All the above statements are correct

Velocity is decreasing with time

Beyond M, the body has negative velocity

constant velocity and a varying speed

zero speed and non-zero acceleration

non-zero speed and zero acceleration

constant speed and a varying velocity

MCQ Questions for Class 9 Physics Motion Quiz 7

Acceleration is defined as

0%
the rate of change of velocity with respect to time.
0%
the rate of change of position with respect to time.
0%
the rate of change of momentum with respect to time.
0%
None of the above

Explanation

Acceleration is defined as the rate of change of velocity with respect to time.

$acceleration = \dfrac{change\ in\ velocity}{time}$

The rate of change of velocity of an object with respect to time is called ..........

0%
Momentum
0%
Displacement
0%
Acceleration
0%
Impulse

Explanation

Acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

$\text{acceleration} = \dfrac{\text{Change in the velociity}}{\text{time}}$

What does the graph represents:

0%
uniform speed
0%
non-uniform speed
0%
both A and B
0%
none of these

Explanation

$\textbf{Step 1-Equation of distance and time}$

Relationship between distance and time is linear. So, its equation will be in form of:

$D = k t$ , where k is any constant, D is distance and t is time.

$\textbf{Step 2-Calculation of velocity}$

On differentiating this equation with respect to time, at left hand side it will be speed as differentiation of distance with respect to time ..

$V = k$

So, speed for the given graph is constant or uniform.

Answer:

Hence, option A is the correct answer.

A person walks along an east-west street and a graph of his displacement from home is shown in figure. His average velocity for the whole time interval is

0%
$0\;ms^{-1}$
0%
$23\;ms^{-1}$
0%
$8.4\;ms^{-1}$
0%
None of above

Explanation

Net displacement of the particle is zero i.e

$S = 0$ over the whole time.

$\therefore$ Average velocity

$v_{avg} = \dfrac{S}{t} = 0$

The cheetah is the fastest land animal that can attain a maximum speed of

$112~ km/h$ . Express this speed in

$m/s$ .

0%
$31 ~m/s$ (approx.)
0%
$30 ~m/s$ (approx.)
0%
$32 ~m/s$ (approx.)
0%
$29 ~m/s$ (approx.)

Explanation

Given,

The maximum speed by the cheetah,

$v=112\ km/h$

We know,

$1\ km=1000\ m$

$1\ h=60\ min$

$1\min=60\ s$

$\therefore 1\ h=60\times 60=3600\ s$

$v=\dfrac{112\times 1000}{3600}$

$v=31.1\ m/s$

What distance was covered by the bus at 9.45 a.m.?

0%
70 m
0%
80 m
0%
90
0%
Can't be determined

Figure shows the position of a particle moving on the x-axis as a function time

0%
the particle has come to rest 6 times
0%
the maximum speed is at t=6 s
0%
the velocity remains positive for t=0 to t=6 s
0%
the average velocity for the total period show in negative

A bullet loses

$\dfrac{1}{20}$ of its velocity in passing through a plank. What is the least number of planks required to stop the bullet?

0%
9
0%
10
0%
11
0%
12

Explanation

let the width of a plank be d and no of planks be n

then final speed after one plank is

$v = u - \dfrac{u}{20} = \dfrac{19u}{20}$

$v^2 = u^2 +2aS$

Velocity decreasing so acceleration is negative.

$(\dfrac{19u}{20})^2 = u^2 -2a\times d$

$a=\dfrac{39u^2}{800d}$

number of planks required to stop the bullet

$V^2 = U^2 + 2aS$

$0 = u^2 - 2\times \dfrac{39u^2}{800d} \times nd$

$n = \dfrac{400}{39} = 10.25$

$\therefore \text{ There are } 11$ full planks.

If an object is moving eastward and slowing down, then the direction of its velocity vector is

0%
Eastward
0%
Westward
0%
Neither eastward nor westward
0%
Not enough information to tell

A student determined to test the law of gravity for himself walks off a sky scraper 320 m high with a stopwatch in hand and starts his free fall (zero initial velocity). 5 second later, superman arrives at the scene and dives off the roof to save the student. What must be the maximum height of the skyscraper so that even superman cannot save him.

