always act on two different bodies

explains the way the force acts on a body

defines the force qualitatively

defines the force quantitatively

MCQ Questions for Class 11 Engineering Physics Laws Of Motion Quiz 1

A uniform rod

$AB$ is suspended from a point

$X$ , at a variable distance

$x$ from

$A$ , as shown. To make the rod horizontal, a mass

$m$ is suspended from its end

$A$ . A set of

$(m,x)$ values is recorded. The appropriate variables that give a straight line, when plotted, are:

0%
$m,\cfrac { 1 }{ x }$
0%
$m,\cfrac { 1 }{ { x }^{ 2 } }$
0%
$m,x$
0%
$m,{ x }^{ 2 }$

Explanation

Let, 'l' and 'M' be the length and mass of rod.

for equilibrium, taking moment about X, we get

$m \times x = M \times (\dfrac{l}{2} - x)$

this can be written as,

$m = (\dfrac{Ml}{2}) \dfrac{1}{x} - M$

comparing with straight line equation,

y = mx + c

we can say that,

$m \propto \dfrac{1}{x}$

A beaker of radius r is filled with water (refractive index

$\dfrac{4}{3}$ ) up to a height H as shown in the figure on the left. The beaker is kept on a horizontal table rotating with angular speed

$\omega$ . This This makes the water surface curved so that the difference in the height of water level at the center and at the circumference of the beaker is

$h (h << H, h << r)$ , as shown in the figure on the right. Take this surface to be approximately spherical with a radius of curvature R. Which of the following is/are correct? (g is the acceleration due to gravity)

0%
$R = \dfrac{h^2 + r^2}{2h}$
0%
$R = \dfrac{3r^2 }{2h}$
0%
Apparent depth of the bottom of the beaker is close to $\dfrac{3H}{2} \left(1 + \dfrac{\omega^2 H}{2g} \right)^{-1}$
0%
Apparent depth of the bottom of the beaker is close to $\dfrac{3H}{4} \left(1 + \dfrac{\omega^2 H}{4g} \right)^{-1}$

Explanation

$9th \ \Delta ABC$

$(R - h)^2 + r^2 = R^2$

$R^2 + h^2 - 2Rh + r^2 = R^2$

$h^2 + r^2 = 2Rh$

$\dfrac{h^2 + r^2}{2h} = R$ ___(i)

$r >>> h$

$R = \dfrac{r^2}{2h}$

$\because h = \dfrac{w^2 r^2}{2g}$

$R = \dfrac{r^2 2g}{2w^2 r^2}$

$R = \dfrac{g}{w^2}$ ___(ii)

$\dfrac{\mu_1}{V} - \dfrac{\mu_2}{u} = \dfrac{\mu_1 - \mu_2}{R}$

$\dfrac{1}{V} + \dfrac{4}{3u} = \dfrac{1 - \dfrac{4}{3}}{R}$

$\dfrac{1}{V} = -\left(\dfrac{1}{3R} + \dfrac{4}{3(H - h)}\right)$

$\dfrac{1}{V} =-\left[\dfrac{1}{3R} + \dfrac{4}{3H}\right]$

Put the value of equation (ii)

$h <<H$

$\dfrac{1}{V} = -\left[\dfrac{4}{3H} \left[1 + \dfrac{3H}{4} \dfrac{w^2}{39} \right]\right]$

$V = -\left[\dfrac{3H}{4} \left(1 + \dfrac{w^2 H}{4}\right)^{-1}\right]$

1993847_1950083_ans_8029739bdb4642a99c141fc467c553ca.png

What is the minimum velocity with which a body of mass

$m$ must enter a vertical loop of radius

$R$ so that it can complete the loop?

0%
$\sqrt{gR}$
0%
$\sqrt{2gR}$
0%
$\sqrt{3gR}$
0%
$\sqrt{5gR}$

Explanation

To just complete the circle, at the highest point, tension is zero and the gravitational force provides the necessary centripetal force.

