MCQ Questions for Class 9 Physics Motion Quiz 10

Odometer is to mileage as compass is to

0%
speed
0%
hiking
0%
needle
0%
direction

A bullet when fired into a target loses half of its velocity after penetrating 20 cm. Further distance of penetration before it comes to rest is :

0%
6.66 cm
0%
3.33 cm
0%
12.5 cm
0%
10 cm
0%
5 cm

Explanation

Given, Initial velocity,

$U=v$
Final velocity,

$V=\frac {v}{2}$
Distance,

$s=20 cm$

Let, the further distance of penetration before it comes to rest be x.

$V^2 =U^2 -2as$

$\left ( \frac {V}{2} \right )^2=V^2 -2a\times 20$

$40 a=V^2-\frac {V^2}{4}$

$40 a=\frac {3V^2}{4}$ ...(i)

and

$V^2 =u^2 +2as$

$v^2=0+2a\times (20 +x)$

$v^2 = 2\times \frac {3}{160}v^2\times (20+x)$ [From Eq. (i)]

$1=\frac {3}{80}(20+x)$

$\frac {80}{3}=20 +x$

$x=\frac {80}{3}-20$

$x=\frac {80-60}{3}$

$x=\frac {20}{3}=6.66cm$

Two cars started moving with initial velocities

$u$ and

$2u$ . For the same deceleration, their respective stopping distances are in the ratio

0%
$1 : 1$
0%
$1 : 2$
0%
$1 : 4$
0%
$2 : 1$
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$4 : 1$

Explanation

$\text{For the first car}:$

v12=u2−2asFor first car$${ v }_{ 1 }^{ 2 }={ u }^{ 2 }-2as$$

Initial velocity =

$u$

Final velocity =

$0$

Deceleration =

$a$

Using the equation of motion:

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

$0=u-2a{ S }_{ 1 }$

$\Rightarrow { S }_{ 1 }=\dfrac { u }{ 2a }$

$\text{For the second car}:$

Initial velocity =

$2u$

Final velocity =

$0$

Deceleration =

$a$

Using the equation of motion:

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

$0={ \left( 2u \right) }^{ 2 }-2a{ S }_{ 2 }$

$\Rightarrow { S }_{ 2 }=\dfrac { 2u }{ a }$

$\therefore \dfrac { { S }_{ 1 } }{ { S }_{ 2 } } =\dfrac { \dfrac { u }{ 2a } }{ \dfrac { 2u }{ a } }$

$\therefore \dfrac { { S }_{ 1 } }{ { S }_{ 2 } } =\dfrac { 1 }{ 4 } = 1:4$

A body of mass

$2kg$ has an initial velocity of

$3m/s$ along

$OE$ and it is subjected to a force of

$4N$ along of direction perpendicular to

$OE$ as shown in figure. The distance travelled by the body in

$4s$ from

$O$ will be

0%
$12m$
0%
$20m$
0%
$28m$
0%
$48m$

Explanation

Let

$OE$ and

$OY$ be along

$X$ and

$Y$ axis respectively,
then

${ u }_{ x }=3m/s,{ a }_{ x }=0\quad$
and

${ u }_{ y }=0,{ a }_{ y }=\cfrac { 4 }{ 2 } =2m/{ s }^{ 2 }$
at

$t=4s$ , let

${S}_{x}$ and

${S}_{y}$ be the displacement along x-axis and y-axis respectively, then

${ S }_{ x }={ u }_{ x }=\cfrac { 1 }{ 2 } { a }_{ x }{ t }^{ 2 }=3\times 4+\cfrac { 1 }{ 2 } (0){ (4) }^{ 2 }=12m$
and

${ S }_{ y }={ u }_{ y }t+\cfrac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }=0\times 4+\cfrac { 1 }{ 2 } (2){ (4) }^{ 2 }=16m$
so, the resultant displacement

$S$ is given as

$S=\sqrt { { { S }_{ x } }^{ 2 }+{ { S }_{ y } }^{ 2 } } =20m$

Fill in the blank:

The _______ circular motion describes as the motion of an object in a circular path with a _______ speed.

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Non-uniform, constant
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Uniform, constant
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Non-uniform, varying
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Uniform, Varying

The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on the planet B. A man jumps to a height of 2m on the surface A. What is the height of jump by the same person on the plane B?

