Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

MCQ Questions for Class 9 Physics Force And Laws Of Motion Quiz 14

Which of the following statements is not true regarding the Newton's third law of motion?

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To every action there is always an equal and opposite reaction.
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Action and reaction act on the same body.
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There is no cause-effect relation action and reaction.
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Action and reaction forces are simultaneous forces.

Explanation

Action and reaction act on the same body is not true regarding Newton's third law of motion. Action and reaction pairs always act on different bodies.

Hence,

option

$B$ is the correct answer.

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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect and Reason is correct

Explanation

$\text { Net force of block } B=\sqrt{N_{1}^{2}+\left(m_{B} \cdot g\right)^{2}} \\$

$\text { Since it is not equal to } N_{1} \text { , therefore statement } I \\$

$\text { i.e. Assertion is wrong. The force exerted by block } \\$

$\text { A on block } B \text { is not equal to net force on block } B \text { . } \\$

$\text { we know that Newton's third Law states that } \\$

$\text { for every action, there is an equal and opposite } \\$

$\text { reaction. So the force exerted by block } A \text { on } B \text { is } \\$

$\text { equal in magnitude to force exerted by block B } \\$

$\text { on } A \text { . } \\$

$\text { Answer: Assertion is incorrect and Reason is correct. }$

1997856_984460_ans_3d75d797fca0455a99d44fd36580ee43.jpeg

Complete the statement of the first law of motion. "A body at rest stays at ____ and a body in motion stays in ____ unless an ____ is applied"

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Motion; Rest; External Force
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Rest; Motion; External Force
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Rest; Motion; Internal Force
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None of these

A force of 2 N is applied on a particle for 2 sec, the change in momentum will be -

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2 Ns
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4 Ns
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6 Ns
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9 Ns

Write true or false for the following statements:

$1 \,N$ that force which produces acceleration of

$1 \,m \,s^{-2}$ in a body of mass

$1 \,g.$

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True
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False

Explanation

False,

One Newton is the force required to accelerate one kilogram of mass at the rate of

$1 m/s^2$ acceleration in the direction of the applied force.

The momentum of a body of given mass is proportional to its:

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speed
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volume
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density
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shape

Two small spheres of masses

${M}_{1}$ and

${M}_{2}$ are suspended by weightless insulating threads of lengths

${L}_{1}$ and

${L}_{2}$ , the spheres carry charges

${Q}_{1}$ and

${Q}_{2}$ respectively. The spheres are suspended such that they are in level with another and the threads are inclined to the vertical at angles of

${\theta}_{1}$ and

${\theta}_{2}$ as shown, which one of the following conditions is essential, if

${ \theta }_{ 1 }={ \theta }_{ 2 }:$

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${ M }_{ 1 }\neq { M }_{ 2 }$ but ${ Q }_{ 1 }={ Q }_{ 2 }$
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${ M }_{ 1 }= { M }_{ 2 }$
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${ Q }_{ 1 }={ Q }_{ 2 }$
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${ L }_{ 1 }={ L }_{ 2 }$

Explanation

For a single change system:-

We can write from

$FBD$ ,

$T\,\cos\theta_1=Mg$

$T\sin\theta_1=Fe$ .

Now for both change,

$Fe$ is equal in magnitude but,

$\theta_1=\theta_2$ necessarily imply.

$M_1=M_2$

But, we can't say

$Q_1=Q_2$ as both change face

$Fe=\dfrac{KQ_1Q_2}{d^2}$ in equal magnitude but opposite direction [Newtons third law] so

$T\,\sin\theta$ component would be equal for both.

1080590_823708_ans_c71d3d6fdcea4a3c988a752d0165b4cd.png

A free-floating astronaut A pushes another free-floating astronaut B in space. The mass of A is greater than that of B. The force exerted by the astronaut A on the astronaut B will be.

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Greater than the force exerted by B on A
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Equal to the force exerted by B on A
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Equal to zero
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Less than the force exerted by B on A

Explanation

According to newton third law for every action there is an equal and opposite reaction.

The force on each other are actually action and reaction. So both forces are equal. So the magnitude of the force exerted by astronaut

$A$ on

$B$ is equal to the force exerted by

$B$ on

$A$ .

Which of the following is not the characteristics of matter?

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The matter has a mass
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Matter does not have inertia
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Matter is affected by gravity
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Matter cannot be destroyed

The momentum of a $225\ g$ ball moving at $35 \ m/s$ is

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$7875 \ kg \,m/s$
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$3.5 \ N$
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$5.0 \ m/s$
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$7.875\ kg \ m/s$

Explanation

Given

$;$

mass

$=225g$

$=0.225kg$

velocity

$=35m/s$

$Momentum = mass \times velocity$

$= 0.225 \times 35 = 7.875\,kg\,m/s$

Hence,

option

$(D)$ is correct answer.

A ball is moving in the direction as shown. What will be the direction of momentum?

