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Physics Motion MCQs with Solutions for CBSE Class 9 | NTSE

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We have provided Physics Motion MCQs with Solutions for Class 9 CBSE & NTSE, to prepare the student for any eventuality and lays a strong foundation for higher level exams.

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1 Introduction of Motion
1.1 Multiple Choice Questions with Answer & Solutions of Motion
2 Also Read

Introduction of Motion

Rest and motion are relative terms which actually depend on the observer.
A body is said to be at rest if its position does not change with time with respect to the observer.
Similarly, A body is said to be in motion if its position changes with time with respect to the observer.
The same body may be at rest with respect to one observer while in motion with respect to some other observer. For example :

Physics Motion
(1) A briefcase in a moving train is at rest with respect to a person sitting in train but in motion with respect to any person standing outside.

(2) Consider a person at O (figure) standing by the roadside who locates a car at the position A. After some time he finds the car at the position B. This change in position of the car gives the impression to the observer that the car is moving.

A fellow passenger in the car finds the driver always sitting next to him while the observer at O finds the driver changing his position from A to B. Therefore, we say that the driver is at rest with respect to the fellow passenger but is in motion with respect to the observer outside. Thus rest and motion are relative terms and are not absolute.

Multiple Choice Questions with Answer & Solutions of Motion

1. Which of the following is not a characteristic of displacement :
(A) It is always positive
(B) It has both magnitude and direction
(C) It can be zero
(D) Its magnitude may be less than or equal to the actual path length of the object
Solution

Ans. (A)
Sol. Displacement has both magnitude & direction and it can be +ve, -ve or Zero.

2. The average velocity of a body is equal to the mean of its initial and final velocities. The acceleration of the body is :
(A) uniform
(B) variable
(C) uniformly variable
(D) zero
Solution

Ans. (A)

Sol. Average Velocity = \left( {\frac{{{\rm{Final}}\,{\rm{Velocity}}\,\,{\rm{ + }}\,\,{\rm{Initial}}\,\,{\rm{Velocity}}}}{{{\rm{Time}}\,\,{\rm{Taken}}}}} \right) {when a = constant}

3. A body whose position with respect to surrounding does not change, is said to be in a state of :
(A) Rest
(B) Motion
(C) Vibration
(D) Oscillation
Solution

Ans. (A)
Sol. When an object is not changing its position w.r.t. its surrounding & time it is said to be at rest.

4. In case of a moving body :
(A) Displacement > Distance
(B) Displacement < Distance
(C) Displacement > Distance
(D) Displacement < Distance
Solution

Ans. (D)
Sol. For a moving body, distance > displacement

5. The position of a particle going along a straight line is x1 = 50m at 10.30 a.m. & x2 = 55m at 10.35 a.m. The velocity of the particle is :
(A) 2m/min
(B) 5m/min
(C) 1m/min
(D) 0.5m/min
Solution

Ans. (C)
Sol. velocity = \frac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \frac{{(55 - 50)}}{5} = 1{\rm{m/min}}

6. A body travels half the distance with speed of 20 m/s and the other half with speed of 30 m/s. The average speed of the body during whole journey is :
(A) zero
(B) 24 m/s
(C) 23 m/s
(D) None of these
Solution

Ans. (B)
Sol. Average speed = \frac{{{\rm{Total}}\,\,{\rm{distance}}}}{{{\rm{Total}}\,\,{\rm{Time}}}}= \frac{d}{{\left( {\frac{{d/2}}{{20}}} \right) + \left( {\frac{{d/2}}{{30}}} \right)}} = 24\,{\rm{m/s}}

7. A 50 m long train passes over a bridge at a speed of 30 km/h. If it takes 36 seconds to cross the bridge, the length of the bridge is :
(A) 100 m
(B) 200 m
(C) 250 m
(D) 300 m
Solution

Ans. (C)
Sol. t = \frac{s}{v} \Rightarrow 36 = \frac{{50 + L}}{{30 \times \left( {\frac{5}{{18}}} \right)}} \Rightarrow \,L = 250\,m

8. A body is moving on a square track of side 1 m. It completes one revolution in 40 seconds. Calculate its displacement in 4 minutes :
(A) \sqrt 2 m 
(B) \sqrt 3 m 
(C) 0 m 
(D) 24 m 
Solution

Ans. (C)
Sol. Object is covering 1m in 10 sec.

