InterviewSolution
Saved Bookmarks
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 particle moves with uniform velocity. Which of the following statements about the motion of the particle is true. (a) its speed is zero. (b) its acceleration is zero. (c) its acceleration is opposite to the velocity. (d) its speed may be variable |
|
Answer» Answer is (b) its acceleration is zero Uniform velocity implies no change in the magnitude and direction. Hence acceleration is zero. |
|
| 2. |
The displacement of a particle is given by √x = t + 1. which of the following statements about its velocity is true.(a) it is zero. (b) it is constant but not zero. (c) it increases with time (d) it decreases with time |
|
Answer» Answer is (c) it increases with time x = t2 + 2t + 1. Hence ν = dx/dt = 2t + 2. It increases with time. |
|
| 3. |
The initial velocity of a particle moving along a straight line is 10 m/s and its retardation is 2 ms-2. the distance moved by the particle in the fifth second of its motion is :(a) 1 m (b) 19 m(c) 50 m (d) 75 m |
|
Answer» Answer is (a) 1m x(nth) = u + 1/2a (2n-1) = 10 + ½ x(-2) (2 x 5-1) = (10 - 9)m = 1m. |
|
| 4. |
A particle moves along X-axis in such a way that its coordinate X varies with the time t according to the expressions x = (8 – 5t + 8t2) meter. The initial velocity of the particle is (a) -5 m/s(b) 8 m/s(c) -8m/s(d) +16 m/s |
|
Answer» Answer is (a) -5 m/s dx/dt = (-5 + 16t) ms-1. At t = 0, dx/dt = -5 ms-1. |
|
| 5. |
The acceleration a of a particle varies with time t as a = bt +c. where b and c are constants. What will be the velocity of the particle which starts from rest after the time t. (a) bt + (1/2)ct2 (b) ct + (1/2)bt2(c) bt + ct2(d) ct + bt2 |
|
Answer» Answer is (b) ct + (1/2)bt2 a = aν/dt = bt + c ν = ∫(bt + c) dt = bt2/2 + ct. |
|
| 6. |
If the displacement of a particle is zero, then what can we say about its distance covered ? (a) it must be zero. (b) it cannot be zero. (c) it is negative (d) it may or may not be zero. |
|
Answer» Answer is (d) it may or may not be zero Displacement can be zero either when the particle is at rest or when it returns to the starting point. |
|
| 7. |
A ball is dropped from the top of tower 100m high. Simultaneously another ball is thrown upward with a sped of 50 ms-1. after what time do they cross each other. (a) 1s (b) 2s (c) 3s (d) 4s. |
|
Answer» Answer is (b) 2s Suppose they cross each other after time t. Then (1/2) x 10 x t2 + [50t - (1/2) + 10 xt2] = 100. This gives t = 2s. |
|
| 8. |
Two bodies are dropped from different heights h1 & h2. The ratio of the times taken by them to reach the ground will be(a) h22 :h12. (b) h1 :h2. (c) √h1 :√h2. (d) none of the above |
|
Answer» Answer is (c) √h1 :√h2. Here h1 = (1/2)gt12 and h2 = (1/2)gt2 |
|
| 9. |
A body is released from the top of a tower of height H meter. After 2 seconds it is stopped and then instantaneously released. What will be its height after next 2 seconds. (a) (H-5) meters. (b) (H-10) meters. (c) (H-20) meters. (d) (H-40) meters. |
|
Answer» Answer is (d) (H-40) meters Distance covered in 2 seconds = (1/2) x 10 x (2)2 = 20m. Total distance covered in 4 seconds = 40m. |
|
| 10. |
A particle is moving on a straight line path with constant acceleration directed along the direction of instantaneous velocity. Which of the following statements about the motion of particle is true. (a) Particle may reverse the direction of motion. (b) distance covered = magnitude of displacement (c) average velocity is less than average speed (d) average velocity = instantaneous velocity. |
|
Answer» Answer is (b) distance covered = magnitude of displacement The Particle is moving on the straight line path and there can be no reversal in the direction of motion. Here distance covered = magnitude of displacement |
|
| 11. |