Hence,

$\dfrac{mv_{top}^2}{R}=mg.............(i)$

Applying conservation of energy at the top and bottom points,

$\dfrac{1}{2}mv_{bottom}^2=mg \times 2R + \dfrac{1}{2}mv_{top}^2.........(ii)$

Substituting

$mv_{top}^2$ from (i) in (ii) and solving,

$v_{bottom}=\sqrt {5gR}$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

A ball is suspended by a thread of length L at the point O on a wall which is inclined to the vertical by

$\alpha$ . The thread with the ball is displaced by a small angle

$\beta$ away from the vertical and also away from the wall. If the ball is released, the period of oscillation of the pendulum when

$\beta > \alpha$ will be

0%
$\displaystyle \sqrt{\frac{L}{g}}\left[\pi+2sin^{-1}\frac{\alpha}{\beta}\right]$
0%
$\displaystyle \sqrt{\frac{L}{g}}\left[\pi-2sin^{-1}\frac{\alpha}{\beta}\right]$
0%
$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}-\pi\right]$
0%
$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}+\pi\right]$

Explanation

The motion of the string can be represented as an angular displacement from the mean position,given as:

$\theta=\beta sin(\omega t)$

Time taken from displacement from mean position to

$\beta$ and back to mean position=

$t_1=\dfrac{T}{2}$

where

$T$ is the time period of the oscillation without the wall barrier.

Thus

$t_1=\dfrac{1}{2}\times 2\pi\sqrt{\dfrac{L}{g}}=\pi\sqrt{\dfrac{L}{g}}$

Time taken to displace from mean position to

$\alpha$ can be found by:

$\alpha=\beta sin(\omega t)$

$\implies t=\dfrac{1}{\omega}sin^{-1}\dfrac{\alpha}{\beta}$

$=\sqrt{\dfrac{L}{g}}sin^{-1}\dfrac{\alpha}{\beta}$

Thus time taken to displace from mean position to

$\alpha$ and back to mean position =

$t_2=2\sqrt{\dfrac{L}{g}}sin^{-1}\dfrac{\alpha}{\beta}$

Thus the time period of the oscillation =

$t_1+t_2=\sqrt{\dfrac{L}{g}}[\pi+2sin^{-1}\dfrac{\alpha}{\beta}]$

An object placed on a ground is in stable equilibrium. If the objects is given a slight push, then initially the position of centre of gravity:

0%
moves nearer to ground
0%
rises higher above the ground
0%
remains as such
0%
may remain at same level

Cleaning of dust from carpet is due to _____.

0%
Inertia of motion
0%
Inertia of rest
0%
Inertia of direction
0%
Momentum

A ball of mass 'm' moves normal to a wall with a velocity 'u' and rebounds with the same speed. The change in momentum of the ball during the rebounding is

0%
2mu towards the wall
0%
2mu away from the wall
0%
zero
0%
mu away from the wall

Explanation

If the collision of the ball and the wall is elastic one then the change in the momentum of the ball after collision is

$= mu-(-mu)=2mu$ away from the wall.

When an athlete takes part in a long jump he runs for a while and then jumps. This is because:

0%
inertia of motion helps him to take a longer jump.
0%
running for a time makes him alert.
0%
it helps him to focus on the point of jump.
0%
he can apply more force for the jump after running.

Statement A: If the force varies with time then the net force is measured by the total change in momentum of the body.
Statement B: Change in momentum and impulsive force are numerically equal.

0%
A & B are correct
0%
A & B are wrong
0%
A is correct and B is wrong
0%
A is wrong and B is correct

Explanation

Force is a rate of change of momentum.

$Force$ is measured as

$rate\ of\ change\ of\ momentum.$ It is not equal to change of momentum.

$F=\dfrac{dM}{dt}$ or,

$F\cdot dt=dM$

Impuse of force,

$J=F\cdot dt=dM$

Hence, Change in momentum and impulsive force are numerically equal.

A is wrong and B is correct.

Which of following Newton's laws reveals the underlying symmetry in the forces that occur in nature?

0%
First law
0%
Second law
0%
Third law
0%
Law of gravitation

When a moving body is suddenly stopped, then:

0%
Frictional force increases.
0%
Friction force remains same as it was while the car was moving.
0%
Friction force reduces but has a non zero value.
0%
Frictional force reduces to zero as it is a self adjusting force.

When a body is in equilibrium, which of the following must be true ?

0%
There is no force acting on it.
0%
The sum of all the forces acting on it must be zero.
0%
The sum of all the forces acting on it is not equal to zero.
0%
There are exactly two forces acting on the body.

Impulse is equal to

0%
momentum
0%
change in momentum
0%
rate of change of momentum
0%
rate of change of force

Explanation

Impulse

$(\vec{J})$ is defined as the change in the momentum.