0%
65 m
0%
85 m
0%
125 m
0%
145 m

Explanation

Assuming superman can move at infinite speed, he would be able to save the student unless he has already fallen onto the pavement in 5 seconds.
Maximum height of the skyscraper so that the student falls on to the pavement in 5 seconds is:

$h = \dfrac{1}{2} gt^{2} = \dfrac{1}{2} (10) (25) = 125m$
Hence, option C is correct.

If a train travelling at

$72$ km/h is to be brought to rest at a distance of

$200$ m then its retardation should be :

0%
$20\ m/s^{2}$
0%
$2 \ s^{2}$
0%
$10 \ m/s^{2}$
0%
$1 \ m/s^{2}$

Explanation

$\displaystyle u=72\frac{km}{h}=72\times\frac{5}{18}=20\;m/s$

Now

$v^{2}=u^{2}+2as$

$0=(20)^{2}+2a(200)$

$0=400+400a$

$400a=-400$

retardation

$=-a=1\;m/s^{2}$

Figure shows the position time graph of a particle moving on the x-axis.

0%
The particle is continuously going in positive x direction
0%
Area under x - t curve shows the displacement of particle
0%
The velocity increases up to a time $t_0$ , and then becomes constant.
0%
The particle moves at a constant velocity up to a time $t_0$ , and then stops.

Explanation

Until time

$t_{0}$ the graph shows a constant slope (which is non zero), which tells us that it is moving with constant velocity.
After time

$t_{0}$ the position doesn't change with time. Hence the body stops.
Option D is correct.

Which of the following four statement is false?

0%
A body can have zero velocity and still be accelerated.
0%
A body can have a constant velocity and still have a varying speed.
0%
A body can have a constant speed and still have a varying velocity.
0%
The direction of the velocity of a body can change when its acceleration is constant.

Which one of the following is travelling with the fastest speed?

0%
An athlete running with a speed of 10 m/s
0%
A bicycle moving with a speed of 200 m/min
0%
A tortoise covering 100 metres in 15 minutes
0%
A motorcycle moving with a speed one km/h

Explanation

To get the correct answer,we need to convert each speed in same unit i.e. m/s
A)An athlete running with a speed of 10 m/s
B)A bicycle moving with a speed of 200 m/min it can be written as

$=\frac{200}{60}m/s$

$=3.33m/s$
C)A tortoise covering 100 metres in 15 minutes

$=\frac{100}{15}m/min$

$=\frac{100}{{15}\times{60}}m/s$

$=0.11m/s$
D)A motorcycle moving with a speed one km/h

$=1km/h$

$=\frac{1000}{{1}\times{60}\times{60}}$

$=0.27m/s$
Comparing A, B, C, and D one can easily say that answer is option A i.e.
An athlete running with a speed of 10 m/s is the fastest.

A stone is dropped from a height of 20 m. How long will it take to reach the ground? (

${g=10 m/s^2}$ ).

0%
2 s
0%
0.5 s
0%
4 s
0%
1 s

Explanation

Given,

$u=0m/s$

$g=10m/s^2$

$h=20m$

3rd equation of motion,

$2gh=v^2-u^2$

$2\times 10\times 20=v^2-0^2$

$v^2=400$

$v=20m/s$

1st equation of motion,

$v=u+gt$

$20=0+10t$

$t=\dfrac{20}{10}=2sec$

The correct option is A.

A truck increases its speed from

$10km/h$ to

$50 km/h$ in

$20$ seconds. Its acceleration is:

0%
$0.55 m/s^2$
0%
$2.55 m/s^2$
0%
$0.75 m/s^2$
0%
$8.65 m/s^2$

Explanation

Given: Initial speed,

$u=10\ km/h=10\times \dfrac{5}{18}m/s$

Final speed,

$v=50\ km/h=50\times \dfrac{5}{18}m/s$

Using equation of motion,

Acceleration,

$a=\dfrac{v-u}{t}$

$=\dfrac{\dfrac{5}{18}\times(50-10)}{20}$

$=\dfrac{5}{18}\times 2$

$=0.55\ m/s^2$

A and B are thrown vertically upward with velocity, 5 m/s and 10 m/s respectively

$(g = 10\ m/s^{2})$ . Find separation between them after one second.