Body is moving on a square track

so in 4 minutes it comes back to its initial position, Hence its displacement = 0

9. A man is running at a speed of 10m/s. Behind him a bullet is fired with a velocity of 500 m/s. What is the distance of the man from shooter, if bullet hits him after 1/2 sec of being fired :
(A) 250m
(B) 245m
(C) 500m
(D) 490m
Solution

Ans. (B)
Sol. Relative speed = 500 – 10 = 490 m/sec.
Time taken = \frac{1}{2}sec.
Distance = Speed × time
= 490 × \frac{1}{2} = 245 m

10. For motion on a straight line path with constant acceleration, the ratio of magnitude of displacement & distance covered is :
(A) = 1
(B) < 1
(C) > 1
(D) < 1
Solution

Ans. (D)
Sol. In straight line path when direction remains same :
distance = displacement
But when direction changes :
distance > displacement
So ratio of displacement to distance < 1

11. In a lift falling freely under gravity, if a man throws a coin in upward direction with velocity u, the motion of the coin with respect to the man will be :
(A) The coin will go to a maximum height & will come back
(B) Coin will go upward with uniform velocity & stick to the roof
(C) Coin will move in downward direction with acceleration g
(D) None of these
Solution

Ans. (B)
Sol. In case of free fall relative acceleration is zero so coin will not fall back as seen by mean therefore, it will go up and stick to the roof.

12. When a person leaves his home for sightseeing by his car, the odometer reads 12352km. When he returns home after two hours the reading is 12416km. What is the average speed of the car during this period?
(A) 30km/hr
(B) 32km/hr
(C) 36km/hr
(D) 40km/hr
Solution

Ans. (B)
Sol. Average speed = \frac{{{\text{Total}}\,{\text{distance}}}}{{{\text{time}}\,\,{\text{taken}}}} 
= \frac{{12416 - 12352}}{2} = 32\,{\text{km/hr}}

13. A hall has the dimensions 10m × 10m × 10m. A fly starting at one corner ends up at a diametrically opposite corner. The magnitude of its displacement is nearly.
(A) 5\sqrt 3 {\text{ }}m 
(B) 10\sqrt 3 \,{\text{m}} 
(C) 20\sqrt 3 {\text{ m}} 
(D) 30\sqrt 3 \,{\text{m}} 
Solution

Ans. (B)
Sol. displacement = \sqrt {{l^2} + {b^2} + {h^2}} = 10\sqrt 3 m

14. A body goes from A to B with a velocity of 20 m/s and comes back from B to A with a velocity of 30 m/s. The average velocity of the body during the whole journey is :
(A) Zero
(B) 24 m/s
(C) 25 m/s
(D) None of these
Solution

Ans. (A)
Sol. Average Velocity = \frac{{{\text{Total}}\,\,{\text{displacement}}}}{{{\text{Total}}\,\,{\text{time}}}} = \frac{0}{t} = 0

15. Choose the wrong statement :
(A) Retardation is a vector quantity
(B) Acceleration due to gravity is a vector quantity
(C) Average speed is a vector quantity
(D) Displacement is a vector quantity
Solution

Ans. (C)
Sol. Acceleration, Retardation & displacement are vector quantities while average speed is a scalar quantity.

