Motion(9th Grade > Physics ) Questions and answers for exam Preparation

9th Grade > Physics

MOTION MCQs

Total Questions : 58 | Page 1 of 6 pages

Question 1. An auto travels the first half time with a uniform speed

u

and the next half time with a uniform speed

v

. Find its average speed.

u+v2
uv2
2uv
uv

Discuss Question

Answer: Option A. -> u+v2
:
A

Given, speed of the auto during the first half of the journey is

u

and the next half of the journey is

v

Let total time taken for the journey be

T

∴

Time taken for the first half of the journey be

\frac{T}{2}

and the next half of the journey be

\frac{T}{2}

Let the average speed be

V_{a v}

We know,

s p e e d = \frac{D i s t a n c e}{t i m e}

∴

D i s t a n c e = s p e e d \times t i m e

During the first half of the journey, distance,

d_{1} = u \times \frac{T}{2}

During the second half of the journey, distance,

d_{2} = v \times \frac{T}{2}

Total distance

= d = d_{1} + d_{2}

= u \times

\frac{T}{2} + v \times \frac{T}{2}

\Rightarrow d = \frac{T}{2} (u + v)

Average speed,

V_{a v}

\frac{T o t a l d i s t a n c e}{T o t a l t i m e t a k e n}

= \frac{d}{T}

= \frac{\frac{T}{2} (u + v)}{T}

= \frac{u + v}{2}

Question 2. The velocity - time graph of a body with a constant acceleration is:

Straight line parallel to velocity axis
Parabola
Hyperbola
Straight line inclined at some angle with respect to time axis

Discuss Question

Answer: Option D. -> Straight line inclined at some angle with respect to time axis
:
D

The Velocity - Time Graph Of A Body With A Constant Accelera...

Since acceleration is constant, the slope of thevelocity-time graph should be a straight line withconstant slope. Thus, the graph whichshows a straight line inclined at some anglewith respect to time axisrepresents the velocity-time graph of a body with a constant acceleration.

Question 3. An artificial satellite is moving in a circular orbit of radius

4200 k m

around the earth. What is its speed if it takes one day to complete one revolution?

4099 kmph
3099 kmph
1099 kmph
2099 kmph

Discuss Question

Answer: Option C. -> 1099 kmph
:
C
Given, the radius of the orbit,

R = 4200 k m

The time taken to revolve around the Earth,

T = 24 h r (1 d a y)

The speed of the satellite,

v = \frac{Circumference of the orbit}{Time taken}

= \frac{2 π R}{T}

= \frac{2 \times 3.14 \times 4200 km}{24 hr}

= 1099 km {hr}^{- 1}

= 1099 kmph

Question 4. A bus decreases its speed from

80 k m p h

60 k m p h

in 5 seconds. Find the acceleration of the bus.

2.22 ms−2
2.11 ms−2
1.11 ms−2
0

Discuss Question

Answer: Option C. -> 1.11 ms−2
:
C
Initial speed,

u = 80 k m p h

80 \times \frac{1000}{60 \times 60} = \frac{800}{36} m s^{- 1}

Final speed,

v = 60 k m p h

60 \times \frac{1000}{60 \times 60} = \frac{600}{36} m s^{- 1}

Time taken

t = 5 s

From the first equation of motion,

a = \frac{v - u}{t}

= \frac{600 - 800}{36 \times 5}

= - 1.11 m s^{- 2}

.
Here sign of acceleration is negative which shows that the speed is decreasing and acceleration is in the opposite direction of the motion.

Question 5. The area below velocity – time graph gives:

acceleration
displacement
average speed
average velocity

Discuss Question

Answer: Option B. -> displacement
:
B
The area under a velocity-time graph gives the displacement of the object.
The area under the curve is velocity x time = displacement.

∵ V e l o c i t y = \frac{Displacement}{Time}

In the graph, the area under the curve is

d i s p l a c e m e n t

= Area of triangle ADE + Area of rectangle ADBC

The Area Below Velocity – Time Graph Gives:

Question 6. In a uniform circular motion, which of the following is true?

Velocity is constant and speed is changing.
Velocity and speed both remain constant.
Speed remains constant and velocity changes continuously.
Neither velocity nor speed remains constant.

Discuss Question

Answer: Option C. -> Speed remains constant and velocity changes continuously.
:
C
In a uniform circular motion, the speedremains constant. However, the direction of the body keeps changing. This is due toan acceleration that is perpendicular to the direction of velocity at every point.Hence,the velocity does not remain constant.

Question 7. A car covers a distance of 300 m in 20 seconds. Calculate the speed of the car.

25 ms−1
20 ms−1
10 ms−1
15 ms−1

Discuss Question

Answer: Option D. -> 15 ms−1
:
D
Given, distance travelled by the car

= 300 m

and the time taken

= 20 s

.
We know,

s p e e d = \frac{d i s t a n c e}{t i m e}

\Rightarrow s p e e d = \frac{300}{20}

15 m s^{- 1}

∴

Speed of car is

15 m s^{- 1}

Question 8. A coin is thrown in a vertically upward direction with a velocity of

5 m s^{- 1}

. If the acceleration of the coin during its motion is

10 m s^{- 2}

in the downward direction, what will be the height attained by the coin and how much time will it take to reach there ?

s=1.5 m, t=0.25 s
s=0.25 m, t=1.5 s
s=1.25 m, t=0.75 s
s=1.25 m, t=0.5 s

Discuss Question

Answer: Option D. -> s=1.25 m, t=0.5 s
:
D
Given, initial velocity,

u = 5 m s^{- 1}

, acceleration,

a = - 10 m s^{- 2}

.
Let '

v

' be the final velocity and '

s

' be the height attained.
At the highest point of motion,

v = 0

Using third equation of motion,

v^{2} = u^{2} + 2 a s,

0 = 5^{2} + [2 \times (- 10) \times s]

\Rightarrow s = 1.25 m

Hence the height attained by the coin is

1.25 m

.
Let the time taken be

t

.
Using the first equationof motion,

v = u + a t

0 = 5 - 10 t

\Rightarrow t = 0.5 s

∴

Time taken by the coin is

0.5 s

Question 9. A

60 m

long train moving on a straight level track passes a pole in

5 s

. Find the speed of the train and the time it will take to cross a

540 m

long bridge.

13 ms−1, 40 s
25 ms−1, 30 s
12 ms−1, 50 s
15 ms−1, 65 s

Discuss Question

Answer: Option C. -> 12 ms−1, 50 s
:
C
Given:
Length of the train,

D = 60 m

,
Time taken to cross the pole,

T = 5 s

,
Length of the bridge

= 540 m

Speed of the train, s = \frac{D}{T}

Speed of the train, s = \frac{60}{5} = 12 m s^{- 1}

Now, distance to be covered to cross a

540 m

long bridge is sum of Length of the train and Length of the bridge.

d = 60 + 540 = 600 m

∴ Time taken = \frac{d}{s}

Time taken = \frac{600}{12} = 50 s

∴

Speed of thetrain is

12 m s^{- 1}

and the time it will take to cross the bridge is

50 s

Question 10. Motion of a satellite in a circular orbit is an example of

non-uniform circular motion
uniform circular motion
uniform linear motion
non-uniform linear motion

Discuss Question

Answer: Option B. -> uniform circular motion
:
B
The motion of a satellite in a circular orbit is uniform circular motion.In uniform circular motion, a body moves in a circle with constant speed butthe velocity changes due to a constant change in direction.