0%
6 m
0%
2/9 m
0%
2/3 m
0%
18 m

Explanation

It is given that acceleration due to gravity on plane. A is 9 times the acceleration due to gravity on planet B i.e.,

$g_A = 9g_B$ ...(i)

From third equation of motion

$v^2 = 2gh$ ...(ii)

At planet

$A, h_A = \dfrac {v_2}{2g_A}$ ...(ii)

At planet

$A, h_B = \dfrac {v_2}{2g_B}$ ...(iii)

Dividing Eq. (ii) by Eq. (iii) we have

$\dfrac {h_A}{h_B} = \dfrac {9g_B} { 9g_A}$

From Eq. (i),

$g_A = 9 g_B$

$\therefore \dfrac {h_A}{h_B} = \dfrac {9g_B} { 9g_A} = \dfrac {1}{9}$

$h_B = 9 h_B = 9 \times 2 = 18 m ( \therefore h_A = 2m )$

The velocity of the object over the interval (2,4) is :

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10 m/s
0%
5 m/s
0%
20 m/s
0%
15 m/s

Explanation

Velocity is defined as the rate of change of position with respect to time.

$v=\dfrac{(s_2-s_1)}{(t_2-t_1)}$

$v=\dfrac{(20-10)}{(4-2)}$

$v=\dfrac{10}{2}$

$v=5\ m/s$

2108104_672558_ans_fce24ef0d19446b38c80fb73dfd59ea0.png

Sheela takes 15 minutes from her house to reach school on her bicycle with a speed of 2 m/s. The distance between her house and the school is

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1.8 km
0%
3 km
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1 km
0%
2.5 km

An object is thrown vertically upward with a speed of

$30m/s$ . The velocity of the object half a second before it reaches the maximum height is

0%
$4.9m/s$
0%
$9.8m/s$
0%
$19.6m/s$
0%
$25.1m/s$

Explanation

Velocity half second before maximum height

$=$ Velocity half second after maximum height (Return journey)
For downward journey :
Initial velocity

$u = 0 \ m/s$
We have

$g = 9.8 \ m/s^2$
Time

$t = \dfrac{1}{2} \ s$
Thus velocity after half second

$v = u+gt$

$v=0+9.8\times \dfrac{1}{2}=4.9m/s$

For a moving body, at any instant of time.

0%
If the body is not moving, the acceleration is necessarily zero
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If the body is slowing, the acceleration is negative
0%
If the body is slowing, the distance is negative
0%
If displacement, velocity, and acceleration at that instant are known, we can find the displacement attained in a given time in the future.

Explanation

$S(t), v(t)$ and

$a$ are given,

$v^{2} - u^{2} = 2aS ... (1)$

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

$v = u + at .... (3)$
from

$(2), \dfrac {S}{t} = u + \dfrac {1}{2}ut$

$v = u + at$ , one can find out

$u$ .
Then, with the help of kinematics, one can find out the displacement attained in a given future time if the equations hold good for that period.
If the body is not moving, acceleration is not necessarily zero, when a particle is moving vertically, at maximum height, velocity is zero, but not acceleration. (a is not true).
(b) if the body is slowing, acceleration is negative.

A scooterist covers a distance of 3 km in 5 minutes. He travels with a speed of

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$100 m s^{-1}$
0%
$360 km h^{-1}$
0%
$100 cm s^{-1}$
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$600 m min^{-1}$

The average speed of an object is defined to be

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One half of the sum of the maximum and the minimum speeds
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Distance it travels multiplied by the time it takes
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The distance it travels divided by the time it takes
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The speed determined over an infinitesimally small time interval
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The value of the speed at the midpoint of the time interval

Explanation

The average speed of an object is defined as the distance traveled divided by the time

$v=\dfrac{D}{t}$

Where, $v$ = speed,

$D$ = distance and

$t$ = time

A car starts moving along a line, first with an acceleration

$a = 5\ ms^{-2}$ starting from rest, then uniformly and finally decelerating at the same rate, comes to rest in the total time of

$25\ seconds (t_{1})$ , then average velocity during the time is equal to

$v = 72\ kmph$ . How long does the particle move uniformly?

