x 10-4 m/s (D) 8.33 x 10-4 m/s Ans: (C) 𝑣𝑑 𝜇= 𝐸 𝑣𝑑 2.5 × 106 = 3×10−10 𝑣𝑑 = 7.5 × 10−4m/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, … (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 × 10-19 C] (A) 1.4 A (B) 1.8 A (C) 1.2 A (D) 1.5 A Ans: (D) 0 𝑟 = 0.72𝐴 𝑓 = 9.4 × 1018 𝑒 𝐼=𝑇 = ef = 1.6 × 10−19 × 9.4 × 1018 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI = 15.05 × 10−1 = 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) 𝑙 Unknown resistance, 𝑅 = 𝑥 100−𝑙 ---(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 μC and 2.8 µC are located at 40 cm apart. The potential at the mid-point 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) 𝑽 = 𝑽𝟏 + 𝑽𝟐 𝟏 𝑸𝟏 𝟏 𝑸𝟐 𝑽= + 𝟒𝝅𝜺𝟎 𝒓𝟏 𝟒𝝅𝜺𝟎 𝒓𝟐 𝐾 𝑉= (1.8 + 2.8) × 10−6 20×10−2 9×109 = 20×10−2 (4.6) × 10−6 = 20.7 × 104 = 21 × 104 V 𝑉 = 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 𝑙 ′ = 1.1𝑙 Then the new area of cross section is given by Volume = constant So 𝐴𝑙 = 𝐴′ 𝑙 ′ 𝐴𝑙 = 𝐴′ (1.1𝑙) 𝐴𝑙 𝐴′ = 1.1𝑙 = 0.91𝐴 𝜌𝑙′ 𝜌(1.1𝑙) New resistance 𝑅 ′ = = 𝐴′ 0.91𝐴 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 𝑅 ′ = 1.21𝑅 New resistance exchanges by 1.21 times but specific resistance remains same. 39. A proton moves with a velocity of 5 x 106 ms-1 through the uniform electric field,𝐸⃗ = 4 x 106[2𝑖̂ + 0.2𝑗̂ + 0.1𝑘̂]Vm-1 and the uniform magnetic field 𝐵 ⃗ = 0.2[𝑖̂ + 0.2𝑗̂ + 𝑘̂]𝑇. The approximate net force acting on the proton is (A) 2.2 x 10-13 N (B) 20 x 10-13 N (C) 5 x 10-13 N (D) 25 x 10-13 N Ans: (B) 𝐹 = 𝑞(𝐸⃗ + 𝜐 × 𝐵 ⃗) = 𝑎[4 × 106 (2𝑖̂ + 0.2𝑗̂ + 0.1𝑘̂ + 106 (−𝑘̂) + 𝑖̂)] 𝐹 = 𝑞 × 106 [9𝑖̂ + 0.8𝑗̂ − 0.6𝑘̂] 𝐹 = 1.6 × 10−13 (√92 + 0. 82 + 0. 62 ) 𝐹 = 1.6 × 10−12 × 9 = 14.4 × 10−13 𝑁 ≈ 20 × 10−13 𝑁 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 10-4 T (B) 9.42 x 10-4 T (C) 3.14 x 10-4 T (D) 6.28 x 10-4 T Ans: (C) 𝜇0 𝑛𝐼 𝐵= 2 4𝜋×10−7 ×200×2.5 𝑁 100 𝐵= (Since, n = = = 200) 2 𝐼 0.5 𝐵 = 3.14 × 10−4 T 41. A galvanometer of resistance 50 𝛺 is connected to a battery of 3 V along with a resistance 2950 𝛺 in series. A full-scale 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 𝛺 (B) 4450 𝛺 (C) 6050 𝛺 (D) 5550 𝛺 Ans: (B) 𝑉 3 Current flowing in galvanometer 𝐼 = 𝑅 = 50+2950 = 10−3 𝐴 𝑔 +𝑅 If current for 30 division is 10-3A 10−3 ×20 2 Then current for 20 