MCQ Questions for Class 9 Physics Motion Quiz 13

When a block of mass

$3\ kg$ is pushed by three persons from the left to the right, a constant acceleration of

$0.2m{s}^{-2}$ is produced in the block. If five persons push the block in the same direction together, then the magnitude of the acceleration of the block will be: (Assume that each person applies equal force)

0%
$0.45m{s}^{-2}$
0%
$0.33m{s}^{-2}$
0%
$0.30m{s}^{-2}$
0%
$1m{s}^{-2}$

Explanation

Let each person applies a force of magnitude

$F$ on the block from the left to the right.

$\underline{\text{When three persons push the block:}}$

Total force on the block =

$3F$

The acceleration of the block (

$a_1$ )=

$0.2\ m/s^2$

The mass of the block is

$m=3\ kg$

From Newton's second law of motion -

$3F=m\times a_1=3\times 0.2=0.6\ N$

$\therefore F=\dfrac{0.6}{3}=0.2\ N$

$\underline{\text{When five persons push the block:}}$

Total force on the block =

$5F=5\times 0.2=1\ N$

The acceleration of the block is

$a_2$

From Newton's second law of motion -

$5F=m\times a_2=3\times a_2$

$\therefore a_2=\dfrac{5F}{3}=\dfrac{1}{3}=0.33\ N$

A particle starts from the origin at

$t=0s$ with a velocity of

$10.0j\ m/s$ and moves in the

$xy-$ plane with a constant acceleration of the

$(8\hat {i}+2j)ms^{-2}$ . Then

$y-$ coordinate of the particle in

$2\ sec$ is

0%
$24\ m$
0%
$16\ m$
0%
$8\ m$
0%
$12\ m$

A balloon is rising vertically up at constant speed

$10\ m/s$ . A stone is dropped from it when the balloon is at a height of

$40\ m$ . Total distance covered by the stone before reaching the ground is

$(take\ g=10\ m/s^{2})$

0%
$40\ m$
0%
$45\ m$
0%
$50\ m$
0%
$60\ m$

A particle starts from rest with constant acceleration. The ratio of speed attained by a particle at

$t = 1\ s,\ 2\ s,\ 3\ s.........$ will be:

0%
$1 : 3 : 5 ........$
0%
$1^2 : 3^2 : 5^2 : ...........$
0%
$\left( 1-\sqrt { 2 } \right) :\left( \sqrt { 2 } \sqrt { 3 } \right) :\left( \sqrt { 3 } -\sqrt { 4 } \right) .......$
0%
$1 : 2 : 3 .......$

A particle is moving on a circular path of radius R with constant speed

$v_0$ . What is the change in magnitude of velocity when it completes

$\frac{{{3^{th}}}}{4}$ of the complete circular path ?

0%
$\sqrt 2 {v_0}$
0%
${v_0}$
0%
Zero
0%
$2 {v_0}$

A water drop falls at regular intervals from a tap

$9$ m above the ground. The fourth drop is leaving the tap at the instant, the first drop touches the ground. How high is the third drop at that instant?

0%
$2$ m
0%
$4$ m
0%
$6$ m
0%
$8$ m

A car is moving on a circular track of radius R covering equal distance in equal intervals of time. This shows that the car has

0%
uniform speed and acceleration of constant magnitude
0%
Non-uniform velocity and acceleration
0%
uniform velocity and zero acceleration
0%
Both (1) and (2)

A rocket of initial mass

$5000 kg$ ejects gets at a constant rate of

$60kg/s$ with a relative speed of

$2050 m/s$ . Acceleration of the rocket 15 second after it is blasted off from the surface of earth will be (

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

0%
$10m/{s}^{2}$
0%
$20m/{s}^{2}$
0%
$30m/{s}^{2}$
0%
$40m/{s}^{2}$

A particle experienced a constant acceleration for

$6$ seconds after starting from rest. If it travels a distance

${ d }_{ 1 }$ in the first two seconds,

${ d }_{ 2 }$ in the next

$2$ seconds, and a distance

${ d }_{ 3 }$ in the last

$2$ seconds, then:

0%
${ d }_{ 1 }:{ d }_{ 2 }:{ d }_{ 3 }=1:1:2$
0%
${ d }_{ 1 }:{ d }_{ 2 }:{ d }_{ 3 }=1:2:3$
0%
${ d }_{ 1 }:{ d }_{ 2 }:{ d }_{ 3 }=1:3:5$
0%
${ d }_{ 1 }:{ d }_{ 2 }:{ d }_{ 3 }=1:5:9$

A particle is moving in x-y plane . At certain instant of time. The components of its velocity and acceleration are as follows

$V_x = 3\ m/s$ ,

$V_y = 4\ m/s$ ,

${ a }_{ x }=2\ m/s^{ 2 }$ and

${ a }_{ y }=1\ m/s^{ 2 }$ . The rate of change of speed at this moment is

0%
$\sqrt { 10 } m/{ s }^{ 2 }$
0%
$\sqrt { 4 } m/{ s }^{ 2 }$
0%
$\sqrt { 5 } m/{ s }^{ 2 }$
0%
$\sqrt { 2 } m/{ s }^{ 2 }$

Is it possible for an object's velocity to increase while its acceleration decreases?