$\therefore$

$\vec{J} = \Delta \vec{P} = m (\vec{v_2} - \vec{v_1})$

A stone tied to a string is rotated in a vertical circle. The minimum speed with which the stone has to be rotated in order to complete the circle :

0%
decreases with increasing the mass of the stone
0%
is independent of the mass of the stone
0%
decreases with increasing the length of the string
0%
is independent of the length of the string

Explanation

$\textbf{Step 1: Energy Conservation [Ref. Fig.]}$

$\text{Dependence on m}$

Let

$u$ be the minimum speed required at

$B$ .

Apply energy conservation between point

$A$ and

$B$ we get

$KE_A + PE_A = KE_B + PE_B$

Taking point

$A$ as zero potential level

$\Rightarrow$

$\dfrac{1}{2} mu^2 = mg (R + R) + \dfrac{1}{2} mv^2$

$....(1)$

From equation

$1$ , we observe that mass

$m$ is cancelled on both sides.

$\Rightarrow u$ is independent of

$m$ .

$\textbf{Steo 2: Solving Equation}$

$\text{Dependence on R}$

At highest point, Tension will be zero for minimum velocity

$\Rightarrow$ gravitational pull will provide necessary centripetal force

$\Rightarrow mg = \dfrac{mv^2}{R}$

$\Rightarrow v^2 = Rg$ .... Putting in equation

$(1)$

We get,

$\dfrac{1}{2} mu^2 = mg (R + R) + \dfrac{1}{2} m (Rg)$

$\Rightarrow u = \sqrt{5Rg}$

$\Rightarrow u$ is proportional to the square root of

$R$ .

Hence option

$B$ is correct.

Action and reaction

0%
always act on two different bodies
0%
are equal in magnitude
0%
act in opposite directions
0%
All the above

In which of the following cases the net force is not equal to zero?

0%
A kite skillfully held stationary in the sky.
0%
A ball falling freely from a height.
0%
A helicopter hovering above the ground.
0%
A cork floating on the surface of water.

The impulse of a body is equal to :

0%
Rate of change of its momentum.
0%
Change in its momentum.
0%
The product of force applied on it and the time of application of the force.
0%
Both (B) and (C)

Which of the following statements are correct regarding linear momentum of a body?

0%
It is a measure of quantity of motion contained by the body.
0%
Change in momentum is the measurement of impulse.
0%
Impulse and acceleration act in same direction to the change in momentum.
0%
Incase of uniform circular motion, the linear momentum is conserved.

Which of the following statements are true?

0%
When the mass of a body is doubled then the momentum of a body is also doubled, provided the body maintains the same velocity.
0%
We feel pain in the hand on hitting the wall, this is a consequence of Newton's third law of motion.
0%
A table cloth can be pulled from a table without dislodging the dishes. This is due to inertia of rest.
0%
Momentum is a vector quantity.

Explanation

Option A is correct .

as P = mv,

$P \space \alpha \space m$ if v is constant.

Option B is correct.

The pain in hand is due to reaction from the wall. It will be same in magnitude as that of applied force according to Newton's third law, for every action, there is an equal and opposite reaction.

Option C is correct.

When table cloth is pulled very fast, due to inertia dishes does not change its state.

Option 4 is correct.

P = mv

as v (velocity) is vector, momentum is also vector.

The average force necessary to stop a hammer with 25 N-s momentum in 0.05s expressed in N is :

0%
500
0%
125
0%
50
0%
250

Explanation

Impulse

$=$ Force

$\times$ time

$25 = F \times 0.05$

$F = \displaystyle \frac{25}{0.05} = 500 N$

....................... is a self adjusting force.

0%
Magnetic Force
0%
Electrostatic Force
0%
Frictional Force
0%
All

Explanation

$\textbf{Part1: Definition}$

$\bullet$ Self-adjusting force is defined as a force which can adjust itself as required by the situation. It can increase or decrease itself with the change made in the system.

$\textbf{Part2: }$

$\bullet$ If the surface of two bodies are in contact with each other and we apply a force on one of them, static friction will appear, even if the body is at rest.

$\bullet$ The static friction will try to make the body to remain at rest. If we apply an external force on the body the static friction will adjust the magnitude of friction force and try to make the object to remain at rest.