16. The displacement -time graph of a particle moving along a straight line is shown in the figure. The velocity is negative at the point:

(A) C displacement -time graph of a particle moving along a straight line
(B) D   
(C) E
(D) F
Solution

Ans. (C)
Sol. Slope of displacement time graph gives velocity

17. A dog runs 120 m away from its master in a straight line in 9.0 s, and then runs halfway back in one-third the time. Calculate (a) its average speed and (b) its average velocity :
(A) Average speed = 15m/s ; Average velocity = 5m/s away from master
(B) Average speed = 5m/s ; Average velocity = 5m/s away from master
(C) Average speed = 15m/s ; Average velocity = 5m/s towards master
(D) Average speed = 5m/s ; Average velocity = 5m/s towards master
Solution

Ans. (A)
Sol. Average speed = \frac{{{\text{Total}}\,\,{\text{Distance}}}}{{{\text{Total}}\,{\text{time}}}} = \frac{{120 + \frac{{120}}{2}}}{{9 + 3}} = \frac{{180}}{{12}} = 15\,{\text{m/s}} 
Average velocity = \frac{{{\text{Total}}\,{\text{Displacement}}}}{{{\text{Total}}\,{\text{time}}}} = \frac{{120 - 60}}{{9 + 3}} = \,5\,{\text{m/s}}

18. A particle covers each \frac{1}{3} of the total distance with speed V1, V2 and V3 respectively. Find the average speed of the particle:
(A) \frac{{{V_1}{V_2}{V_3}}}{{{V_1}{V_2} + {V_2}{V_3} + {V_1}{V_3}}}  
 (B) \frac{{2{V_1}{V_2}{V_3}}}{{{V_1}{V_2} + {V_2}{V_3} + {V_1}{V_3}}} 
(C) \frac{{3{V_1}{V_2}}}{{{V_1}{V_2} + {V_2}{V_3}}} 
(D) \frac{{3{V_1}{V_2}{V_3}}}{{{V_1}{V_2} + {V_2}{V_3} + {V_1}{V_3}}} 
Solution

Ans. (D) 
Sol. Average speed = \frac{{{\text{Total}}\,{\text{Distance}}}}{{{\text{Total}}\,{\text{Time}}}} = \frac{{3d}}{{\frac{d}{{{v_1}}} + \frac{d}{{{v_2}}} + \frac{d}{{{v_3}}}}} = \frac{{3{V_1}{V_2}{V_3}}}{{{V_1}{V_2} + {V_2}{V_3} + {V_3}{V_1}}}

19. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h.
Finding the market closed, he instantly turns and walks home with a speed of 7.5 km/h. The average speed of the man over the interval of time 0 to 40 minute is equal to :
(A) 5 km/h 
(B) \frac{{25}}{4}{\text{km/h}}  
(C) \frac{{30}}{4}{\text{km/h}} 
(D) \frac{{45}}{8}{\text{km/h}} 
Solution

Ans. (D)
Sol. Average speed = \frac{{{\text{Total}}\,{\text{distance}}}}{{{\text{Total}}\,{\text{time}}}} = \frac{{2.5\,{\text{km}}\,{\text{ + }}\,\left( {{\text{7}}{\text{.5 × }}\frac{{{\text{10}}}}{{{\text{60}}}}} \right)\,km}}{{(40/60)}} = \frac{{45}}{8}\,{\text{km/hr}}{\text{.}}

20. A Cat runs 200 m away from a wall in a straight line in 12 sec. and then runs halfway back in two-third the time. Calculate its average speed
(A) 10m/s
(B) 15m/s
(C) 20m/s
(D) 5m/s
Solution

Ans. (B)
Sol. Average speed = \frac{{{\text{Total}}\,{\text{Distance}}}}{{{\text{Total}}\,{\text{Time}}}} = \frac{{200 + 100}}{{12 + 8}} = 15\,{\text{m/s}}

21. Motion represented in the following graph is :
(A) uniform Motion represented in the following graph is
(B) Accelerated
(C) Retarded
(D) None of these
Solution

Ans. (B)
Sol. Slope of disp-time graph gives velocity, as shown in fig. Velocity is increasing so object is in accelerated motion.