0%
$5\ seconds$
0%
$2.5\ hours$
0%
$1.5\ hours$
0%
$15\ seconds$

Explanation

Let the initial velocity $u=0$

The time taken and the distance traveled during accelerated motion and retarding motion will be same as the acceleration is same

For accelerated motion

$v=u+at=5t$

And for retardation

$0=v-5t$

$v=5t$

As the velocity v will become u for retardation

Total distance traveled

$=$ area of $v-t$

$=$ area $I+$ area $II+$ area $III$

$2\left( \cfrac { 1 }{ 2 } 5{ t }^{ 2 } \right) +5tx=$ distance traveled during motion

$x=$ time during which particle remains in uniform motion

$5{ t }^{ 2 }+5tx=$ speed $x$ time

$5{ t }^{ 2 }+5tx=20\times 25$

${ t }^{ 2 }+tx=100\rightarrow 1$

Also the total time is given $25$ seconds

Thus $2t+x=25$

Thus $x=25-2t\rightarrow 2$

Substituting $2$ in $1$

${ t }^{ 2 }+t(25-2t)=100$

${ t }^{ 2 }-25t+100=0$

$t=5$ or $t=20$

Thus the time will be $t=5$ sec as $t=20$ s is not possible as it will exceed the given time

The particle remains in motion for

$5$ seconds

Which of the following is not a unit of speed?

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Meter/second
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Second/meter
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Kilometer/hour
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Meter/minute.

In the following displacement (x) vs time (t) graph, at which among the points P, Q and R is the object's speed increasing?

0%
P only
0%
R only
0%
Q and R only
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P, Q, R

A car travels from rest with a constant acceleration

$'a'$ for

$t$ seconds. What is the average speed of the car for its journey, if the car moves along a straight road ?

0%
$v=\dfrac{at^2}{2}$
0%
$v=2at^2$
0%
$v=\dfrac{at}{2}$
0%
None

Explanation

Initial speed of car

$u = 0$
Distance covered by car in

$t$ seconds is given by second equation of motion,

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

$\implies \ S = \dfrac{1}{2}at^2$

$(\because u=0)$

Average speed of car =

$\dfrac{ \text{total distance}}{\text{total time} }$

$v_{avg} = \dfrac{S}{t} = \dfrac{\dfrac{1}{2}at^2}{t}=\dfrac{at}{2}$

So option C is correct.

In uniform circular motion, direction of velocity goes on changing continuously, however the magnitude of velocity is constant.

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True
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False

In uniform circular motion, direction of velocity is along the _______ drawn to the position of particle on the circumference of the circle.

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normal
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tangent
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can be both
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none of these

The figure shows the

$x-t$ plot of a particle in one-dimensional motion. Two different equal intervals of time are shown. Let

$v_1$ and

$v_2$ be speed in time intervals

$1$ and

$2$ respectively. Then

0%
$v_1 > v_2$
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$v_2 > v_1$
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$v_1 = v_2$
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Data is insufficient

Explanation

We know,

The slope of the $x-t$ graph represents the speed. The average speed of a particle shown in the $x-t$ graph is obtained from its slope in a particular interval of time. So, the higher the slope of the graph, the higher is the average speed.

Now see the graph,

It is clear from the graph that the magnitude of the slope is more in interval- $1$ than in the interval- $2$ .

Therefore, the speed of the particle in $2$ will be less than the $1$ .

Thus $v_{1} > v_{2}$

Revolution of the electron around the nucleus of an atom is an example of uniform circular motion.

0%
True
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False

$150\ m$ long train takes

$10\ s$ to pass a man who is going in the same direction at the speed of

$2\ kmph$ . What is the speed of the train?

0%
$52\ kmph$
0%
$56\ kmph$
0%
$84\ kmph$
0%
Data inadequate

Explanation

Let the speed of the train be

$x\ kmph$

Length of the train

$= 150 m = \dfrac{150}{1000} = \dfrac{3}{20} km$
and time taken to cross the man =

$10$ seconds =

$\dfrac{10}{60 \times 60}$
Relative speed =

$(x - 2)\ kmph$
(Both are moving in same direction)
We know that,

$\dfrac{\text{Length of the train}}{\text{Relative speed}} =$ Time taken to cross the man
Relative speed =

$\dfrac{3 \times 360}{20 \times 1}$

$x - 2 = \dfrac{3 \times 360}{20 \times 1}$

$x = 54 + 2 = 56\ kmph$
Hence, speed of train is

$56\ kmph$ .

Figure shows the displacement (x) -time (t) graph of the particle moving on the x-axis.