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$W$
0%
$N$
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$S$
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$E$

Explanation

Since momentum

$p=mv$

Direction of momentum is same as the direction of velocity. Hence, the direction of momentum of ball in the given case is East.

A fly strikes the windshield of a moving truck as shown. What can you say about the forces: the force on the fly and the force on the truck.

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Force on the fly $>$ Force on the truck
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Force on the fly $<$ Force on the truck
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Force on the fly $=$ Force on the truck
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None of these

Club the cases shown under balanced or unbalanced forces (all arrows represent same magnitude of force)

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Balanced forces- $A,B$ and Unbalanced forces- $C,D$
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Balanced forces- $C,D$ and Unbalanced forces- $A,B$
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Balanced forces- $A,C$ and Unbalanced forces- $B,D$
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Balanced forces- $B,D$ andUnbalanced forces- $A,C$

Explanation

In cases

$B$ and

$D$ , forces are equal and opposite. While, in cases

$A$ and

$C$ , the forces are equal but not opposite.

Find net force on the object shown in the figure. Will the object move?

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$F_{net} = 0$ ; No
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$F_{net} \neq 0$ ; No
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$F_{net} = 0$ ; Yes
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$F_{net} \neq 0$ ; Yes

Explanation

Total force towards right

$= 20+6 = 26$ units
Total force towards left

$= 13+13 = 26$ units
So, net force

$(F_{net}) = 26-26 = 0$
Hence, forces are balanced and the object will not move.

A rocket with a lift-off mass

$3.5\times {10}^{4}\ kg$ is blasted upwards with an initial acceleration of

$10m{s}^{-2}$ . Then the initial thrust of the blast is:

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$3.5\times {10}^{5}N$
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$7.0\times {10}^{5}N$
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$14.0\times {10}^{5}N$
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$1.75\times {10}^{5}N$

Explanation

According to the second law of Newton-

$F = ma$

$F$ ; Force or thrust

$m$ ; the mass of the object

$a$ ; acceleration of the object

$\Rightarrow F = (3.5 \times 10^4) \times 10$

$\Rightarrow F = 3.5 \times 10^5\ N$

A gun weighing

$10\ kg$ fires a bullet of

$30\ g$ with a velocity

$330\ m/s$ With what velocity does the gun recoil?

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$2\ m/s$
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$1\ m/s$
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$0\ m/s$
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$3\ m/s$

Explanation

Given,

Mass of the gun,

$m_g=10\ kg$

Mass of the bullet,

$m_b=30\ g=0.03\ kg$

Velocity of the bullet,

$v_b=330m/s$

Velocity of the gun,

$v_g$

By applying the conservation of linear momentum,

$m_gv_g=m_bv_b$

$10\times v_g=0.03\times 330$

$v_g=1\ m/s$

Find net force on the object shown in the figure and answer if the forces are balanced or unbalanced

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$(F_{net}) = 0$ ; Balanced Forces
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$(F_{net}) \neq 0$ ; Balanced Forces
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$(F_{net}) = 0$ ; Unbalanced Forces
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$(F_{net}) \neq 0$ ; Unbalanced Forces

Explanation

According to given fig.

Net horizontal force

$= 20-20 = 0$ Unit
Net vertical force

$= 13-6 = 7$ units
So, net force

$= 7$ units (downward)
Hence,

$F_{net}\neq 0$ ; forces are unbalanced,

Option D

The velocity acquired by a mass

$m$ in travelling a certain distance

$d$ starting from rest under the action of a constant force is directly proportional to:

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$\sqrt {m}$
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$m^{o}$
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$\dfrac {1}{\sqrt {m}}$
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$m$

Explanation

Force,

$F=ma$

$F\,\propto \,m$

Hence, force is directly proportional to $m$

If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be

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Going up with increasing speed
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Going down with increasing speed
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Going up with uniform speed
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Going down with uniform speed

Explanation

If the tension in the cable of the elevator is the same as the weight of the elevator, it means that the net force on the system is zero. Therefore, the net acceleration of the system will also be zero.

This means that the elevator is moving up or down with uniform speed.

Option

$C \ \&\ D$ are correct.

Due to the application of a force on a body of mass

$100\ kg$ that is initially at rest, the body moves with an acceleration of

$20{ms}^{-2}$ in the direction of the force. Find the magnitude of the force.

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$1000N$
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$2000N$
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$3000N$
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$4000N$

Explanation

Apply Newton's second law

$F = ma$

$F$ is net applied foorce

$m$ mass of the object

$a$ acceleration of the object

$\Rightarrow F= 100\times 20$

$\Rightarrow F=2000 N$

$5\ kg$ block rests on horizontal surface. Which of the following option is correct? (

$g = 10 \mathrm { m } \mathrm { s } ^ { -2 }$ )

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The normal force on the block equals its weight because of Newton's second law
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The normal force on the block is equal to the weight not because of Newton's third law
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The normal force on the block equals its weight because of Newton's third law
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The normal force on the block is $50\ N$ .