22. The variation of velocity of a particle moving along a straight line is shown in the figure. The distance travelled by the particle in 4s is: The variation of velocity of a particle moving along a straight line is shown in the figure
(A) 25m
(B) 30m
(C) 55m
(D) 60m
Solution

Ans. (C)
Sol. Area under V-t graph gives displacement
dispplacement = \frac{1}{2}(1)\,(20) + 20 \times 1 + \{ 10 \times 1 + \frac{1}{2}(1)(10)\} + 10 \times 1 
= 10 + 20 + 15 + 10 = 55 m

23. For the velocity time graph given below : For the velocity time graph given

(A) Displacement in 20s is 75 m
(B) Distance travelled in 20s is 125 m
(C) The body has taken only one turn
(D) All the above
Solution

Ans. (D)
Sol. Area under V-t graph gives displacement so,
S = \frac{1}{2}(15 + 5)\,(10) - \frac{1}{2}\,(5)\,(10) = 75\,{\text{m}} 
D = \frac{1}{2}(15 + 5)\,(10) + \frac{1}{2}(5)\,(10) = 125\,{\text{m}} 
and veloctiy becomes zero at t = 15s and body continuoues to move so body has taken 1 turn.

24. Between two stations a train accelerates uniformly at first, then moves with constant speed and finally retards uniformly. If the ratios of time taken are 1 : 8 : 1 and the greatest speed is 60 km/hr, find the average speed over the whole journey.
(A) 60 km/hr
(B) 54 km/hr
(C) 30 km/hr
(D) 20 km/hr
Solution

Ans. (B) ind the average speed over the whole journey
Sol. Average speed = \frac{{{\text{Total}}\,{\text{distance}}}}{{{\text{Total}}\,{\text{time}}}} 
= \frac{{\frac{{\text{1}}}{{\text{2}}}(10t + 8t)\,(60)}}{{10t}} = 54\,km/hr

 

 

25. Velocity time graph AB shows that the body has :

(A) A uniform acceleration Velocity time graph AB shows that the body has
(B) A uniform retardation
(C) Uniform speed
(D) Initial velocity OA and moving with uniform retardation
Solution

Ans. (D)
Sol. Slope of velocity-time graph gives acceleration, so in the given graph at t = 0 object has some velocity and after that it moves with uniform retardation.

26. An object is dropped from rest its velocity (v) changes with distance (s) in following form

(A) An object is dropped from rest its velocity

(B) An object is dropped from rest its velocity
(C) An object is dropped from rest its velocity

(D) An object is dropped from rest its velocity
Solution

Ans. (C)
Sol. When an object is dropped from rest it is undergoing constant acceleration
v2 = 0 + 2gS
v \propto \,\sqrt S , so graph would be parabola symmetric to s-axis.

27. The speed of a particle in distance–time graph given below is :

(A) 1/4 m/s The speed of a particle in distance–time graph given
(B) 4 m/s
(C) 10 m/s
(D) zero
Solution

Ans. (B)
Sol. Slope of distance time graph gives speed
speed = \frac{{20}}{5} = 4\,m/s

28. The train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. The total distance moved by the train will be :
(A) 2.9 km
(B) 2 km
(C) 3 km
(D) 2.7 km
Solution

Ans. (D) train starts from rest and moves with a constant acceleration
Sol. distance = \frac{1}{2}(90)\,(60) = 2.7\,{\text{km}}

 

 

 

 

29. The displacement time graph for two particles A and B are straignt lines inclined at angles of 30° and 60° with time axis. The ratio of velocities of VA : VB is : [ tan 30° \frac{1}{{\sqrt 3 }}= , tan 60° = \sqrt 3 ] 
(A) 1 : 2 
(B) 1:\sqrt[{}]{3} 
(C) \sqrt 3 \,:\,1 
(D) 1 : 3
Solution

Ans. (D) 
Sol. VA = tan 30° = \frac{1}{{\sqrt {\text{3}} }} 
VB = tan 60° = \sqrt 3  
\Rightarrow \frac{{{V_A}}}{{{V_B}}} = \frac{1}{3}

30. The area shown in Figure represents : The area shown in Figure represents

(A) momentum
(B) acceleration
(C) displacement
(D) speed
Solution

Ans. (C)
Sol. Area under V-T graph gives displacement

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