0%
No, this is impossible because of the way in which acceleration is defined.
0%
No, because if acceleration is decreasing the object will be slowing down.
0%
No, because velocity and acceleration must always be in the same direction.
0%
Yes, an example would be a falling object in a viscous medium, where the acceleration continuously decreases but velocity increases until a certain point.

The position-time

$(x - t)$ graph for a body thrown vertically upwards from ground is best shown by

In a motion with constant acceleration the velocity is reduced to zero in

$5$ seconds and after covering a distance of

$100\ m$ . The distance covered by the particle in next

$5$ second will be :-

0%
zero
0%
250 m
0%
100 m
0%
500 m

A particle starts from rest with uniform acceleration and its velocity after T seconds is

$u$ . The displacement of the particle in last second of this motion is

0%
$\frac {u}{2T}(2T+1)$
0%
$\frac {u}{2T}(2T-1)$
0%
$\frac {u}{T}(2T+1)$
0%
$\frac {u}{T}(T-1)$

Corresponding to the process shown in figure, what is heat given to the gas in the process ABCA ?

0%
1 J
0%
$\dfrac {3}{2} J$
0%
$\dfrac{1}{2}$ J
0%
0

Starting from rest objects

$1$ falls freely for

$4$ seconds and object

$2$ falls freely for

$8$ seconds .Compared to object

$1$ , object

$2$ falls

0%
half as far
0%
twice as far
0%
four times as far
0%
sixteen times as far

A particle is moving in x-y plane . At certain instant of time. The components of its velocity and acceleration are as follows

$V_x=3\ m/s$ ,

$V_y = 4\ m/s$ ,

${ a }_{ x }=2\ m/s^{ 2 }$ and

${ a }_{ y }=1\ m/s^{ 2 }$ The rate of change of speed at this moment is

0%
$\sqrt { 10 }\ m/{ s }^{ 2 }$
0%
$\sqrt { 4 }\ m/{ s }^{ 2 }$
0%
$\sqrt { 5 }\ m/{ s }^{ 2 }$
0%
$\sqrt { 2 }\ m/{ s }^{ 2 }$

A particle is released from rest from a tower of height

$3h$ The ratio of time taken to fall equal heights

$h , t _ { 1 } : t _ { 2 } : t _ { 3 }$ is

0%
$\sqrt { 3 } : \sqrt { 2 } : 1$
0%
$3 : 2 : 1$
0%
$9 : 4 : 1$
0%
$1 : ( \sqrt { 2 } - 1 ) : ( \sqrt { 3 } - \sqrt { 2 } )$

What is the angle between instantaneous displacement and acceleration during the retarded motion

0%
zero
0%
$\pi$
0%
$\frac{\pi}{2}$
0%
$\frac{\pi}{4}$

A conveyor belt is moving horizontally at a speed of

$4 \ m/sec$ with a box of mass

$20 \ kg$ on it. The box suddenly drops off the conveyer belt to the ground and it takes

$0.1 \ second$ for the box to come to rest. Considering the height of the conveyer belt to be negligible, the distance moved by the box in the conveyor belt is:

0%
$0$
0%
$0.2$
0%
$0.4$
0%
$0.8$

In a damped oscillator, the mass of the oscillator is m = 200g, force constant is K = 90N/m and the damping constant b is 40 g/s. Find the time taken for its mecharical energy to drop to half its initial value

0%
0.3s
0%
0.208 s
0%
6.93s
0%
3.46s

A body moving with constant acceleration covers 24 m in the

$4^{th}$ second and 36 m in the

$6^{th}$ second. The initial velocity of the body is

0%
$12 ms^{-1}$
0%
$3 ms^{-1}$
0%
$4 ms^{-1}$
0%
$5 ms^{-1}$

In the distance-time graph of

$3$ cars

$A, B$ and

$C$ which car has the highest speed and lowest speed.

