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Page 1 16 th March. 2021 | Shift 2 SECTION – A 1. A mosquito is moving with a velocity v ? = 0.5 t 2 ? i 3tj 9k ? ? ? ? m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2s? (1) 1 2 tan fromx axis 3 ? ? ? ? ? ? ? ? (2) 1 5 tan fromx axis 2 ? ? ? ? ? ? ? ? (3) 1 2 tan fromy axis 3 ? ? ? ? ? ? ? ? (4) 1 5 tan fromy axis 2 ? ? ? ? ? ? ? ? Sol. Bonus Given 2 ˆ ˆ ˆ v 0.5t i 3tj 9k ? ? ? ? t 2 v ? ? = ˆ ˆ ˆ 2i 6j 9k ? ? ? angle made by direction of motion of mosquito will be – cos -1 2 11 ? ? ? ? ? ? from x-axis = tan -1 117 2 ? ? ? ? ? ? ? ? cos -1 6 11 ? ? ? ? ? ? from y-axis = tan -1 85 9 ? ? ? ? ? ? ? ? cos -1 9 11 ? ? ? ? ? ? from z-axis = tan -1 40 9 no. option is matching 2. Statement I: A cyclist is moving on an unbanked road with a speed of 7 kmh –1 and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s 2 ) Statement-II: If the road is banked at an angle of 45°, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh -1 without slipping. In the light of the above statements, choose the correct answer from the options given below. (1) Both statement I and statement II are false (2) Both statement I and statement II are true (3) Statement I is correct and statement II is incorrect (4) Statement I is incorrect and statement II is correct Sol. (2) On a horizontal ground, v max = ?Rg = ? ? 0.2 2 9.8 = 1.97 m/s 1.97 × 18 5 = 7.12 km/hr= 7.2 km/hr Statement-2 v max = ? ? ? ? ? ? ? ? ? ? ? ? tan gr 1 tan = ? ? 12 2 9.8 0.8 = 19.5 km/hr v min = ? ? ? ? ? ? ? ? ? ? ? ? tan rg 1 tan = ? ? 0.8 2 9.8 1.2 = 12.01 km/hr Page 2 16 th March. 2021 | Shift 2 SECTION – A 1. A mosquito is moving with a velocity v ? = 0.5 t 2 ? i 3tj 9k ? ? ? ? m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2s? (1) 1 2 tan fromx axis 3 ? ? ? ? ? ? ? ? (2) 1 5 tan fromx axis 2 ? ? ? ? ? ? ? ? (3) 1 2 tan fromy axis 3 ? ? ? ? ? ? ? ? (4) 1 5 tan fromy axis 2 ? ? ? ? ? ? ? ? Sol. Bonus Given 2 ˆ ˆ ˆ v 0.5t i 3tj 9k ? ? ? ? t 2 v ? ? = ˆ ˆ ˆ 2i 6j 9k ? ? ? angle made by direction of motion of mosquito will be – cos -1 2 11 ? ? ? ? ? ? from x-axis = tan -1 117 2 ? ? ? ? ? ? ? ? cos -1 6 11 ? ? ? ? ? ? from y-axis = tan -1 85 9 ? ? ? ? ? ? ? ? cos -1 9 11 ? ? ? ? ? ? from z-axis = tan -1 40 9 no. option is matching 2. Statement I: A cyclist is moving on an unbanked road with a speed of 7 kmh –1 and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s 2 ) Statement-II: If the road is banked at an angle of 45°, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh -1 without slipping. In the light of the above statements, choose the correct answer from the options given below. (1) Both statement I and statement II are false (2) Both statement I and statement II are true (3) Statement I is correct and statement II is incorrect (4) Statement I is incorrect and statement II is correct Sol. (2) On a horizontal ground, v max = ?Rg = ? ? 0.2 2 9.8 = 1.97 m/s 1.97 × 18 5 = 7.12 km/hr= 7.2 km/hr Statement-2 v max = ? ? ? ? ? ? ? ? ? ? ? ? tan gr 1 tan = ? ? 12 2 9.8 0.8 = 19.5 km/hr v min = ? ? ? ? ? ? ? ? ? ? ? ? tan rg 1 tan = ? ? 