$\bullet$ It has a limit between $0$ and $\mu N$ where N is the normal reaction and $\mu$ is coefficient of friction . It opposes the externally applied force between these limits.

So, we can say that static friction is a self-adjusting force because it does not have any fixed magnitude, but it will adjust the force according to the applied force.

Hence , option (c) is correct.

Newton's first law of motion gives the concept of inertia.

0%
True
0%
False

If two balls each of mass 0.6 kg moving in opposite directions with speed of 4 m s

$^{-1}$ collide and rebound with same speed, then the impulse imparted to each ball due to other is:

0%
$0.48 kg m s^{-1}$
0%
$0.53 kg m s^{-1}$
0%
$0.81 kg m s^{-1}$
0%
$0.92 kg m s^{-1}$

Explanation

Impulse

$= F \times t=$ change in momentum

$=$ 0.06 (4-(-4))

$=$ 0.48 kg m s

$^{-1}$

The time, in which a force of 2 N produces a change as momentum of 0.4 kg m s

$^{-1}$ in the body is :

0%
0.2s
0%
0.02s
0%
0.5s
0%
0.05s

Explanation

$F=\displaystyle \frac{\Delta P}{t}$

$2 = \displaystyle \frac{0.4}{t}$ (or)

$t = 0.2 s$

A hammer of mass 5 kg moving with a speed of 2 m s

$^{-1}$ strikes the head of a nail driving it 20 cm into the wall. Find the impulse.

0%
10 N s
0%
20 N s
0%
30 N s
0%
40 N s

Explanation

Impulse

$=$ m(v - u)

$=$ 5 (0 - 2)

$=$ 10 N s

A ball tied to a string is swung in a vertical circle. Which of the following remains constant?

0%
Tension in the string.
0%
Speed of the ball.
0%
Centripetal force.
0%
Earth's pull on the ball.

Explanation

Earth's pull on the ball remains constant

$(W=mg)$ .
Tension in the string changes along the vertical circular path.
Speed of the ball changes as sum of kinetic energy and potential energy remains constant during motion.
Speed changes, hence, centripetal force changes.

Newton's third law:

0%
defines the force qualitatively
0%
defines the force quantitatively
0%
explains the way the force acts on a body
0%
gives the direction of force

The linear momentum of a ball of mass

$50 g$ is

$0.5 kg { m s }^{ -1 }$ . Find its velocity.

0%
$0.01$ ${ m s }^{ -1 }$
0%
$1$ ${ m s }^{ -1 }$
0%
$5$ ${ m s }^{ -1 }$
0%
$10$ ${ m s }^{ -1 }$

Explanation

We know that

$p= mv$
where

$p =0.5 kg ms^{-1}$ linear momentum

$m=50g = 0.05kg$ mass,

$v =$ velocity

$\therefore v = \dfrac{p}{m}$

$\Rightarrow =\dfrac{0.5}{0.05} = 10 m/s$

'The action and reaction are equal in magnitude'. Is this statement true and which law is related with it ?

0%
Yes, Newton's second law.
0%
Yes, Newton's third law.
0%
No, Newton's third law.
0%
Yes, Newton's first law.

A metal ball does not rebound when struck on a wall, whereas a rubber ball of same mass when thrown with the same velocity on the wall rebounds. From this it is inferred that:

0%
change in momentum is same in both.
0%
change in momentum in rubber ball is more.
0%
change in momentum in metal ball is more.
0%
initial momentum of metal ball is more than that of rubber ball.

A body of mass

$5 kg$ is moving with velocity

$2 { ms }^{ -1 }$ . Calculate its linear momentum.

0%
$2.5$ $kg$ ${ ms }^{ -1 }$
0%
$10$ $kg$ ${ ms }^{ -1 }$
0%
$5$ $kg$ ${ ms }^{ -1 }$
0%
$20$ $kg$ ${ ms }^{ -1 }$

Explanation

We know that

$p= mv$

where

$p =$ linear momentum

$m=5kg$ mass,

$v =2 m/s$ velocity

$p = 5 \times 2 = 10 kg ms^{-1}$

Momentum is most closely related to :

0%
force
0%
impulse
0%
KE
0%
power

Explanation

Force is defined as the rate of change of momentum.

It is given by

$\dfrac {\text{change of momentum}}{time}$ .

Impulse is defined as the change of momentum. It is given by

${Force} \times {time}$ .