The distance covered by a particle moving along a straight line path with uniform acceleration is given by : x = 6 + 6t + 10t2 metre. What is the initial velocity of the particle.(a) 6 ms-1(b) 7 ms-1(c) 8 ms-1 (d) 9 ms-1 |
|
Answer» Answer is (b) 7 ms-1 Velocity = dx/dt = 7 + 20t. At t = 0, initial velocity is 7 ms-1. |
|
| 12. |
A particle starts with initial velocity 10ms-1. It covers a distance of 20 m along a straight into two seconds. What is the acceleration of the particle. (a) zero (b) 1 ms-2(c) 10 ms-2 (d) 20 ms-2 |
|
Answer» Answer is (a) zero Since the particle covers a distance of 20 m in 2 seconds & its initially velocity is 10 ms-1, so that particle moves with the constant velocity. |
|
| 13. |
A particle starts from the rest & moving with constant acceleration covers a distance x1 in the 3rd second x2 in the 5th second. The ratio x1/x2 =(a) 3/5 (b) 5/9 (c) 9/25 (d) 25/81 |
|
Answer» Answer is (b) 5/9 Use xn = ν0 + (a/2)(2n-1) |
|
| 14. |
A particle starts from rest and moves with constant acceleration a. what is the nature of the graph between the time (t) and the displacement.(a) straight line (b) instant velocity (c) average speed (d) average velocity |
|
Answer» Answer is (b) instant velocity Here x = (1/2)at2. It is of the form y = mx2. Therefore, the graph is a symmetric parabola. |
|
| 15. |
The time elapsed is plotted along X-axis and the acceleration is plotted along the Y-axis. The area between the graph and the X-axis gives.(a) Average velocity (b) distance covered (c) differences in velocities (d) difference in accelerations |
|
Answer» Answer is (c) differences in velocities The area under a-t graph gives change in velocity. |
|
| 16. |
Which of the following statement is false for motion with uniform velocity(a) The motion is along a straight line path.(b) The motion is always in the same direction(c) Magnitude of displacement < distance covered(d) Average velocity is equal to the instantaneous velocity. |
|
Answer» Answer is (c) Magnitude of displacement < distance covered For motion with uniform velocity, the magnitude of displacement = distance covered. |
|
| 17. |
A particle starts with velocity u and moves with constant acceleration a .what is the nature of the graph between the time (t) and the displacement (x).(a) straight line(b) symmetric parabola(c) Asymmetric parabola(d) Rectangular hyperbola. |
|
Answer» Answer is (c) Asymmetric parabola Here x = ut + (1/2)at2. It is of the form y = ax + bx2. So the graph is a symmetric parabola |
|
| 18. |
The velocity graph of a motion starting from rest with uniform acceleration is a straight line: (a) Parallel to time - axis. (b) parallel to velocity axis (c) not passing through origin (d) having none of the above characteristics. |
|
Answer» Answer is (d) having none of the above characteristics The velocity time graph of motion with uniform acceleration is inclined to both the time and velocity axis and passes through the origin. |
|
| 19. |
The slope of the velocity time graph for retarded motion is: (a) Positive (b) negative (c) zero (d) can be +ve, -ve or zero. |
|
Answer» Answer is (b) negative Slope of time graph gives the acceleration or retardation. For retarded motion, the angle made by the graph with time axis is more than π/2. Hence slope is negative. |
|
| 20. |
A particle moving with uniform speed can possess :(a) radial acceleration(b) tangential acceleration(c) both radial and tangential accelerations(d) neither radial nor tangential acceleration |
|
Answer» Answer is (a) radial acceleration Since speed is uniform, acceleration can be centripetal. Or radial when the particle is moving along a circular path |
|
| 21. |
A car accelerates from rest for time t1 at constant rate a1 and then it retards at the constant rate a2 for time t2 and comes to rest. t1/t2 =(a) a1/a2 (b) a2/a1 (c) a12/a22 (d) a22/a12 |
|
Answer» Answer is (b) a2/a1 v = a1t1 = a2t2 hence t1/t2 = a2/a1 |
|