0%
The particle is at rest
0%
The particle is continuously going along x-direction
0%
The velocity of the particle increases upto 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

The slope of the Displacement- time graph provides the value of the velocity of a moving object. From the graph, it is visible that, till time $t_{0}$ object is having a constant slope and hence its velocity will be constant. Whereas after $t_{0}$ slope is zero and hence the object is not moving at all. Thus option D is most suitable and hence is the correct option.

The correct option is D.

Which of the following statements is incorrect?

0%
Path length is a scalar quantity whereas displacement is a vector quantity.
0%
The magnitude of displacement is always equal to the path length traversed by an object over a given time interval.
0%
The displacement depends only on the end points whereas path length depends on the actual path followed.
0%
The path length is always positive whereas displacement can be positive, negative and zero.

Explanation

Displacement is the smallest distance between initial and final is a vector quantity it may be

$+ve$ ,

$-ve$ , zero according to direction. (in figure displacement is

$d$ ) But path length is a scalar quantity and this is equal to total distance cover, it is always positive in figure path length

$=x_{1}+x_{2}+x_{3}$
1007319_936216_ans_540bc09b156340d992329951a550859e.png

Which of the following graphs represents the position-time graph of a particle moving with negative velocity?

Explanation

$velocity = \frac{\left | Distance x \right |}{time t}=\frac{\mathrm{d} x}{\mathrm{d} t}$

$\int \mathrm{d} x=-v\mathrm{d} t$

$x=-vt+c$

Hence graph shown in B is the correct option.

1007377_936251_ans_ffcb3347d83d4e1891657d8047afaf7b.png

A ball is travelling with uniform translatory motion. This means that

0%
It is at rest
0%
The path can be straight line or circular and the ball travels with uniform speed.
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All parts of the ball have the same velocity (magnitude and direction) and the velocity is constant.
0%
The centre of the ball moves with constant velocity and the ball spins about its centre uniformly.

The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are as shown in the figure. Choose the incorrect statement regarding these graphs.

0%
A lives closer to the school than B
0%
A starts from the school earlier than B
0%
A walks faster than B
0%
A and B reach home at the same time.

Explanation

$\textbf{Explanation:}$

$\textbf{Step 1: Positions of their respective houses from school}$

Since the slope of OP is lesser OQ, it infers that point P is closer to O than Q.

⸫ A lives closer to a school than B. Hence, (a) is true.

$\textbf{Step 2: Their starting times from school}$

From the graph, for line OP, x=0; t=0

For line OQ, when x=0; t has a positive finite value which means that A started from school earlier than B.

Hence, (b) is also true.

$\textbf{Step 3: Their respective reaching times}$

Point P and Q lie at the same vertical line on the abscissa (t) which denotes that both A and B reach their respective homes at the same time ‘t’.

Hence, (d) is also true

$\textbf{Step 4: Their respective walking speeds}$

The pace of the children is indicated by the slope of the x-t graph as their motion is uniform motion.
From the graph, it can be seen that the slope of OQ is greater than OP which means that pace of B is greater than A.

Hence, (c) is false.

Final Answer : $\textbf{Hence, the correct option is (c) A walks faster B}$

A football is kicked into the air vertically upwards with velocity

$u$ . The velocity of the ball at the highest point is:

0%
$u$
0%
$2u$
0%
$zero$
0%
$4u$

Explanation

When football is kicked upwards in air

net acceleration of the football is given by

$F=mg$

$a=\dfrac{F}{m}$

$a=\dfrac{mg}{m}=g=9.8m/s^{2}$

now as it moves up its velocity will decrease due to gravity which acts opposite to the motion.now as the ball moves high velocity continuously decreases and finally velocity will become

$ZERO$ as it will reach the highest point.

so the velocity of football at the top position will be

$ZERO$

The displacement-time graph of a moving particle is as shown in the figure. The instantaneous velocity of the particle is negative at the point

0%
C
0%
D
0%
E
0%
F

Explanation

$\textbf{Explanation}$

The velocity is

$v = dx / dt$

The instantaneous velocity is given by the slope of the displacement time graph.

