# Obj-2 Rest and Motions Kinematics HC Verma Solutions Ch-3

**Obj-2 Rest and Motions** Kinematics HC Verma Solutions Ch-3 Vol-1 Concept of Physics for Class-11. **Solution of Objective -2 (MCQ-2) **Questions of Ch-3 **Rest and Motions** Kinematics** **(Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

** Obj-2 (MCQ-2 ) Rest and Motions** Kinematics HC Verma Solutions Ch-3 Vol-1 Concept of Physics for Class-11

Board | ISC and other board |

Publications | Bharti Bhawan Publishers |

Chapter-3 | Rest and Motions Kinematics |

Class | 11 |

Vol | 1st |

writer | H C Verma |

Book Name | Concept of Physics |

Topics | Solution of Objective-2 (MCQ-2) Questions |

Page-Number | 50 , 51 |

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Question for Short Answer

Objective-I

Objective-II (Currently Open)

Exercise

### HC Verma Solutions of Ch-3 ** Rest and Motions** Kinematics **Obj-2 (MCQ-2) Questions **Vol-1 Concept of Physics for Class-11

(Page-50)

**Question-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-1**

The options** (a)**** **and** (d) **are correct

**Explanation:**

Displacement is zero because the initial and final positions are the same.

**Question-2**

A particle moves along the X-axis as x = u (t − 2 s) + 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-2**

The options** (c)**** **and** (d) **are correct

**Explanation:**

x = u(2 s − 2 s) + a (2 s − 2 s)

^{2}= 0 (origin)

**Question-3**

(a) 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

(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-3**

The options** (a)**** , (b) **and** (c) **are correct

**Explanation:**

Average velocity = Displacement / Total time

Displacement ≤ Distance

∴ Average velocity ≤ Average speed In uniform circular motion, speed is constant but velocity is not.

In one complete circle of uniform motion, average velocity is zero. Instantaneous velocity is never zero in the interval.

**Question-4**

(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-4**

The options** (b) **and** (d) **are correct

**Explanation:**

Velocity and acceleration are vector quantities that can be changed by changing direction only (keeping magnitude constant).

**Question-5**

(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-5**

The options **(a), (b) **and** (d) **are correct

**Explanation:**

**Question-6**

(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-6**

The options **(b), (c) **and** (d) **are correct

**Explanation:**

(c) Since the acceleration is zero from t = 0 s to t = 10 s, change in velocity is 0.

Velocity in this interval = Initial velocity = 0

Also,

Speed in this interval = Initial speed = 0

(d) From t = 0 s to t = 10 s, speed is zero.

Here, velocity is zero and initial velocity is zero.

So, the change in velocity is zero; i.e., acceleration is zero.

**Question-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-7**

The option** (a) **is correct

**Explanation:**

The magnitude of the velocity of a particle is equal to its speed.

(a) Velocity being a vector quantity has magnitude as well as direction, and magnitude of velocity is called speed.

**(b) Average velocity** =

Total displacement / Total time taken

**Average speed** =

Total distance travelled / Total time taken

Distance ≥ Displacement

∴ Average speed ≥ Average velocity

The magnitude of average velocity in an interval is not always equal to its average speed in that interval.

(c) If speed is always zero, then the distance travelled is always zero. Hence, the total distance travelled and the average speed will be zero.

(d) If the speed of a particle is never zero, the distance travelled by the particle is never zero. Hence, the average speed will not be zero.

**Question-8**

The velocity-time plot for a particle moving on a straight line is shown in the figure.

(a) The particle has a constant acceleration.

(b) The particle has never turned around.

(c) The particle has zero displacement.

(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-8**

The options **(a) **and** (d) **are correct

**Explanation:**

(a) The slope of the v–t graph gives the acceleration. For the given graph, the slope is constant. So, acceleration is constant.

(b) From 0 to 10 seconds, velocity is in positive direction and then in negative direction. This means that the particle turns around at t = 10 s.

(c) Area in the v–t curve gives the distance travelled by the particle.

Distance travelled in positive direction ≠ Distance travelled in negative direction

∴ Displacement ≠ Zero

(d) The area of the v–t graph from t = 0 s to t = 10 s is the same as that from t = 10 s to t= 20 s. So, the distance covered is the same. Hence, the average speed is the same.

**Question-9**

In 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 = 6s.

(c) The velocity remains positive for t = 0 to t = 6s.

(d) The average velocity for the total period shown is negative.

**Answer-9**

The option **(a)**** **is correct

**Explanation:**

(a) The slope of the x–t graph gives the velocity. Here, 6 times the slope is zero. So, the particle has come to rest 6 times.

(b) As the slope is not maximum at t = 6 s, the maximum speed is not at t = 6 s.

(c) As the slope is not positive from t = 0 s to t = 6s, the velocity does not remain positive.

(d) Average velocity =

For the shown time (t = 6 s), the displacement of the particle is positive. Therefore, the average velocity is positive.

(Page-51)

**Question-10**

The accelerations of a particle as seen from two frames S_{1} and S_{2} have equal magnitude 4 m/s^{2}.

(a) The frames must be at rest with respect to each other.

(b) The frames may be moving with respect to each other.

(c) The acceleration of S_{2} with respect to S_{1} may either be zero of 8 m/s^{2}.

(d) The acceleration of S_{2} with respect to S_{1} may be anything between zero and 8 m/s^{2}.

**Answer-10**

The option **(d)**** **is correct

**Explanation:**

—: End of **Obj-2 (MCQ-2) Rest and Motions** Kinematics HC Verma Solutions :–

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