0%
$A , C$
0%
$C,A$
0%
$A,B$
0%
$B,A$

A block of weight

$4 kg$ is resting on a smooth horizontal plane. If it is struck by a jet of water at the rate of

$2 kg/s$ and at the speed of

$10 m/s$ , then the initial acceleration of the block is

0%
$15$ $m/s^2$
0%
$10$ $m/s^2$
0%
$2.5$ $m/s^2$
0%
$5$ $m/s^2$

A body of mass 5 kg having a velocity of 80

$ms^{-1}$ experiences a force in the opposite direction. The body is brought to rest in 8 sec. What is the retardation produced by the force?

0%
$10ms^{-2}$
0%
$15ms^{-2}$
0%
$20ms^{-2}$
0%
$25ms^{-2}$

Explanation

Apply first equation of motion-

$v = u + at$

$v = 0$ ; final velocity that is zero

$u = 80\ ms^{-1}$ ; initial velocity

$a$ ; acceleration or retardation

$t$ ; time taken

$\Rightarrow 0 = 80 + 8a$

$\Rightarrow a = \dfrac{-80}{8}$

$\Rightarrow a = -10\ ms^{-2}$

A body starting with initial velocity

$u$ comes back to the starting point in

$t$ seconds. Then the acceleration is

0%
-2 u/t
0%
2 /t
0%
t/2u
0%
u/t

A car accelerates uniformly from

$18 km/h$ to

$36 km/h$ in

$5s$ . The distance covered by the car will be

0%
$1 m$
0%
$18 m$
0%
$37.5 m$
0%
none of theses

The acceleration time graph of a particle moving along a straight line is as shown in the figure. After what time the particle acquires

$0$ velocity.

Note :- The particle initially starts at rest.

0%
$12 \ s$
0%
$8 \ s$
0%
$5 \ s$
0%
$16 \ s$

A ship moves at 40 km/h due north and suddenly moves towards east through

$9\mathring { 0 }$ and continuous to move with the same speed. Then the change in its velocity is

0%
$zero$
0%
$40\sqrt { 2 }$ km/h north-east
0%
$40\sqrt { 2 }$ km/h south-east
0%
$None$

A particle moves with constant acceleration for 6 seconds after starting from rest. The distance travelled during the first three consecutive 1 seconds interval is in the ratio?

0%
1 : 1 : 1
0%
1 : 2 : 3
0%
1 :3 : 5
0%
1 : 5 : 9

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

0%
The particle has a constant acceleration
0%
The particle has never turned around
0%
The average speed in the interval $0$ to $10$ s in the same as the average speed in the interval $10$ s to $20$ s
0%
Both (a) and (c) are correct

Explanation

Option A is true because in a velocity-time graph, the slope (which shows the rate of change of velocity) denotes acceleration. In this graph the slope is constant, so acceleration is constant.

Option B is not true because at

$t=10 s$ the velocity changes its sign from positive to negative, which means it has turned around.

Option C is true because in

$0$ to

$10 s$ speed changes from

$10$ to

$0 \,m/s$ and in

$10$ to

$20 s$ speed changes from

$0$ to

$10 \,m/s$ . So average speed will be the same.

Consider the above graph, which of the following correctly represents the final and initial velocity (

$v$ and

$u$ ) respectively?

0%
$OA$ , $OC$
0%
$OA$ , $BD$
0%
$BC$ , $OA$
0%
$OA$ , $BC$

Explanation

The following details are obtained from the graph above:
The initial velocity of the body,

$u$ =

$OA$
The final velocity of the body,

$v$ =

$BC$

Which of the following information is correct from the above graph?

0%
$BC = BD + DC$
0%
$v = BD + DC$
0%
$BD + OA$
0%
All of these

Explanation

From the graph, we know that

$BC = BD + DC$ ,

Therefore,

$v = BD + DC$

$v = BD + OA$ (since

$DC = OA)$

Hence option

$D$ is correct.

In the above graph,

$BC$ is the ________ and

$OC$ is the _______.

0%
Total time, final velocity
0%
Final velocity, total time
0%
Total time, inital velocity
0%
Initial velocity, final velocity

Explanation

$BC$ is the final velocity and

$OC$ is the total time represented as

$t$ , hence option

$B$ is correct.

In the above graph, the velocity of the body changes from

$A$ to

$B$ in a time

$t$ at a _______ rate.

0%
Uniform
0%
Non-uniform
0%
Zig-zag
0%
None of these

Explanation

The velocity of the body changes from A to B in time t at a uniform rate since the slope is a straight inclined line, hence option

$A$ is correct.

Rectilinear motion can be uniform but circular motion can't

0%
True
0%
False

CBSE Questions for Class 9 Physics Motion Quiz 13 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Motion Quiz 13 - MCQExams.com

0:0:1

Practice Class 9 Physics Quiz Questions and Answers

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