0%
5 m
0%
12 m
0%
3 m
0%
7 m

Explanation

$s_A = ut - (1/2)gt^2$

$= 5t - \displaystyle \frac{1}{2} \times 10 \times t^{2}$

$= 5 \times 1 - 5 \times 1^{2}$

$= 5 - 5 = 0\ m$

$s_B = ut - \displaystyle \frac{1}{2} gt^{2}$

$= 10 \times 1 - \displaystyle \frac{1}{2} \times 10 \times 1^{2}$

$= 10 - 5 = 5\ m$
Therefore,

$s_B - s_A =\, separation\, = 5m$

A moving bus is brought to rest within 30 seconds by applying brakes. What is its initial velocity, if the retardation due to brakes is 2

$ms^{-2}$ ?

0%
$60\ ms^{-1}$
0%
$15\ ms^{-1}$
0%
$40\ ms^{-1}$
0%
$30\ ms^{-1}$

Explanation

Given:

Time,

$t=30\ sec$

Retardation,

$a=-2\ m/s^2$

Final velovity,

$v=0$

From equation of motion

$\Rightarrow v=u+at$

$\Rightarrow 0=u-2\times 30$

$\Rightarrow u=60 \ m/s$

A car is moving with a uniform velocity of 40 km h

$^{-1}$ in straight line. Its acceleration after 1 hour is:

0%
40 km h $^{-1}$
0%
20 km h $^{-1}$
0%
30 km h $^{-1}$
0%
zero

A ball starts rolling on a horizontal surface with an initial velocity of

$1 \ m/s.$ Due to friction, its velocity decreases at the rate of

${0.1 m/s^2}$ . How much time will it take for the ball to stop?

0%
1 s
0%
100 s
0%
10 s
0%
0.1 s

Explanation

In the given question, we have

$u= \ 1 \ m/s$

$a = - 0.1 \ m/s^2$ (negative sign indicates that the direction of acceleration is opposite to the initial velocity)
As the ball finally stops,

$v=0$

$t=?$
According to first equation of motion,

$v = u + at$

$\Rightarrow 0 = 1 + (-0.1)t$

$\Rightarrow 1 = 0.1 t$
So

$t=10 \ sec$

Option C is correct.

A car falls off a bridge and drops to the ground in

$0.5 s$ . Let

${g=10m/s^2}$ . How high is the bridge from the ground?

0%
$1.25 m$
0%
$2.5 m$
0%
$5 m$
0%
$10 m$

Explanation

Answer is A.

In this case, the distance traveled by the car off to the ground is equal to the total height of the bridge.
So, the total distance traveled by the falling car to the ground is calculated as follows.

$s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$ where,

$s$ - distance covered

$u$ - initial velocity which is 0 in this case

$a$ - acceleration due to gravity which is

$10m/{ s }^{ 2 }$

$t$ - 0.5 seconds

Substituting these values in the given equation we get,

$s=0\times 0.5+\dfrac { 1 }{ 2 } \times 10\times 0.5^{ 2 }\quad =\quad 1.25m.$

Hence, the height of the bridge is 1.25 m.

A boy running at an average speed of

$4 km/ hr$ reaches school from his home in

$\dfrac{1}{2}\ hr$ . The distance of the school from his home is

0%
$2 km$
0%
$8 km$
0%
$4 km$
0%
$6 km$

Explanation

Given:

Average speed of the boy

$v =4\ km/h$
Time taken

$t =\dfrac{1}{2}$ hour

Distance

$D=?$

We know that

$Speed(v)=\dfrac{Distance(D)}{Time(t)}$

$\Rightarrow Distance(D)=4\times\dfrac{1}{2}$

$\Rightarrow Distance(D)=2\ km$

When a bus driver travelling at a speed of 20m/s applied brakes and brings the bus to rest in 10 seconds, then retardation will be

0%
${+ 2 m/s^2}$
0%
${+ 0.5 m/s^2}$
0%
${-2 m/s^2}$
0%
${-0.5 m/s^2}$

$A$ and

$B$ are arguing about uniform acceleration.

$A$ states that acceleration means "the longer you go."

$B$ states that acceleration means "the further you go." Who is right?

0%
$A$
0%
$B$
0%
Both $A$ and $B$
0%
None of these

A driver is driving his car along a road as shown in Figure. The driver makes sure that the speed-o-meter reads exactly 40 km/h. What happens to the velocity of the car from P to Q ?

