InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A body is dropped from a height h, after the third rebound it rises to h/2. What is the coefficient the restitution ?(A) (1/2) (B) (1/2)1/2(C) (1/2)1/4 (D) (1/2)6 |
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Answer» Answer is (D) (1/2)6 [hn/ h] = e2n. That is (h/2)/ h = e6. This gives e = (1/2)6. |
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| 2. |
If a body of mass 200 g falls from a height 200 m and its total P.E is converted into K.E. at the point of contact of the body with earth surface, then what is the decrease in P.E of the body at the contact ?(A) 200 J(B) 400 J(C) 600 J(D) 900 J |
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Answer» Answer is (B) 400 J ∆Up = mgh = 0.2 x 10 x 200 |
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| 3. |
The kinetic energy acquired by a mass m in travelling a certain distance d, starting from rest, under the action of a constant force is directly proportional to (A) √m (B) independent of m. (C) 1/√m. (D) m. |
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Answer» Answer is (B) independent of m. K.E = (1/2) mν2, & ν2 = 2ax = 2(F/m)x. Hence K.E = Fx |
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| 4. |
A shell of mass M is moving with a velocity V. it explodes into two parts, and one of the parts of mass m is left stationary. What will be the velocity of the other part.(A) MV/ (M - m)(B) MV/ ( M + m)(C) mV/ (M - m)(D) mV/ (M + m) |
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Answer» Answer is (A) MV/ (M - m) Apply conservation of momentum MV = (M - m)ν. |
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| 5. |
A bullet is fired on a wooden block. Its velocity reduces by 50% on penetrating 30 cm. How long will it penetrate further before coming to east. (A) 30 cm (B) 15 cm (C) 10 cm (D) 5 cm |
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Answer» Answer is (C) 10 cm Velocity reduces to 50%. So the kinetic energy left with the bullet will be 25% of the initial value and it will travel further for 10 cm. |
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| 6. |
An automobile engine of mass M accelerates and a constant power P is applied by the engine The instantaneous speed of the engine will be(A) [Pt/M]1/2. (B) [(2)t/M]1/2.(C) [Pt/2M]1/2.(D) [PT/4M]1/2. |
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Answer» Answer is (B) [(2)t/M]1/2. Fν = P Or M(dν/dt)ν = P That is ∫ν dν = ∫(P/M) dt. Hence ν = [2Pt/ M]1/2. |
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| 7. |
A bullet is fired and gets embedded in block kept on table, then if tale is frictionless. (A) Only K.E. is conserved (B) Only momentum is conserved (C) Both are conserved (D) None of these. |
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Answer» Answer is (B) Only momentum is conserved It is a case of inelastic collision. |
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| 8. |
A bomb explodes into two fragments of masses 3 kg and 1 kg. total kinetic energy of the fragments is 6 MJ. What is the ratio of the kinetic energies of smaller mass to the larger mass ? (A) 1/3. (B) 3. (C) 1/9. (D) 9. |
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Answer» Answer is (B) 3. Here MV = mν Squaring both sides and simplifying we get [(1/2) MV2]M = (1/2)(mν2)m. That is (1/2)mν2/ (1/2)Mν2 = M/m. |
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| 9. |
A ball is dropped from height 10. Ball in embedded in sand 1m and stops.(A) Only momentum remains conserved.(B) Only kinetic energy remains conserved. (C) Both momentum and K.E. are conserved.(D) Neither K.E. nor momentum is conserved. |
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Answer» Answer is (A) Only momentum remains conserved. It is a case of inelastic collision. |
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| 10. |
A ball of mass M moving with velocity ν collides elastically with wall and rebounds .The change in momentum of the ball will be :(A) Zero(B) Mν (C) 2 Mν(D) 4 Mν |
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Answer» Answer is (C) 2 Mν ∆p = Mν - (- Mν) = 2 Mν |
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| 11. |
Which of the following statements is false?(A) Momentum is conserved in all types of collisions. (B) Energy is conserved in all types of collisions. (C) During elastic collisions conservative forces are involved. (D) Work energy theorem is not applicable to inelastic collisions. |
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Answer» Answer is (D) Work energy theorem is not applicable to inelastic collisions. Work energy theorem is always applicable. |
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| 12. |
Two bodies of masses m and 4m are moving with equal kinetic energy. The ratio of their linear momentum is (A) 4 :1(B) 1 : 1(C) 1 : 2(D) 1 : 4 |
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Answer» Answer is (C) 1 : 2 K = (1/2) Mν2 = p2/2M That is P = √2 kM. hence P1/P2 = √[2KM/(2K x 4 m)] = √(1/4) = (1/2). |
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| 13. |
If momentum is increased by 20% then K.E increases by(A) 44%(B) 55%(C) 66%(D) 77% |
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Answer» Answer is (A) 44% K = p2/2m K’ = [(1200/100)P]2/2m = 1.44 p2/2m = 1.44 K ∆K = K’ – K = 0.44 K = 44 % K |
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| 14. |
A body of mass 2 kg is thrown up vertically with K.E. of 490 joules. If the acceleration due to gravity is 9.8 m/s2, the height at which the K.E of the body becomes half of its original value is given by :(A) 50m (B) 12. 5m(C) 25m (D) 10m. |
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Answer» Answer is (B) 12. 5m When he kinetic energy is halved, the potential energy s equal to the kinetic energy. That is mgh = (490/2)J = 245 J. this gives h =12.5 m. |
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| 15. |
A bullet of mass 10 g leaves a rifle at a initial velocity of 100 m/s. and strikes the earth at the same level with a velocity of 500 m/s. The work done in joules overcoming the resistance of air will be(A) 375 (B) 3750 (C) 5000 (D) 500 |
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Answer» Answer is (B) 3750 W = ∆E = (1/2) x 0.01 [(1000)2 – (500)2] = 3750 J |
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