0.8 2 9.8 1.2 = 12.01 km/hr 3. Calculate the time interval between 33% decay and 67% decay if half-life of a substance is: (1) 40 minutes (2) 20 minutes (3) 60 minutes (4) 13 minutes Sol. (2) T 1/2 = 20 min ? ? ln2 = 20 min ? ? ? ? ? ? ln2 20(min) ? ? N t = N 0 e - ?t t 0 N N = 1 t e ? ? ? 0.67 = 1 t e ? ? ? ln (0.67) = - ?t 1 ? ln ? ? ? ? ? ? 100 67 = ?t 1 ? t 1 = ? ? ? ? ? ? ? 100 ln 20(min) 67 (ln2) Similarly, t 2 = ? ? ? ? ? ? ? 100 ln 20(min) 34 (ln2) t 2 – t 1 = 19.57 min ? 20 min. 4. A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance h = 9.8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is: (take g = 9.8 ms –2 ) M m h v (1) 831.4 m/s (2) 841.4 m/s (3) 811.4 m/s (4) 821.4 m/s Sol. (1) P i = 0.01 × u + 0 = P f = 6 × v v = 0.01u 6 using energy conservation 1 2 × 6 × 2 u 600 ? ? ? ? ? ? = 6 × 9.8 × 9.8 × 10 -2 u = 6 × 98 × 2 = 588 2 m/s Page 3 16 th March. 2021 | Shift 2 SECTION – A 1. A mosquito is moving with a velocity v ? = 0.5 t 2 ? i 3tj 9k ? ? ? ? m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2s? (1) 1 2 tan fromx axis 3 ? ? ? ? ? ? ? ? (2) 1 5 tan fromx axis 2 ? ? ? ? ? ? ? ? (3) 1 2 tan fromy axis 3 ? ? ? ? ? ? ? ? (4) 1 5 tan fromy axis 2 ? ? ? ? ? ? ? ? Sol. Bonus Given 2 ˆ ˆ ˆ v 0.5t i 3tj 9k ? ? ? ? t 2 v ? ? = ˆ ˆ ˆ 2i 6j 9k ? ? ? angle made by direction of motion of mosquito will be – cos -1 2 11 ? ? ? ? ? ? from x-axis = tan -1 117 2 ? ? ? ? ? ? ? ? cos -1 6 11 ? ? ? ? ? ? from y-axis = tan -1 85 9 ? ? ? ? ? ? ? ? cos -1 9 11 ? ? ? ? ? ? from z-axis = tan -1 40 9 no. option is matching 2. Statement I: A cyclist is moving on an unbanked road with a speed of 7 kmh –1 and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s 2 ) Statement-II: If the road is banked at an angle of 45°, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh -1 without slipping. In the light of the above statements, choose the correct answer from the options given below. (1) Both statement I and statement II are false (2) Both statement I and statement II are true (3) Statement I is correct and statement II is incorrect (4) Statement I is incorrect and statement II is correct Sol. (2) On a horizontal ground, v max = ?Rg = ? ? 0.2 2 9.8 = 1.97 m/s 1.97 × 18 5 = 7.12 km/hr= 7.2 km/hr Statement-2 v max = ? ? ? ? ? ? ? ? ? ? ? ? tan gr 1 tan = ? ? 12 2 9.8 0.8 = 19.5 km/hr v min = ? ? ? ? ? ? ? ? ? ? ? ? tan rg 1 tan = ? ? 0.8 2 9.8 1.2 = 12.01 km/hr 3. Calculate the time interval between 33% decay and 67% decay if half-life of a substance is: (1) 40 minutes (2) 20 minutes (3) 60 minutes (4) 13 minutes Sol. (2) T 1/2 = 20 min ? ? ln2 = 20 min ? ? ? ? ? ? ln2 20(min) ? ? N t = N 0 e - ?t t 0 N N = 1 t e ? ? ? 0.67 = 1 t e ? ? ? ln (0.67) = - ?t 1 ? ln ? ? ? ? ? ? 100 67 = ?t 1 ? t 1 = ? ? ? ? ? ? ? 100 ln 20(min) 67 (ln2) Similarly, t 2 = ? ? ? ? ? ? ? 100 ln 20(min) 34 (ln2) t 2 – t 1 = 19.57 min ? 20 min. 4. A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance h = 9.8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is: (take g = 9.8 ms –2 ) M m h v (1) 831.4 m/s (2) 841.4 m/s (3) 811.4 m/s (4) 821.4 m/s Sol. (1) P i = 0.01 × u + 0 = P f = 6 × v v = 0.01u 6 using energy conservation 1 2 × 6 × 2 u 600 ? ? ? ? ? ? = 6 × 9.8 × 9.8 × 10 -2 u = 6 × 98 × 2 = 588 2 m/s 16 th March. 