0%
Velocity remains constant
0%
Velocity first increases then decreases
0%
Velocity first decreases then increases
0%
Nothing can be decided

In Fig. 2.8

0%
Retardation is uniform
0%
Velocity is decreasing with time
0%
Beyond M, the body has negative velocity
0%
All the above statements are correct

A body cannot have a

0%
zero speed and non-zero acceleration
0%
non-zero speed and zero acceleration
0%
constant velocity and a varying speed
0%
constant speed and a varying velocity

Which of the following decreases in motion along a straight line with constant retardation?

0%
speed
0%
acceleration
0%
displacement
0%
none of these

Explanation

Option A:

By first equation of motion,

$v=u-at$

Since it is retardation, then

$a \rightarrow negative$

Let us assume the moving body has some initial speed

$u$ in

$+x$ axis.

So we can see, as time

$t$ increases, final speed

$v$ decreases.

So option A is correct.

Option B:

Acceleration can't decrease because it is given in the question itself that it is constant retardation, i.e, constant acceleration opposite in direction to the speed.

Option C:

Displacement will also keep on increasing towards

$+x$ axis as long as final speed

$v$ doesn't become zero.

The slope of a distance-time graph of a moving body gives its

0%
Speed
0%
Displacement
0%
Velocity
0%
Acceleration

Explanation

The speed can be determined from the slope of the Distance-time graph.
Slope =

$=\frac{rise}{run} =\frac{distance}{time}$

A van is moving on the road at

${8ms^{-1}}$ .It accelerates at

${2ms^{-2}}$ for a distance of 9m. How fast is it moving now?

0%
${36 ms^{-1}}$
0%
${16 ms ^{-1}}$
0%
${18 ms^{-1}}$
0%
${10 ms ^{-1}}$

Explanation

Initial speed

$u \space = \space 8 \space m/s$ .

Acceleration

$a \space = \space 2 \space m/{s}^{2}$ .

Distance covered while accelerating

$s \space = \space 9 \space m$ .

Let the final speed of the van be

$v$ .

Using the fundamental kinetics equation,

${v}^{2} \space = \space {u}^{2} \space + \space 2as$ .

${v}^{2} \space = 64 + 2\times2\times9$ .

We get

${v}^{2} \space = 64 + 36$ .

${v}^{2} \space = 100$ .

Thus

$v \space = \space 10 \space m/s$ .

A stone tied to a string is whirled in a circle. As it is revolving, the rope suddenly breaks.Then :

0%
The stone flies off tangential to the circle of rotation
0%
The stone moves radially inward
0%
The stone moves radially outward
0%
The motion of the stone depends upon its velocity

A body whose speed in a particular direction is constant :

0%
must be accelerating
0%
must be retarding
0%
has a constant velocity in that direction
0%
all the above

The displacement of a body starting from rest, and moving with an acceleration of

$1 \ { ms }^{ -2 }$ at the end of 5 s is ____ m.

0%
12.5
0%
25
0%
7.5
0%
15

Explanation

Given :

$a = 1 \ m/s^2$

$t= 5 \ s$
Initial speed of the body

$u = 0 \ m/s$
Displacement of the body can be calculated using 2nd equation of motion

$S= ut+\dfrac{1}{2}at^2$

$\Rightarrow S = 0+\dfrac{1}{2}\times 1\times 5^2 = 12.5 \ m$

Choose the incorrect statement:

0%
the speedometer of a car measures its instantaneous speed
0%
the velocity of a body is always greater than the speed of that body.
0%
the position-time graph of a body moving with variable velocity is a curve.
0%
velocity-time graph of a uniform motion is a straight line parallel to time-axis.

A driver has to apply sudden brakes to stop his vehicle moving at

$72 \ {km}/{hr}$ and stops within

$2 \ seconds$ . The retardation produced is then

0%
$20 \ {m}/{{s}^{2}}$
0%
$10 \ {m}/{{s}^{2}}$
0%
$15 \ {m}/{{s}^{2}}$
0%
$18 \ {m}/{{s}^{2}}$

Explanation

Using the first equation of motion,

$v = u + at$
Here,

$v = 0$

$u = 72 \times \dfrac{1000}{3600} = 20 \ m/s$

$t = 2s$

Hence,

$0 = 20 + a\times 2$

$a = -\displaystyle\frac{20}{2} = -10 \ {m}/{{s}^{2}}$

Since, sign of

$a$ is negative. Hence, retardation with magnitude

$10 \ m/s^2$ is produced.