Kinetic energy is the energy possessed by an object because of its motion. It is given by

$\dfrac 12mv^2$ . It can't be calculated by momentum alone. Power is the rate of work done with respect to time. it is not related to momentum.

Among force and impulse, There is a lesser calculation for impulse. Also, it has same dimensions as momentum. Hence that is most closely related.

A gun requiring the support of shoulder shows intensive recoil when fired with:

0%
the butt held tightly with the shoulder.
0%
the butt held loosely against the shoulder.
0%
butt held loosely or tightly against the shoulder.
0%
a bullet of greater mass.

Explanation

Let

$m$ be the mass of the bullet and

$M$ be the mass of the gun. If

$v$ is the velocity of bullet then applying Linear Momentum Conservation

$MV = mv$ i.e.,

$V = \displaystyle \frac{mv}{M}$
Thus,

$V$ is the velocity of recoil of gun when held loosely.
But when the gun is held tightly with the shoulder, the mass of the man supports the gun thus reducing the speed of gun. Hence, the gun experiences intensive recoil when fired with the butt held loosely with the shoulder.

The SI unit of momentum is :

0%
$kg ms^{-1}$
0%
$kg m s^{-2}$
0%
$kg m s^{2}$
0%
$kg m^{-1}s^{-1}$

Frictional force always reduces:

0%
Speed
0%
Velocity
0%
Acceleration
0%
Time

The maximum force of friction when the body is just begining to move is known as the

0%
Limiting friction
0%
Rolling friction
0%
Kinetic friction
0%
(1) and (2) both

If the net force on an object is zero, then the object:

0%
is in equilibrium
0%
is not in equilibrium
0%
may or may not be in equilibrium
0%
always has zero mass

State whether given statement is True or False.
Force never exists in pairs.

0%
True
0%
False

Frictional force always opposes:

0%
The state of rest
0%
The state of motion
0%
None of the above
0%
Both the states

A force of

$10 N$ acts on a body of mass

$20 kg$ for

$10 s$ .Change in its momentum is:

0%
$5 kg m/s$
0%
$100 kg m/s$
0%
$200 kg m/s$
0%
$1000 kg m/s$

Explanation

$\Delta P = F\Delta t = 10 \times 10 = 100 kg m/s$

Where,

$\Delta$ P = Change in momentum,

F = Applied force

$\Delta$ t = time for which force has been applied on the body

You are on a frictionless horizontal plane. You can get off the plane :

0%
by jumping
0%
by throwing some pieces of stone
0%
by rolling your body on the surface
0%
by running on the plane

State whether given statement is True or False.
An object moving on a frictionless horizontal surface should move with constant velocity

0%
True
0%
False

The force of friction on a stationary body on a rough surface________

0%
will always be zero
0%
might be zero
0%
will be equal to the force of gravity
0%
will be positive

The momentum of a 10 kg body moving at 36 km/h is:

0%
100 kg.m/s
0%
360 kg. m/s
0%
3.6 kg.m/s
0%
36000 kg .m /s

Explanation

Given :

$m = 10$ kg

$v = 36$ km/hr

$= 36\times \dfrac{5}{18} = 10$ m/s

Momentum of the body

$P = mv = 10\times 10 = 100$ kg.m/s

Momentum is a measure of

0%
Weight
0%
Mass
0%
Quantity of motion
0%
Velocity

The combined effect of mass and velocity is taken into account by a physical quantity called :

0%
torque
0%
moment of force
0%
momentum
0%
moment of momentum

Explanation

Mass ad velocity both are considered to be quantities which defines motion.

The combination of both is called quantity of motion.

It is momentum.

$P = mv$

A metallic ball strikes a wall and falls down whereas a tennis ball having the same mass and velocity bounces back. The conclusion from this is that :

0%
Both suffer equal change in momentum
0%
The tennis ball suffers a greater change in momentum
0%
Metallic ball suffers a greater change in momentum
0%
The momentum of the tennis ball is less than that of the metallic ball

Explanation

For metallic ball,

$\triangle P_M =0-mu=-mu$

For tennis ball,

$\triangle P_T= -mu-mu = -2mu$

$\therefore Option\ B\ is\ correct.$

Impulse is a scalar quantity. State true or false.

0%
True
0%
False

Newton's first law of motion is also called Galileo's law of:

0%
Conservation of momentum
0%
Inertia
0%
Friction
0%
Conservation of mass

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.