2021 | Shift 2 5. What will be the nature of flow of water from a circular tap, when its flow rate increased from 0.18 L/min to 0.48 L/min? The radius of the tap and viscosity of water are 0.5 cm and 10 –3 Pa s, respectively. (Density of water: 10 3 kg/m 3 ) (1) Remains steady flow (2) Unsteady to steady flow (3) Steady flow to unsteady flow (4) Remains turbulent flow Sol. (3) The nature of flow is determined by reynolds no. R e = vD ? ? if R e < 1000 ? flow is steady 1000 < R e < 2000 ? flow becomes unsteady R e > 2000 ? flow is turbulent (R e ) initial = 10 3 × 3 2 2 0.18 10 (0.5 10 ) 60 ? ? ? ? ? ? ? × 2 3 1 10 10 ? ? ? = 382.16 (R e ) final = 10 3 × 3 2 2 0.48 10 (0.5 10 ) 60 ? ? ? ? ? ? ? × 2 3 1 10 10 ? ? ? = 1019.09 6. The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index? Assume the radii of curvature of the faces of lens are R 1 and R 2 respectively. (1) Zero (2) 1 2 1 2 R R R –R (3) Infinite (4) 1 Sol. (3) 1 f = 2 1 n 1 n ? ? ? ? ? ? ? 1 2 1 1 R R ? ? ? ? ? ? ? 1 f = 0 f = infinite. 7. For the given circuit, comment on the type of transformer used. P ~ L O A D S i 1 i L 0.11A 60 W V 2 220 V (1) Step down transformer (2) Auxilliary transformer (3) Step-up transformer (4) Auto transformer Sol. (3) P s = V s × I s 60 = V s × 0.11 V s = 60 0.11 = 6000 11 , V p = 220 volt V s > V p so it is step up transformer Page 4 16 th March. 2021 | Shift 2 SECTION – A 1. A mosquito is moving with a velocity v ? = 0.5 t 2 ? i 3tj 9k ? ? ? ? m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2s? (1) 1 2 tan fromx axis 3 ? ? ? ? ? ? ? ? (2) 1 5 tan fromx axis 2 ? ? ? ? ? ? ? ? (3) 1 2 tan fromy axis 3 ? ? ? ? ? ? ? ? (4) 1 5 tan fromy axis 2 ? ? ? ? ? ? ? ? Sol. Bonus Given 2 ˆ ˆ ˆ v 0.5t i 3tj 9k ? ? ? ? t 2 v ? ? = ˆ ˆ ˆ 2i 6j 9k ? ? ? angle made by direction of motion of mosquito will be – cos -1 2 11 ? ? ? ? ? ? from x-axis = tan -1 117 2 ? ? ? ? ? ? ? ? cos -1 6 11 ? ? ? ? ? ? from y-axis = tan -1 85 9 ? ? ? ? ? ? ? ? cos -1 9 11 ? ? ? ? ? ? from z-axis = tan -1 40 9 no. option is matching 2. Statement I: A cyclist is moving on an unbanked road with a speed of 7 kmh –1 and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s 2 ) Statement-II: If the road is banked at an angle of 45°, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh -1 without slipping. In the light of the above statements, choose the correct answer from the options given below. (1) Both statement I and statement II are false (2) Both statement I and statement II are true (3) Statement I is correct and statement II is incorrect (4) Statement I is incorrect and statement II is correct Sol. (2) On a horizontal ground, v max = ?Rg = ? ? 0.2 2 9.8 = 1.97 m/s 1.97 × 18 5 = 7.12 km/hr= 7.2 km/hr Statement-2 v max = ? ? ? ? ? ? ? ? ? ? ? ? tan gr 1 tan = ? ? 12 2 9.8 0.8 = 19.5 km/hr v min = ? ? ? ? ? ? ? ? ? ? ? ? tan rg 1 tan = ? ? 0.8 2 9.8 1.2 = 12.01 km/hr 3. Calculate the time interval between 33% decay and 67% decay if half-life of a substance is: (1) 40 minutes (2) 20 minutes (3) 60 minutes (4) 13 minutes Sol. (2) T 1/2 = 20 min ? ? ln2 = 20 min ? ? ? ? ? ? ln2 20(min) ? ? N t = N 0 e - ?t t 0 N N = 1 t e ? ? ? 