division is = 3 × 10−3 𝐴 30 Let the series resistance = R 2 3 × 10−3 = 3 50 + 𝑅 Then 𝑅 = 4950𝛺 42. A circular coil of wire of radius ‘r’ has ‘n’ turns and carries a current ‘I’. The magnetic induction ‘B’ at a point on the axis of the coil at a distance √3r from its centre is 𝜇0 𝑛𝐼 𝜇0 𝑛𝐼 𝜇0 𝑛𝐼 𝜇0 𝑛𝐼 (A) (B) (C) (D) 16𝑟 4𝑟 32𝑟 8𝑟 HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI Ans: (A) 𝑛𝜇0 𝑖𝑎2 𝐵= 2(𝑎2 +𝑥 2 )3/2 𝑛𝜇 𝐼𝑟 2 0 = 2(𝑟 2+3𝑟 2 )3/2 𝑛𝜇 𝐼𝑟 2 0 = 2×8×𝑟 3 𝜇0 𝑛𝐼 𝐵= 16𝑟 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) 𝑉2 Power 𝑃 = 𝑅 As the resistance bulb is constant 𝛥𝑃 2𝛥𝑉 ∴ = 𝑃 𝑉 𝛥𝑃 % decrease in power = × 100 𝑃 2𝛥𝑣 = × 100 𝑉 = 2 × 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 × 10−3 Wb, then self-induction of the solenoid is (A) 2.0 henry (B) 1.0 henry (C) 4.0 henry (D) 2.5 henry Ans: (B) 𝜑 = 𝐿𝑖 𝑁(𝜑𝑒𝑎𝑐ℎ𝑡𝑢𝑟𝑛 ) = 𝐿𝑖 500(4 × 10−3 ) = 𝐿(2) 2000 × 10−3 𝐿= = 1𝐻 2 45. A fully charged capacitor ‘C’ with initial charge ′𝑞0 ′is connected to a coil of self inductance ‘L’ at t = 0. The time at which the energy is stored equally between the electric and the magnetic field is 𝜋 (A) 𝜋√𝐿𝐶 (B) 4 √𝐿𝐶 (C) 2𝜋√𝐿𝐶 (D) √𝐿𝐶 Ans: (B) Change on the capacitor at any time t 𝑞 = 𝑞0 𝑐𝑜𝑠 𝜔 𝑡 ……………(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 𝑞2 1 𝑞 2 0 = 2 ( 2𝐶 ) 2𝐶 𝑞0 𝑞= √2 From equation (1) 𝑞0 = 𝑞0 𝑐𝑜𝑠 𝜔 𝑡 √2 1 𝑐𝑜𝑠 𝜔 𝑡 = √2 1 𝜔𝑡 = 𝑐𝑜𝑠 −1 ( ) √2 𝜋 1 𝜔𝑡 = (𝑆𝑖𝑛𝑐𝑒 𝜔 = ) 4 √𝐿𝐶 𝜋 𝑡 = √𝐿𝐶 4 46. A magnetic field of flux density 1.0 Wb m-2 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 bm-2 N = 80 A = 0.01 m2 dt = 0.2 −𝑑𝜑 𝜖= 𝑑𝑡 𝑁𝐴𝐵 =− 𝑑𝑡 80×0.01×1 𝜖=− =4V 0.2 47. An alternating current is given by 𝑖 = 𝑖1 𝑠𝑖𝑛 𝜔 𝑡 + 𝑖2 𝑐𝑜𝑠 𝜔 𝑡. The r.m.s current is given by 𝑖 2 +𝑖22 𝑖 2 +𝑖22 𝑖1 +𝑖2 𝑖1 −𝑖2 (A) √ 1 2 (B) √ 1 (C) (D) √2 √2 √2 Ans: ( 𝑖 = 𝑖1 𝑠𝑖𝑛 𝜔 𝑡 + 𝑖2 𝑐𝑜𝑠 𝜔 𝑡 = (√𝑖12 + 𝑖22 ) (𝑠𝑖𝑛(𝜔𝑡 + 𝜑)) √𝑖12 +𝑖22 = √2 𝑖1 𝑖2 𝑖 = √𝑖12 + 𝑖22 ( 𝑠𝑖𝑛 𝜔 𝑡 𝑐𝑜𝑠 𝜔 𝑡) √𝑖12 +𝑖22 √𝑖12 +𝑖22 i = √𝑖12 + 𝑖22 (sin(ωt + φ)) where tan φ = i2 1 √𝑖12 +𝑖22 𝑖𝑟𝑚𝑠 = √2 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 iron-ore 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 𝛺, L = 0.9 H, C= 2.0 𝜇𝐹 and w = 1000 rad/sec., then impedance of the circuit is (A) 500 𝛺 (B) 400 𝛺 (C) 1300 𝛺 (D) 900 𝛺 Ans: (A) 𝑅 = 300𝛺, 𝐿 = 0.9𝐻, 𝐶 = 2𝜇𝐹, 𝜔 = 1000𝑟𝑎𝑑/𝑠 