Since slope is negative at point

$E$ , instantaneous velocity is negative at

$E$

Hence, Option C is correct

A particle is released from rest from a tower of height 3h. The ratio of the intervals of time to cover three equal heights h is:

0%
$t_1 : t_2 : t_3$ = 3 : 2 : 1
0%
$t_1 : t_2 : t_3$ = $1 : (\sqrt2 - 1) : (\sqrt3 -2)$
0%
$t_1 : t_2 : t_3$ = $\sqrt3 : sqrt2 : 1$
0%
$t_1 : t_2 : t_3$ = $1 : (\sqrt2 - 1) : (\sqrt3 - \sqrt2)$

Explanation

For

$A$ to

$B-$

$h=(a)(t_1)+\cfrac{1}{2}gt^2_1\quad\quad\quad [2^{nd} \text{ equation of motion.}]$

$t_1=\sqrt{\cfrac{2h}{g}}$ ..................(1)

From

$A$ to

$C-$

$2h=(0)(t_2)+\cfrac{1}{2}gt^2_2$

$t_2=\sqrt{\cfrac{4h}{g}}$ ..................(2)

Similarly,

$t_3=\sqrt{\cfrac{6h}{g}}$ ........(3

)

Times taken from

$A$ to

$B=t_1=\sqrt{\cfrac{2h}{g}}$

Times taken from

$B$ to

$C=t_2-t_1=\sqrt{\cfrac{4h}{g}}-\sqrt{\cfrac{2h}{g}}= \ \, \sqrt{\cfrac{2h}{g}}(\sqrt{2}-1)$

Time taken from

$C$ to

$D=t_3-t_2=\sqrt{\cfrac{6h}{g}}-\sqrt{\cfrac{4h}{g}}=\sqrt{\cfrac{2h}{g}}(\sqrt{3}-\sqrt{2})$

$\therefore$ Ration

$=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

Option- D

Which of the following is not true for displacement?

0%
It cannot be zero.
0%
Its magnitude is greater than the distance travelled by the object.
0%
It can be zero
0%
Its magnitude is less than the distance travelled by the object.

Multiple Correct Answers Type:
A car accelerates from rest at a constant rate of

$2{\text{ m}}{{\text{s}}^{ - 2}}$ for some time. Then it retards at a constant rate of

$4{\text{ m}}{{\text{s}}^{ - 2}}$ and comes to rest. It remains in motion for 6 s. find maximum velocity and total distance traveled.

0%
Its maximum speed is $8{\text{ m}}{{\text{s}}^{ - 1}}$
0%
Its maximum speed is $6{\text{ m}}{{\text{s}}^{ - 2}}$
0%
It travelled a total distance of 24 m.
0%
It travelled a total distance of 18 m.

A particle is acted upon by a force of constant magnitude. Which is always perpendicular to the velocity of the particle? The motion of the particle takes place in a plane. It follows that

0%
its velocity is constant
0%
its acceleration is constant
0%
its kinetic energy is constant
0%
it moves in a circular path

Explanation

The given case is of uniform circular motion in which the speed

$v$ remains constant as the force

$F$ is always perpendicular to the direction of velocity. A force can bring a change in magnitude of velocity only when force is acting along velocity.

Since kinetic energy $KE=\dfrac{1}{2}mv^2$

So kinetic energy is also constant as

$v$ is constant.

Direction of velocity and acceleration (also force) keep on changing, although their magnitude remains constant.

So option

$(C)$ and

$(D)$ are correct.

1944890_1010298_ans_4f38b61f8fc94a64b04dcf27e89c2345.png

Which parameters shown below are common between uniform circular motion and uniform linear motion

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

A body starts from rest under uniform acceleration. If distance covered by it in

$3^{rd}$ second is

$3m$ , then the distance covered in

$5^{th}$ second will be

0%
$5m$
0%
$3m$
0%
$\cfrac{27}{5}m$
0%
$\cfrac{24}{5}m$

Explanation

$s = u + \cfrac{a}{2} (2n -1)$

As,

$3 = 0 + \cfrac{a}{2} [2 \times 3 -1]$

$\therefore a = \cfrac{6}{5}$

Now,

$s' = 0 + \cfrac{6}{2\times 5} (2\times 5 -1)$

$= \cfrac{6}{2 \times 5} \times 9 = \cfrac{54}{10}$

$= \cfrac{27}{5}m$

A body dropped from a height

$h$ with initial velocity zero, strikes the ground with a velocity

$3m/s$ . Another body of same mass dropped from the same height

$h$ with an initial velocity of

$4m/s$ . The final velocity of second mass, with which it strikes the ground is

0%
$5m/s$
0%
$12m/s$
0%
$3m/s$
0%
$4m/s$

Explanation

First body initial speed

$u=0$

Final speed

$=3 m/s$

Second body initial speed

$=4 m/s$

Body is dropped from height

$h$

$a= g= 10m/{s}^{2}$

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

${3}^{2}=0+20 h$

$9=20h$

$h=\frac{9}{20}$

Second object is also dropped from same height.