0.67 = 1 t e ? ? ? ln (0.67) = - ?t 1 ? ln ? ? ? ? ? ? 100 67 = ?t 1 ? t 1 = ? ? ? ? ? ? ? 100 ln 20(min) 67 (ln2) Similarly, t 2 = ? ? ? ? ? ? ? 100 ln 20(min) 34 (ln2) t 2 – t 1 = 19.57 min ? 20 min. 4. A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance h = 9.8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is: (take g = 9.8 ms –2 ) M m h v (1) 831.4 m/s (2) 841.4 m/s (3) 811.4 m/s (4) 821.4 m/s Sol. (1) P i = 0.01 × u + 0 = P f = 6 × v v = 0.01u 6 using energy conservation 1 2 × 6 × 2 u 600 ? ? ? ? ? ? = 6 × 9.8 × 9.8 × 10 -2 u = 6 × 98 × 2 = 588 2 m/s 16 th March. 2021 | Shift 2 5. What will be the nature of flow of water from a circular tap, when its flow rate increased from 0.18 L/min to 0.48 L/min? The radius of the tap and viscosity of water are 0.5 cm and 10 –3 Pa s, respectively. (Density of water: 10 3 kg/m 3 ) (1) Remains steady flow (2) Unsteady to steady flow (3) Steady flow to unsteady flow (4) Remains turbulent flow Sol. (3) The nature of flow is determined by reynolds no. R e = vD ? ? if R e < 1000 ? flow is steady 1000 < R e < 2000 ? flow becomes unsteady R e > 2000 ? flow is turbulent (R e ) initial = 10 3 × 3 2 2 0.18 10 (0.5 10 ) 60 ? ? ? ? ? ? ? × 2 3 1 10 10 ? ? ? = 382.16 (R e ) final = 10 3 × 3 2 2 0.48 10 (0.5 10 ) 60 ? ? ? ? ? ? ? × 2 3 1 10 10 ? ? ? = 1019.09 6. The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index? Assume the radii of curvature of the faces of lens are R 1 and R 2 respectively. (1) Zero (2) 1 2 1 2 R R R –R (3) Infinite (4) 1 Sol. (3) 1 f = 2 1 n 1 n ? ? ? ? ? ? ? 1 2 1 1 R R ? ? ? ? ? ? ? 1 f = 0 f = infinite. 7. For the given circuit, comment on the type of transformer used. P ~ L O A D S i 1 i L 0.11A 60 W V 2 220 V (1) Step down transformer (2) Auxilliary transformer (3) Step-up transformer (4) Auto transformer Sol. (3) P s = V s × I s 60 = V s × 0.11 V s = 60 0.11 = 6000 11 , V p = 220 volt V s > V p so it is step up transformer 8. Red light differs from blue light as they have: (1) Same frequencies and same wavelengths (2) Different frequencies and different wavelengths (3) Different frequencies and same wavelengths (4) Same frequencies and different wavelengths Sol. (2) Red light and blue light have different wavelength and different frequency. 9. The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? (m p = 1.00727u m e = 0.00055u) (1) (1860) 2 : 1 (2) 43 : 1 (3) 1860 : 1 (4) 41.4 : 1 Sol. (2) ? e = 12.27 V Å ? p = 0.286 V Å ? ? e p = 12.27 0.286 = 43 10. A charge Q is moving dl ? ? ? distance in the magnetic field B ? ? . Find the value of work done by B ? ? . (1) Infinite (2) 1 (3) –1 (4) Zero Sol. (4) F qV B ? ? ? ? ? P = F.V ? ? = 0 ? W = 0 11. Amplitude of a mass spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to half of its initial value? (ln 2 = 0.693) (1) 15.01 s (2) 17.32 s (3) 0.034 s (4) 34.65 s Sol. (4) A = A 0 bt 2m e ? bt 2m = ln2 = 0.693 t = 2m b × 0.693 t = 2× 500 20 × 0.693 t = 50 × 0.693 = 34.6 sec. 12. A resistor develops 500 J of thermal energy in 20s when a current of 1.5A is passed through it. If the current is increased from 1.5A to 3A, what will be the energy developed in 20s. (1) 500 J (2) 1000 J (3) 2000 J (4) 1500 J Page 5 16 th March. 