Explanation

Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equal the size of the force on the second object

Weight,

$W=mg$

$\Rightarrow W=5\times 10$

$\Rightarrow W=50\, N$

Hence, The normal force on the block equals its weight because of Newton's third law equal to $50\ N$

Which of the following objects experience balanced forces?

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A ball dropping vertically
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A car accelerating uniformly from rest
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A trolley moving at a constant velocity down an inclined plane
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A plane flying in a circle with constant speed

Explanation

In case(c) Trolley is moving at constant speed so no change in velocity means acceleration is zero.

$a=0$

Now according to Newton's second law

$F=ma$

$F=m\times0$

$F=0$

So it experience no net force or the forces on it are balanced.

According to Newton's third law action is always equal and opposite to the reaction, a horse can pull a cart because it applies a ______.

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Force on cart
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Force on ground
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Both A and B
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None of these

A force of 15N acts separately on two bodies of masses 3kg and 5kg. The ratio of the acceleration produced in the two cases will be

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$5:3$
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$3:5$
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$8:15$
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$15:8$

Explanation

Given,

$F=15N$

$m_1=3kg$

$m_2=5kg$

We know that,

$F=ma$

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

The force acted on the bodies is same,

So,

$\dfrac{a_1}{a_2}=\dfrac{m_2}{m_1}=\dfrac{5}{3}$

Ratio,

$a_1:a_2=5:3$

The correct option is A.

Fill in the blanks.
At least _____________ objects must be present to experience a gravitational force.

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1
0%
2
0%
3
0%
4

When a body is stationary

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there is no force acting on it
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the forces acting on it are not in contact with it
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the combination of forces acting on it balance each other
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None of the above

Explanation

Consider a stationary block on a table. Gravity force (weight $W=mg$ ) acts downwards and normal reaction force from table acts upwards.

Hence, the combination of forces acting on it balance each other.

1065830_1179741_ans_69260c69689048fdb2d9f8413df85b07.png

When two bodies A and B interact with each other, A exerts a force of

$10N$ on B towards east. What is the force exerted by B on A?

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A exerts $10N$ force on B towards West
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B exerts $20N$ force on A towards East
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B exerts $10N$ force on A towards West
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A exerts $20N$ force on B towards East

A person holding a rifle (mass of person and rifle together is 100 kg) stands on a smooth surface and fires 10 shots horizontal, in 2 s. Each bullet has a mass of 10 g with a muzzle velocity of

$800{ ms }^{ -1 }.$ The final velocity acquired by the person and the average force exerted on the person are

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$-0.8{ ms }^{ -1 },16N$
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$-0.8{ ms }^{ -1 },8N$
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$-1.6{ ms }^{ -1 },16N$
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$-1.6{ ms }^{ -1 },8N$

Explanation

$P ( initial ) = P ( final )$

$0 = n \times m \times u + ( M - n \times m ) \times v$

where:

$n = 10, m = 10 g = 0.01 kg, u = 800 m/s, M = 100 kg$

$0 = 10 \times 0.01 kg \times 800 m/s + ( 100 kg - 10 \times 0.01 kg ) \times v$

$v = - \dfrac{80 kgm/s} {99.9 kgm/s}$

$v = -0.8 m/s$

Then :

$F =\dfrac{\Delta P}{\Delta t}= \dfrac{ 10 \times 0.01 kg \times 800 m/s }{ 5 s} = 16 N$

In the graph below is given the variation of force with time. Find out the net change in momentum of the object.

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24 kg m/s
0%
10 kg m/s
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6 kg m/s
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3 kg m/s

From a stationary tank of mass

$125000$ pound, a small shell of mass

$25$ pound is fired with a muzzle velocity of

$1000\ ft/sec$ . The tank recoils with a velocity of ..........

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$0.1\ ft/sec$
0%
$0.2\ ft/sec$
0%
$0.4\ ft/sec$
0%
$0.8\ ft/sec$

Explanation

Given:

Mass of the tank

$M=125000\ pound$

Mass of the shell

$m=25\ pound$

Velocity of the shell

$v=100\ ft/sec$

Recoil velocity of the tank

$V=?$

Since There is no external force acting, we can apply conservation of momentum.

According to the conservation of momentum,

$\text{Initial momentum of (tank+shell)} =\text{Final momentum of (tank + shell)}$

$(M+m)\times 0 =M\times V + m\times v$

$\Rightarrow 0 = 125000\times V + 25\times 1000$

$\Rightarrow V=-\dfrac{25000}{125000}=-\dfrac{1}{5}=-0.2\ ft/sec$

The -ve sign indicates that the tank will recoin in the opposite direction to the direction of motion of the shell.

CBSE Questions for Class 9 Physics Force And Laws Of Motion Quiz 14 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Force And Laws Of Motion Quiz 14 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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