OBJECTIVE-II
1. Consider the motion of the tip of the minute hand of a clock. In one hour
(a) the displacement is zero
(b) the distance covered is zero
(c) the average speed is zero
(d) the average velocity is zero.
Answer: (a), (d)
Explanation: In one hour the tip of the minute hand comes to the same position, so its displacement and average velocity are zero. Hence (a) and (d) is correct. But in this time the tip covers a distance equal to the circumference of a circle of radius equal to the length of the minute hand. So (b) and (c) are incorrect.
2. A particle moves along the X-axis as
x=u(t-2s)+a(t-2 s)² .
(a) the initial velocity of the particle is u
(b) the acceleration of the particle is a
(c) the acceleration of the particle is 2a
(d) at t=2 s particle is at the origin.
Answer: (c), (d)
Explanation: The velocity at any instant is given by v=dx/dt=u+2a(t-2)
initial velocity is the velocity at t=0, putting this value we get v=u-4a which is not zero. So (a) is incorrect.
Acceleration is given by dv/dt =2a, So (b) is incorrect and (c) is correct.
Putting t=2 s in the given displacement equation we get x=0, so (d) is also correct.
3. Pick the correct statement:
(a) The average speed of a particle in a given time is never less than the magnitude of the average velocity
(b) It is possible to have a situation in which |dv/dt| ≠ 0 but d|v|/dt=0
(c) The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval.
(d) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval. (Infinite accelerations are not allowed.)
Answer: (a), (b), (c)
Explanation: Since the distance covered by a particle in a given time will never be less than the magnitude of displacement, so the average speed is never less than the magnitude of the average velocity. (a) is correct.
|dv/dt| is the magnitude of the acceleration and d|v|/dt is the rate of change of speed. Consider the case of uniform circular motion. Here speed of the particle is constant, so d|v|/dt=0. But magnitude of the acceleration |dv/dt| ≠ 0. (b) is also correct.
As we have seen in question no 1, the average velocity of the tip of the minute hand is zero in the time interval of one hour, but its instantaneous velocity is never zero in this interval. So (c) is also correct.
If the average velocity of a particle moving on a straight line is zero in a time interval, it means its displacement is zero. It can only be possible if the particle has returned back to its original position at least once in this time interval. So at the instant when it reverses the direction of its velocity, the instantaneous velocity of the particle will definitely be zero. So (d) is incorrect.
4. An object may have
(a) varying speed without having varying velocity
(b) varying velocity without having varying speed
(c) nonzero acceleration without having varying velocity
(d) nonzero acceleration without having varying speed
Answer: (b), (d)
Explanation: If the velocity is constant, speed can not vary. So (a) is not true.
In a uniform circular motion velocity varies while speed remains constant and since the velocity varies acceleration is non zero. So (b) and (d) both are true.
The rate of change of velocity is acceleration, so without having varying velocity, the acceleration is zero. So (c) is not true.
5. Mark the correct statements for a particle going on a straight line:
(a) If the velocity and acceleration have opposite sign, the object is slowing down.
(b) If the position and velocity have opposite sign, the particle is moving towards the origin.
(c) If the velocity is zero at an instant, the acceleration should also be zero at that instant.
(d) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.
Answer: (a), (b), (d)
Explanation: If the acceleration and velocity of the particle have opposite signs, it has a retarding effect. As is the case of a particle thrown vertically upward. So (a) is true. (b) is also obviously true.
(c) is not true. Consider the case of a particle thrown vertically upward. At its highest point, the velocity is zero but acceleration is not zero.
Acceleration is defined as the rate of change of velocity. If in a time interval velocity is zero, it means in this interval velocity does not change. So acceleration at any instant in this time interval is zero. Hence (d) is true.
6. The velocity of a particle is zero at t=0.
(a) The acceleration at t=0 must be zero.
(b) The acceleration at t=0 may be zero
(c) If the acceleration is zero from t=0 to t=10 s, the speed is also zero in this interval.
(d) If the speed is zero from t=0 to t=10 s, the acceleration is also zero in this interval.
Answer: (b), (c), (d)
Explanation: (a) is not true. Consider the case of a particle dropped vertically downward, its velocity at t=0 is zero but acceleration is not zero.