2021 | Shift 2 SECTION – A 1. A mosquito is moving with a velocity v ? = 0.5 t 2 ? i 3tj 9k ? ? ? ? m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2s? (1) 1 2 tan fromx axis 3 ? ? ? ? ? ? ? ? (2) 1 5 tan fromx axis 2 ? ? ? ? ? ? ? ? (3) 1 2 tan fromy axis 3 ? ? ? ? ? ? ? ? (4) 1 5 tan fromy axis 2 ? ? ? ? ? ? ? ? Sol. Bonus Given 2 ˆ ˆ ˆ v 0.5t i 3tj 9k ? ? ? ? t 2 v ? ? = ˆ ˆ ˆ 2i 6j 9k ? ? ? angle made by direction of motion of mosquito will be – cos -1 2 11 ? ? ? ? ? ? from x-axis = tan -1 117 2 ? ? ? ? ? ? ? ? cos -1 6 11 ? ? ? ? ? ? from y-axis = tan -1 85 9 ? ? ? ? ? ? ? ? cos -1 9 11 ? ? ? ? ? ? from z-axis = tan -1 40 9 no. option is matching 2. Statement I: A cyclist is moving on an unbanked road with a speed of 7 kmh –1 and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s 2 ) Statement-II: If the road is banked at an angle of 45°, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh -1 without slipping. In the light of the above statements, choose the correct answer from the options given below. (1) Both statement I and statement II are false (2) Both statement I and statement II are true (3) Statement I is correct and statement II is incorrect (4) Statement I is incorrect and statement II is correct Sol. (2) On a horizontal ground, v max = ?Rg = ? ? 0.2 2 9.8 = 1.97 m/s 1.97 × 18 5 = 7.12 km/hr= 7.2 km/hr Statement-2 v max = ? ? ? ? ? ? ? ? ? ? ? ? tan gr 1 tan = ? ? 12 2 9.8 0.8 = 19.5 km/hr v min = ? ? ? ? ? ? ? ? ? ? ? ? tan rg 1 tan = ? ? 0.8 2 9.8 1.2 = 12.01 km/hr 3. Calculate the time interval between 33% decay and 67% decay if half-life of a substance is: (1) 40 minutes (2) 20 minutes (3) 60 minutes (4) 13 minutes Sol. (2) T 1/2 = 20 min ? ? ln2 = 20 min ? ? ? ? ? ? ln2 20(min) ? ? N t = N 0 e - ?t t 0 N N = 1 t e ? ? ? 0.67 = 1 t e ? ? ? ln (0.67) = - ?t 1 ? ln ? ? ? ? ? ? 100 67 = ?t 1 ? t 1 = ? ? ? ? ? ? ? 100 ln 20(min) 67 (ln2) Similarly, t 2 = ? ? ? ? ? ? ? 100 ln 20(min) 34 (ln2) t 2 – t 1 = 19.57 min ? 20 min. 4. A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance h = 9.8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is: (take g = 9.8 ms –2 ) M m h v (1) 831.4 m/s (2) 841.4 m/s (3) 811.4 m/s (4) 821.4 m/s Sol. (1) P i = 0.01 × u + 0 = P f = 6 × v v = 0.01u 6 using energy conservation 1 2 × 6 × 2 u 600 ? ? ? ? ? ? = 6 × 9.8 × 9.8 × 10 -2 u = 6 × 98 × 2 = 588 2 m/s 16 th March. 2021 | Shift 2 5. What will be the nature of flow of water from a circular tap, when its flow rate increased from 0.18 L/min to 0.48 L/min? The radius of the tap and viscosity of water are 0.5 cm and 10 –3 Pa s, respectively. (Density of water: 10 3 kg/m 3 ) (1) Remains steady flow (2) Unsteady to steady flow (3) Steady flow to unsteady flow (4) Remains turbulent flow Sol. (3) The nature of flow is determined by reynolds no. R e = vD ? ? if R e < 1000 ? flow is steady 1000 < R e < 2000 ? flow becomes unsteady R e > 2000 ? flow is turbulent (R e ) initial = 10 3 × 3 2 2 0.18 10 (0.5 10 ) 60 ? ? ? ? ? ? ? × 2 3 1 10 10 ? ? ? = 382.16 (R e ) final = 10 3 × 3 2 2 0.48 10 (0.5 10 ) 60 ? ? ? ? ? ? ? × 2 3 1 10 10 ? ? ? = 1019.09 6. The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index? Assume the radii of curvature of the faces of lens are R 1 and R 2 respectively. (1) Zero (2) 1 2 1 2 R R R –R (3) Infinite (4) 1 Sol. (3) 1 f = 2 1 n 1 n ? ? ? ? ? ? ? 