If a trolley starts from rest with an acceleration of 2

$m/s^2$ , the velocity of the body after 4 s would be

0%
2 m/s
0%
8 m/s
0%
4 m/s
0%
6 m/s

Explanation

Given:

initial velocity

$u=0$ (Since the body is starting from rest)

acceleration

$a=2\ m/s^2$

time taken

$t=4\ s$

final velocity

$v=?$

According to the first equation of motion

$v = u + at$

$v = 0 +2\times4 = 8\ m/s$

The SI unit for the resultant velocity is

0%
m/s
0%
km/s
0%
cm/s
0%
min/s

Explanation

velocity is given by

$\dfrac{displacement}{time}$

SI unit of displacement is meter and that of time is sec.

so SI unit of velocity is meter/sec.

A body has an acceleration of

$\displaystyle -4\ { ms }^{ -2 }$ . Then its retardation is equal to:

0%
$-4\ { ms }^{ 2 }$
0%
$4\ { ms }^{ -2 }$
0%
Zero
0%
Can't be determined

Explanation

The change in the velocity of an object per unit time is known acceleration.

Retardation is negative acceleration.

Given,

Acceleration,

$a=-4\ m/s^2$

Retardation

$=-a=-(-4) =4\ m/s^2$

A driver is driving his car along a road is shown in Fig. The driver makes sure that the speedometer reads exactly 40 km/h. What happens to the velocity of the car from P to Q?

0%
Velocity remains constant
0%
Velocity first increases then decreases
0%
Velocity first decreases then increases
0%
Nothing can be decided

A body is thrown up with an initial velocity

$u$ and covers a maximum height of

$h$ , then

$h$ is equal to

0%
$\displaystyle \frac{u^2}{2g}$
0%
$\displaystyle \frac{u}{2g}$
0%
2ug
0%
none of these

Explanation

Relation between Displacement and the Velocity is

${ v}^{ 2 }={u}^{ 2 }+2as$

When we throw particle upward final velocity of it becomes zero because of retardation by gravity.

Here,

$s=h, a=-g$

Now our relation becomes:

$0={ { u } }^{ 2 }-2gh$

$h=\dfrac { {u}^{ 2 } }{ 2g }$

Option A is correct.

Figure shows the displacement-time graphs (a) and (b) for a body moving in a straight path drawn on the same scales. Then

0%
slope of line in (a) is greater than the slope of line in (b)
0%
slope of line in (b) is greater than the slope of line in (a)
0%
slope of line in (a) is equal to the slope of line in (b)
0%
nothing can be said about the slopes

Explanation

slope of any graph is given by

$dy/dx=Tan\theta$

means change of function y with respect to function x.

$\theta$ is angle made by the graph with x axis.

as angle for a is greater then that of b so as

$Tan\theta$ is increasing between 0 to 90, so

$Tan\theta _a$ >

$Tan\theta _b$

so slope of a is greater then that of b.

beat possible answer is optionA.

Figure shows the displacement-time graphs (a) and (b) for a body moving in a straight path drawn on the same scales. Then the velocity in case of (b) is

0%
more than the velocity in case of (a)
0%
less than the velocity in case of (a)
0%
equal to the velocity in case of (a)
0%
square of the velocity in case of (a)

What does the speed of a body describes about its motion ?

0%
Direction
0%
State
0%
Type
0%
Rapidity

Explanation

$speed =\frac{distance}{time}$ Speed tells us how fast or slow the distance is being covered by the moving body .Hence it denotes it's rapidity(Means how rapid the moving object is).

The SI unit of speed is:

0%
$m/s$
0%
$km/s$
0%
$cm/s$
0%
None of these

Explanation

The distance moved by an object in a unit time is known as speed.

The SI unit of speed is

$m/s$ .

The velocity of a particle increases from u to v in a time t during which it covers a distances S. If the particle has a uniform acceleration, which one of the following equations does not apply to the motion?