therefore,

$h=\frac{20}{9}$

$u =4 m/s$

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

$v=5 m/s$

A particle moves on the

$x-$ axis. When the particle's acceleration is positive and increasing then

0%
It must speeding up
0%
It must be slowing down
0%
It may speeding up
0%
None of the these

For the V-t graph shown displacement during 15 seconds is

0%
25 m
0%
50 m
0%
zero
0%
75m

Explanation

The total displacement of a particle is equal to the area under the velocity-time graph.

Net Displacement =Area of triangle

$\Rightarrow d =\dfrac {1}{2}base \times hight$

$\Rightarrow d = \dfrac12(15 \times 10)$

$\Rightarrow d =75\ m$

A force of

$5\ N$ acts on a

$15\ kg$ body initially at rest. The work done by the force during the first second of motion of the body is

0%
$5\ J$
0%
$6\ J$
0%
$\dfrac {5}{6}\ J$
0%
$75\ J$

Explanation

Initial speed

$u = 0$

Acceleration

$a = \dfrac{F}{m} = \dfrac{5}{15} = \dfrac{1}{3} \ m/s^2$

Distance covered in one second

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

where

$t = 1 \ s$

Or,

$S = 0+\dfrac{1}{2}(\dfrac{1}{3})(1)^2 = \dfrac{1}{6} \ m$

Work done

$W = FS = 5\times \dfrac{1}{6} = \dfrac{5}{6} \ J$

A heavy stone is thrown from a cliff of height

$h$ with speed

$v$ . The stone will hit the ground with maximum speed if it is thrown:

0%
vertically downward
0%
vertically upward
0%
horizontally
0%
the speed does not depend on the inital direction

Explanation

Correct Answer: Option (D)

According to the Equation $v^{2}-u^{2}=2gh$ ,Where v is final Velocity, u is initial velocity, g is acceleration due to gravity and h is the height of the cliff

As 2gh and u is constant, the final velocity(v) is constant irrespective of the direction.

A scooterist travels at 30 km/h along straight path for 20 min. What is distance?

0%
1.5 km
0%
6 km
0%
10 km
0%
90 km

Explanation

If the scooter is travelling at precisely

$30 km/h$ in a perfectly straight line.

We know

$1$ hour=

$60$ minutes

Therefore to calculate the distance traveled in the started fraction of the hour.

$\therefore 20$ minutes is

$= \frac{{20}}{{60}} = \frac{1}{3}$ of an hour or

$3.3333$ %

$30\,km \times \frac{1}{3} = 10\,km$

Therefore

$10km$ is the distance traveled in

$20$ minutes of travelling at a speed of

$30 km/h$

A train starting from a railway station and moving with uniform acceleration, attain a speed of

$40km {h^{- 1}}$ in

$10$ minutes. Its acceleration is:

0%
$18.5m {s^{- 2}}$
0%
$1.85cm {s^{- 2}}$
0%
$18.5cm {s^{- 2}}$
0%
$1.85m {s^{- 2}}$

Explanation

$acc = \dfrac{{{v_2} - {v_1}}}{t} = \dfrac{{{v_2} - 0}}{t} = \dfrac{{40 - 0}}{{10 \times 60}}$

$= \dfrac{{40 \times 1000 \times 100}}{{3600 \times 10 \times 60}}cm{s^{ - 2}}$

$= 1.85cm{s^{ - 2}}$

Particle is dropped from the height of 20m on horizontal ground. There is wind blowing due to which horizontal acceleration of the particle becomes 6m

${ s }^{ -2 }$ . Find the horizontal displacement of the particle till it reaches ground.

0%
6m
0%
10m
0%
12m
0%
24m

Explanation

The correct opption is C.