1 2 1 1 R R ? ? ? ? ? ? ? 1 f = 0 f = infinite. 7. For the given circuit, comment on the type of transformer used. P ~ L O A D S i 1 i L 0.11A 60 W V 2 220 V (1) Step down transformer (2) Auxilliary transformer (3) Step-up transformer (4) Auto transformer Sol. (3) P s = V s × I s 60 = V s × 0.11 V s = 60 0.11 = 6000 11 , V p = 220 volt V s > V p so it is step up transformer 8. Red light differs from blue light as they have: (1) Same frequencies and same wavelengths (2) Different frequencies and different wavelengths (3) Different frequencies and same wavelengths (4) Same frequencies and different wavelengths Sol. (2) Red light and blue light have different wavelength and different frequency. 9. The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? (m p = 1.00727u m e = 0.00055u) (1) (1860) 2 : 1 (2) 43 : 1 (3) 1860 : 1 (4) 41.4 : 1 Sol. (2) ? e = 12.27 V Å ? p = 0.286 V Å ? ? e p = 12.27 0.286 = 43 10. A charge Q is moving dl ? ? ? distance in the magnetic field B ? ? . Find the value of work done by B ? ? . (1) Infinite (2) 1 (3) –1 (4) Zero Sol. (4) F qV B ? ? ? ? ? P = F.V ? ? = 0 ? W = 0 11. Amplitude of a mass spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to half of its initial value? (ln 2 = 0.693) (1) 15.01 s (2) 17.32 s (3) 0.034 s (4) 34.65 s Sol. (4) A = A 0 bt 2m e ? bt 2m = ln2 = 0.693 t = 2m b × 0.693 t = 2× 500 20 × 0.693 t = 50 × 0.693 = 34.6 sec. 12. A resistor develops 500 J of thermal energy in 20s when a current of 1.5A is passed through it. If the current is increased from 1.5A to 3A, what will be the energy developed in 20s. (1) 500 J (2) 1000 J (3) 2000 J (4) 1500 J 16 th March. 2021 | Shift 2 Sol. (3) I 2 RT = H 1 2 H H = ? ? ? ? ? ? 2 1 2 I I ? H 2 = 500 × ? ? ? ? ? ? 2 3 3 / 2 = 2000 J 13. Calculate the value of mean free path ( ?) for oxygen molecules at temperature 27°C and pressure 1.01 × 10 5 Pa. Assume the molecular diameter 0.3nm and the gas is ideal. (k = 1.38 × 10 –23 jK –1 ) (1) 102 nm (2) 32 nm (3) 58 nm (4) 86 nm Sol. (1) Given : P = 1.01 × 10 5 Pa d = 0.3 nm k = 1.38 × 10 -23 Jk -1 T = 27°C ? ? ? mean = 2 1 N 2 d V ? ? ? ? ? ? ? and PV = nKT ? ? ? mean = 2 KT 2 d p ? ? 23 18 5 1.38 10 300 2 3.14 0.09 10 1.01 10 ? ? ? ? ? ? ? ? ? ? 102 nm 14. A bimetallic strip consists of metals A and B. It is mounted rigidly as shown. The metal A has higher coefficient of expansion compared to that of metal B. When the bimetallic strip is placed in a cold bath, it will: A B (1) Not bend but shrink (2) Neither bend nor shrink (3) Bend towards the right (4) Bend towards the left Sol. (4) ? A > ? B length of both strips will decrease ?L A > ?L BRead More
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