0%
2 S = (v + u) t
0%
$a = \displaystyle \frac{v - u}{t}$
0%
$v^2 = u^2 - 2 aS$
0%
$S= \displaystyle \left[ u + \frac{1}{2} at \right] t$

Explanation

$\dfrac{dv}{dt}=a$

by integrating once we get

$v-u=at$ i.e. option B.

also

$\dfrac{ds}{dt}=v$ so integrating once more we get

$S=ut+\frac{1}{2}at^2$ i.e. option D

by using

$v=u+at$ and $$ expression of S can be converted to A.

now eleminating t from equation in option A and option B we will get

and expression

$v^2-u^2=2as$ but in option C where

$v^2-u^2=-2aS$ so expression in option C is not correct so answer is option C.

A stone projected vertically upwards reaches to the level of window

$10\ m$ from the ground. Find the magnitude of the velocity of the stone at the time of its projection.

0%
$\displaystyle { 7\ ms }^{ -1 }$
0%
$\displaystyle { 36\ ms }^{ -1 }$
0%
$\displaystyle { 28\ ms }^{ -1 }$
0%
$\displaystyle { 14\ ms }^{ -1 }$

Explanation

Height attained by the stone thrown vertically upwards

$h = 10 \ m$

Velocity of projection

$u = \sqrt{2gh}$

$u$ : Initial velocity

$g$ : Acceleration due to gravity

$h$ : Height attained by the object

$\Rightarrow \ u = \sqrt{2\times 10\times 9.8}$

$\Rightarrow u = 14\ m/s$

An automobile travelling with the speed of

$\displaystyle 72\ km {h}^{-1}$ can be stopped within a distance of

$20\ m$ , by applying brakes. Determine the retardation.

0%
$\displaystyle -100 {ms} ^{-2}$
0%
$\displaystyle -10 {ms} ^{-2}$
0%
$\displaystyle 5 {ms} ^{-2}$
0%
$\displaystyle -5 {ms} ^{-2}$

Explanation

Initial velocity of automobile

$u = 72 \ km/h = 72\times \dfrac{5}{18} = 20 \ m/s$

Final velocity

$v = 0 \ m/s$

Distance covered to stop

$S = 20 \ m$

Using

$v^2 - u^2 = 2aS$

$\therefore$

$0 - 20^2 = 2\times a\times 20$

$\implies \ a = -10 \ m/s^2$

In the above shown figure, the velocity

0%
increases between point O and A
0%
increases between point A and B
0%
decreases between points A and B
0%
is zero throughout

The ratio of magnitude of velocity of projection in the upward motion to the magnitude of velocity of the body on reaching the point of projection in its downward motion is_____ .

0%
9 : 1
0%
1 : 1
0%
2 : 1
0%
4 :1

Explanation

At the same horizontal level, the velocities of an object thrown up have equal magnitude but are opposite in direction. Thus the ratio is

$1:1$

Let us assume an object was thrown up with a velocity

$u_1$ and it reaches a height

$h.$ Its final velocity

$v_1=0.$

Using third equation of motion,

$v_1^2-u_1^2=2(-g)h$

$u_1^2=2gh \Rightarrow u_1=\sqrt{2gh}$ upwards

Now it comes down again through the same height

$h$ . Let us assume its final velocity is

$v_2.$ We know its initial velocity for the descend will be

$u_2=0$

$v_2^2-u_2^2=2(-g)(-h)$

$v_2^2=2gh \Rightarrow v_2= \sqrt{2gh}$ downwards

So we see that the magnitude of

$u_1$ and

$v_2$ is same.

Tripling the speed of a motor car multiplies the distance needed for stopping it by

0%
3
0%
6
0%
9
0%
some other number

Explanation

Hint: Use kinematics equation of motion

Step1:Relation between velocity and distance is

$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{aS}$

If the final velocity is Zero then the relation will become

$\mathrm{u}=\sqrt{2 \mathrm{a} \mathrm{S}}$

Now if we want to multiply the velocity $\mathbf{3}$ time then we have to multiply the distance $\boldsymbol{9}$ time to make the equation balanced.

$\textbf{Hence option C correct}$

CBSE Questions for Class 9 Physics Motion Quiz 7 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Motion Quiz 7 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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