MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 9 Physics Motion Quiz 13 - MCQExams.com
CBSE
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)
Report Question
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
Report Question
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 $$(take\ g=10\ m/s^{2})$$
Report Question
0%
$$40\ m$$
0%
$$45\ m$$
0%
$$50\ m$$
0%
$$60\ m$$
Explanation
undefined
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:
Report Question
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 ?
Report Question
0%
$$\sqrt 2 {v_0}$$
0%
$${v_0}$$
0%
Zero
0%
$$2 {v_0}$$
Explanation
The magnitude of velocity is the speed.
As the particle is moving in the circular path with the constant speed. Thus the magnitude of velocity will not change whatsoever. Only the direction will be different at different positions on the track.
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?
Report Question
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
Report Question
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 ($$g = 10m/{s}^{2}$$?)
Report Question
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:
Report Question
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
Report Question
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?
Report Question
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
Report Question
0%
0%
0%
0%
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 :-
Report Question
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
Report Question
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 ?
Report Question
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
Report Question
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
Report Question
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
Report Question
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
Report Question
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:
Report Question
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
Report Question
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
Report Question
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.
Report Question
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
Report Question
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?
Report Question
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
Report Question
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
Report Question
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.
Report Question
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
Report Question
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?
Report Question
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?
Report Question
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?
Report Question
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?
Report Question
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 _______.
Report Question
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.
Report Question
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
Report Question
0%
True
0%
False
Explanation
Circular motion is the motion of an object that moves at a fixed distance from a fixed point and if it covers equal distances if speed is constant, it will undergo uniform motion, hence the given statement is false.
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 9 Physics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page