𝑍 = √𝑅 2 + (𝑋𝐶 − 𝑋𝐶 )2 1 1 𝑋𝐶 = 𝜔𝐶 = 103 ×2×10−6 = 0.5 × 103 = 500𝛺 𝑋𝐿 = 𝜔𝐿 = 103 × 0. = 0.9 × 103 = 900𝛺 ∴ 𝑍 = √3002 + (900 − 500)2 = √9 × 104 + 42 × 104 = √25 × 102 = 500𝛺 50. Which of the following radiation of electromagnetic waves has the highest wavelength? (A) IR – rays (B) Microwaves (C) X-rays (D) UV-rays Ans: (B) Microwaves 51. The power of a equi-concave lens is – 4.5 D and is made of a material of R.I. 1.6, the radii of curvature of the lens is (A) – 2.66 cm (B) 115.44 cm (C) – 26.6 cm (D) +36.6 cm Ans: (C) P = -4.5 D, n = 1.6, R=? 1 1 1 = (𝑛 − 1) (𝑅 + 𝑅) 𝑓 1 1 𝑃 = (𝑛 − 1) ( + ) 𝑅 𝑅 2 −4.5 = 0.6 (𝑅) 1.2 𝑅 = −4.5 = −26.6cm 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 𝑖 = 4𝐴 Angle of deviation 𝛿 = 2𝑖 − 𝐴 3 = 2 (4 𝐴) − 𝐴 3 𝐴 = 2𝐴 − 𝐴 = = 300 2 53. A convex lens of focal length ‘f ’is placed somewhere in between an object and a screen. The distance between the object and the screen is ‘x’. If the numerical value of the magnification produced by the lens is ‘m’, then the focal length of the lens is (m+1)2 𝑥 (𝑚−1)2 𝑥 mx mx (A) (B) (C) (m+1)2 (D) (m-1)2 𝑚 𝑚 Ans: (C) 𝜐 𝜐 𝜐 + (−𝑢) = 𝑥 (Since 𝑚 = |𝑢| = −𝑢 ∴ 𝜐 = −𝑚𝑢) −𝑚𝑢 − 𝑢 = 𝑥 ∴ −𝑢(𝑚 + 1) = 𝑥 −𝑥 ∴ 𝑢 = 𝑚+1 𝑚𝑥 𝜐 = 𝑚+1 1 1 1 𝑚+1 𝑚+1 𝑚+1 Now, =𝜐−𝑢= + = (𝑚 + 1) [ 𝑚𝑥 ] 𝑓 𝑚𝑥 𝑥 𝑚𝑥 ∴𝑓= (𝑚+1)2 54. A series resonant ac circuit contains a capacitance 10-6 F and an inductor of 10-4 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 𝜐 = 2𝜋√𝐿𝐶 1 1 1 𝜐 = 2𝜋 [ ] = 2𝜋 √1010 𝐻𝑧 √(10−4 )(10−6 ) HKS PU COLLEGE HASSAN CREATIVE PU COLLEGE KARKALA CREATIVE PU COLLEGE UDUPI 105 𝜐= 𝐻𝑧 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. = (𝜇 − 1) (𝑅 − 𝑅2 ), 𝜇is the refractive. Since red light 𝑓 has least 𝜇, 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 ∝ 𝑟 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) 𝐷𝜆 Fringe width, 𝛽 = 𝑑 ∴ratio of fringe width for red and violet 𝛽𝑟𝑒𝑑 𝜆𝑟𝑒𝑑 = 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 𝛽𝑣𝑖𝑜𝑙𝑒𝑡 𝛽𝑟𝑒𝑑 = 2𝛽𝑣𝑖𝑜𝑙𝑒𝑡 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 ❖ Dr. Gananath Shetty B Mr. Sharath Rai A N ❖ Mr. Abhishek Kumar Mr. Amith Guha ❖ Mr. Thirumala Reddy Mr. Karthik M B ❖ Mrs. Nidhi B Shetty Mr. Mohammed Fanu ❖ Mr. Lakshman Bajila Mr. Joel M F ❖ 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