(b) is true because we do not have information about the velocity in next instants. If in next instants the velocity still remains zero, the acceleration will be zero.
(c) is true because, if acceleration is zero that means there is no change in velocity. The velocity remains zero in this interval.
(d) is true because if the speed remains zero from t=0 to t= 10 s, that means the particle is at rest in this interval, so the acceleration is also zero in this interval.
7. Mark the correct statements :
(a) The magnitude of the velocity of a particle is equal to its speed.
(b) The magnitude of average velocity in an interval is equal to its average speed in that interval.
(c) It is possible to have a situation in which the speed of a particle is always zero but the average speed is not zero.
(d) It is possible to have a situation in which the speed of the particle is never zero but the average speed in an interval is zero.
Answer: (a).
Explanation: At any instant, the magnitude of the velocity is also its speed. So (a) is correct.
(b) is incorrect because velocity is displacement in unit time while speed is distance covered in unit time.
(c) is incorrect because if speed is zero in a time interval, that means the particle has not moved in this interval and no distance traveled. So average speed has to be zero.
(d) is incorrect because if the speed of the particle is never zero, means it has traveled some distance. So the average speed in the interval will not be zero.
8. The velocity-time plot for a particle moving on a straight line is shown in the figure (3-Q4).
 |
| Figure for Q-8 |
(a) The particle has a constant acceleration.
(b) The particle has never turned around.
(c) The particle has zero displacements.
(d) The average speed in the interval 0 to 10 s is the same as the average speed in the interval 10 s to 20 s.
Answer: (a), (d).
Explanation: (a) is true because in a velocity-time graph the slope (which shows the rate of change of velocity) denotes acceleration. In this graph the slope is constant, so acceleration is constant.
(b) is not true because at t=10 s the velocity changes its sign from positive to negative, means it has turned around.
(c) is not true because displacement in the positive direction of v is given by the area of the graph above the time axis and displacement in the negative direction of v is given by the area of the graph below the time axis. In this case, later is greater than former not equal. So displacement is not zero.
(d) is true because in 0 to 10 s speed changes from 10 to 0 m/s and in 10 to 20 s speed changes from 0 to 10 m/s. So average speed will be the same.
9. Figure (3-Q5) shows the position of a particle moving on the X-axis as a function of time.
(a) The particle has come to rest 6 times.
(b) The maximum speed is at t=6 s.
(c) The velocity remains positive for t=0 to t=6 s.
(d) The average velocity for the total period shown is negative.
 |
| Figure for Q-9 |
Answer: (a).
Explanation: Slope of the curve at any point gives the velocity of the particle at that point. In the given graph the curve has three troughs and three crests. At these points, the slope of the curve (tangent is horizontal to time axis) is zero. So at six points velocity is zero, means it has come to rest.
(b) is not true because the slope of the curve is not maximum at t=6 s.
(c) is not true because the gradient of the slope is not positive always between t=0 to t=6 s.
(d) is not true because the final displacement is not less than the initial.
10. The accelerations of a particle as seen from two frames S1 and S2 have equal magnitude 4 m/S².
(a) The frame must be at rest with respect to each other.
(b) The frames may be moving with respect to each other but neither should be accelerated with respect to the other.
(c) The acceleration of S2 with respect to S1 may either be zero or 8 m/s².
(d) The acceleration of S2 with respect to S1 may be anything between 0 and 8 m/s²
Answer: (d).
Explanation: The acceleration of S2 with respect to S1 is given by
a= ap,s1 - ap,s2, This vector sum can be easily understood by the following figure:--
Explanation to the answer for question number 10.
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CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 18 - Geometrical Optics
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
Sir in Q.no 9 isn't the final displacement less than the intial displacement. Hmmn we can actually see it with a cm scale that the x-coordinate of final displacement is slightly below than that of initial displacement. Or am i misinterpreting the info. Can you plz make it clear.
ReplyDeleteP.S. loved your work it's been aaa great help
Thank you Sahil Baghel for your interest in these solutions.
ReplyDeleteWhen it is not very clear to the naked eyes whether it is more or less, it is better to assume it equal. If you assume it equal even then the explanation in (d) is ok.