Given,

$height =20m$

$Acceleration=6m/s^2$

We know

$u=0,g=10m/s^2,s=20m$

So,

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

$s=\dfrac{1}{2}gt^2$

$20=\dfrac{1}{2}10t^2$

$t=2s$

Since

$a=6m/s^2$ in the horizonttal,

Thus,

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

$s=\dfrac{1}{2}at^2$

$=\dfrac{1}{2}g\times2^2$

$=12m$

A ball is dropped downwards. After 2 second another ball is dropped downwards from the same point. What is the distance between them after 3 second?

0%
25 m
0%
20 m
0%
50 m
0%
9.8 m

A particle moves for

$8\ s$ . It first accelerates from rest and then retards to rest. If the retardation be

$3$ times the acceleration, then the time for which it accelerates will be

0%
$2s$
0%
$3s$
0%
$4s$
0%
$6s$

Explanation

Let

$t$ be the time of acceleration

Then retardation time

$= 8 - t$

total time

$= 8\ s$

Let

$V_1$ be velocity after acceleration

let acceleration be

$a$

$\begin{array}{l} { V_{ 1 } }=u+at \\ here\, \, u=0 \\ { V_{ 1 } }=at\to \left( i \right) \end{array}$

none retardation

$= 3a$ (given)

Hence,

$\begin{array}{l} V={ V_{ 1 } }-3a\left( { 8-t } \right) & \\ V=0, & as\, \, body\, \, comes\, \, to\, \, rest \\ { V_{ 1 } }=3a\left( { 8-t } \right) & \\ at=3a\left( { 8-t } \right) ,put\, \, { V_{ 1 } }\, \, from\, \, \left( i \right) & \\ t=24-3t & \\ 4t=24 & \\ t=6\, \, { { second } } & \end{array}$

What determine the nature of the path followed by the particle

0%
Speed
0%
Velocity
0%
Acceleration
0%
None

A body starts from the rest with uniform acceleration. If its displacement in the 3rd seconds and 7th second are

$x_1$ and

$x_2$ , then:

0%
$13x_1 = 5 x_2$
0%
$5x_1 = 13 x_2$
0%
$3x_1 = 7 x_2$
0%
$7x_1 = 3 x_2$

Explanation

$s_n$ is the distance covered in

$n{th}$ sec

$s_n=u+\dfrac{a}{2}(2n-1)$

$u=0$

$\dfrac{x_1}{x_2}=\dfrac{2\times 3-1}{2\times 7-1}$

$13x_1=5x_2$

A bullet initially moving with a velocity

$20ms^{-1}$ strikes a target and comes to rest after penetrating a distance

$10 cm$ in the target .Calculate the retardation caused by the target.

0%
$1000m/s^2$
0%
$2000m/s^2$
0%
$3000m/s^2$
0%
$4000m/s^2$

Explanation

Given,

$u=20m/s$

$s=10cm=0.1m$

$v=0m/s$

From 3rd equation of motion,

$2as=v^2-u^2$

$a=\dfrac{v^2-u^2}{2s}=\dfrac{0^2-20\times 20}{2\times 0.1}$

$a=-2000m/s^2$

Negative sign means it is retardation.

The correct option is B.

The diagram drawn above shows the distance-time graphs of four cars A,B, C and D moving on a levelled road. Choose the correct statement.

0%
Car B is moving faster than car C.
0%
Car C is the fastest.
0%
Car A is the slowest.
0%
Car D is moving slower than car C.

A short bar magnet allowed to fall along the axis of horizontal metallic ring. Starting from rest, the distance falls by the magnet in one second may be:

0%
$4.0\ m$
0%
$5.0\ m$
0%
$6.0\ m$
0%
$7.0\ m$

Explanation

When a short bar magnet allowed to fall along the axis of horizontal metallic ring, then it will fall under the influence of gravity.

$\because \,s=ut+\dfrac{1}{2}g{{t}^{2}}$

$here,\,\,u=0\,\,and\,\,t=\,1\,\,\sec$

$s=\dfrac{1}{2}\times 10\times 1$

$=5\,\,m$

The distance falls by the magnet in one second is less than $5\,m$ because when the magnet falls freely through the ring, flux increases and a field is induced whose direction is opposite to that of bar magnet. the magnet experiences a repulsion and hence acceleration of magnet becomes less than g.

A truck traveling due north at

$50 \ km/hr$ turns west and travels at the same speed. What is the change in magnitude of speed?

0%
$0$
0%
$100$
0%
$5$
0%
$50$

CBSE Questions for Class 9 Physics Motion Quiz 10 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Motion Quiz 10 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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