Thnq sir. Can I ask some more questions??? Those will be of H C Verma only, actually kind of some explanation on doubts that I M having in it's worked out example section. It would really be a grt help. I would refer to somewhere else in internet but those places doesn't have qualified persons to answer having mixed answers confusing even more. So can i clear my some doubts for this chap for the book. I totally understand if you find it a hassel to do so. But it would be my sincere request to you, kindly help me with your knowledge if you find it capable to do so.
DeleteThanks a lot sir ...this blog has been of great help ....actually i have doubt in option (b) of Q 5.
ReplyDeleteSir as far my understanding displacement has the same direction as the velocity .so how position and velocity can have opposite signs ? Can u please elaborate this with an example .
And sir in option (d) does the line 'if the velocity is zero for a time interval' means average velocity ? I am not able to understand option d as well .
Dear student, there is a difference between position and displacement. The position vector (or simply position) is the vector with its tail at the origin and tip at the point where the object is. The displacement is the vector with its tip at the point where the object is and the tail at the point where the object was at the considered instant of time.
DeleteThus if the velocity of an object is opposite to the position, then it is moving towards the origin because the position vector is always outwards from the origin. The displacement is along the velocity. Hence the option (b).
For the option (d), let us understand it with an example. Suppose the velocity is zero for a time interval from t = 3 s to t = 10 s. It means the object has stopped between 3 - 10 s. So in between 3 s to 10 s the acceleration is zero at any instant because there is no change in the velocity. Hence the option (d). There is no question of average velocity.
Sir i am finding it hard to understand option d of Q 9 although u have explained in one of the comments .
ReplyDeleteThe average velocity is the final displacement between t=0 to t=8 s divided by the total time.
DeleteThanks a lot sir for your efforts otherwise hcv would have been a nightmare ...but could u please give reasons to eliminate options a , b, c in Q 10
ReplyDeleteIn option (a), the 'must' word is not correct. If it were to be replaced by 'may' then the option (a) would have been correct.
DeleteSince option (d) has been proved OK, so only zero or 8 m/s² in option (c) is not correct.
The statement in (c) that 'neither should be accelerated' is not correct because we see that even after acceleration w.r.t. to each other the acceleration measured in both frames may be the same.
Thanks u sir
ReplyDeleteSir I am having doubt in Q 2 . Why Cant the Q be solved just by comparing the equation given in Q with s = ut + 1/2at^2?
ReplyDeleteIt is the same equation with difference that the particle is not at the origin at t=0.
DeleteSir in Q 3 in option d , u have mentioned that " at the instant when it reverses the direction of its velocity , the instantaneous velocity of the particle will definitely be zero" .sir why on reversing the velocity will be zero ? I mean in daily life we do reverse our vehicles but without making the speed '0'. Also what is the significance of the line " infinite acceleration are not allowed" (as mentioned in the question)?
ReplyDeleteThanks a lot sir for awesome explanation.
When you move along velocity axis from +v to -v, you have to cross the origin at certain instant where the v=0. For the daily life, you may not have felt it but it is true for a particle.
DeleteSince dv/dt =a, an infinite 'a' means very very quick reversal of velocity with very very high magnitude at that instant that may be difficult to explain the zero velocity at that instant.
Sir in Q 3 isn't the option c and d same? For explanation of option C you have given example of a clock . Let's assume the clock is showing 6:00 PM. Then minute hand is at at '12' .as time passes minute hand moves to 1, then 2, then 3 etc thereby increasing the displacement. At '6' the displacement of the minute hand is maximum. Then it starts moving to 7 then , 8 , then 9 and thereby decreasing the displacement. Sing the minute hand in some way retracing the path back in the same way as you have mentioned in example for option d ?
ReplyDeleteOptions (c) and (d) are not the same. In (d) the particle is moving along a straight line.
DeleteSir u am not able to understand the option d of Q 5 . It is mentioned that " if the velocity is zero for a TIME INTERVAL" .there is a possibility that the object started and ended at the starting point in the given time interval . This mean that at any instant of time there was some value of the velocity and it was not zero . Since there was some value of the velocity, there is also a possibility that the value may had been ' varying ' . In that case the acceleration at any ' INSTANT ' would not have been zero.
ReplyDeleteWhat you are thinking is the average velocity in the given time interval but the problem mentions that velocity is zero. Hence the given explanation.
Delete