If a car is moving at a velocity $$v$$ in a circular road with coefficient of friction $$\mu$$ and banking angle $$\theta$$. Find the radius of the road such that friction force doesn't act on the car in the radial direction.

$$r = \dfrac{\sqrt{v}}{g\tan \theta}$$

$$r = \sqrt{\dfrac{v^2}{g\tan \theta}}$$

MCQ Questions for Class 11 Engineering Physics Laws Of Motion Quiz 7

A particle of mass m strikes a wall with speed v at an angle

30^{0}

$30^0$ with the wall elastically as shown in the figure. The magnitude of impulse imparted to the ball by the wall is :

0%
mv
0%
$\dfrac{mv}{2}$
0%
2mv
0%
$\sqrt{3}$ mv

Explanation

Momentum changing to normal = impulse.

Impulse imparted

= 2 m v sin 30 °

$= 2mv \sin 30°$

=mv

1171610_769745_ans_412e252410f04bb08d92b216415adf34.JPG

Two people push a car for 3 sec, with a combined net force of 200 N.The impulse provided to the car............

0%
400 N-sec.
0%
500 N-sec.
0%
600 N-sec.
0%
300 N-sec.

Explanation

Time of action

t = 3 s

$t = 3 \ s$

Force applied

F = 200 N

$F = 200 \ N$

Impulse

I = F \times t = 200 \times 3 = 600 N s

$I = F\times t = 200\times 3 = 600 \ N \ s$

A simple pendulum with bob of mass

m

$m$ and length

x

$x$ is held in position at an angle

1

$1$ and then angle

2

$2$ with the vertical. When released from these positions, speeds with which it passes the lowest positions are

v_{1}

${v}_{1}$ and

v_{2}

${v}_{2}$ respectively. Then,

\frac{v_{1}}{v_{2}}

$\cfrac { { v }_{ 1 } }{ { v }_{ 2 } }$ is

0%
$\cfrac { 1-\cos { { \theta }_{ 1 } } }{ 1-\cos { { \theta }_{ 2 } } }$
0%
$\sqrt { \cfrac { 1-\cos { { \theta }_{ 1 } } }{ 1-\cos { { \theta }_{ 2 } } } }$
0%
$\sqrt { \cfrac { 2gx(1-\cos { { \theta }_{ 1 } } ) }{ 1-\cos { { \theta }_{ 2 } } } }$
0%
$\sqrt { \cfrac { 1-\cos { { \theta }_{ 1 } } }{ 2gx(1-\cos { { \theta }_{ 2 } } ) } }$

Explanation

A simple pendulum with bob of mass m and length x is held in position at an angle 1 and then angle 2 with the vertical. When released from these positions, speeds with which it passes the lowest positions are

v_{1}

$v_1$ and

v_{2}

$v_2$ respectively.

Kinetic Energy = loss in PE

m v_{1}^{2} / 2 = m g L (1 - c o s θ_{1})

$m v_1^2 /2 = m g L (1 - cos θ_1)$

So,

Similarly

m v_{2}^{2} / 2 = m g L (1 - c o s θ_{2})

$m v_2^2 /2 = m g L (1 - cos θ_2)$

Ratio

\frac{v_{1}}{v_{2}} = \sqrt{\frac{(1 - c o s θ_{1})}{(1 - c o s θ_{1})}}

$\dfrac{v_1}{ v_2} =\sqrt{\dfrac{ (1 - cos θ_1)}{ (1 - cos θ_1)}}$ .

If n balls hit elastically and normally on a surface per unit area and all units has mass m and are moving with same velocity u, then force on surface is:

0%
$mun$
0%
$2mun$
0%
$\dfrac{1}{2}mu^2n$
0%
$mu^2n$

A bullet is fired from a gun. The force on the bullet is given by

F = 600 - 2 \times 10^{5} t

$F = 600 - 2\times 10^{5}t$ , where

F

$F$ is in newtons and

t

$t$ in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

0%
$9\ Ns$
0%
Zero
0%
$0.9\ Ns$
0%
$1.8\ Ns$

Explanation

Force at which time zero u

F = 0 = 600 - 2 \times 10^{5} t t = \frac{600}{2 \times 10^{5}} = 3 \times 10^{- 3} s

$F=0=600-2\times { 10 }^{ 5 }t\\ t=\dfrac { 600 }{ 2\times { 10 }^{ 5 } } =3\times { 10 }^{ -3 }s$

Impulse imparted

= \int_{0}^{t} F d t

$=\int _{ 0 }^{ t }{ Fdt }$

= \int_{0}^{t} (600 - 2 \times 10^{5} t) d t = {[600 t - 1 \times 10^{5} t^{2}]}_{0}^{5} = 1.8 - 0.9 = 0.9 N s

$=\int _{ 0 }^{ t }{ (600-2\times { 10 }^{ 5 } } t)dt\\ =\left[ 600t-1\times { 10 }^{ 5 }{ t }^{ 2 } \right] _{ 0 }^{ 3\times { 10 }^{ -3 } }\\ =1.8-0.9\\ =0.9Ns$

A curved road of diameter

1.8 k m

$1.8\ km$ is banked so that no friction is required at a speed of

30 m / s

$30\ m/s$ . What is the banking angle?

0%
$6^{\circ}$
0%
$16^{\circ}$
0%
$26^{\circ}$
0%
$0.6^{\circ}$

Explanation

Radius of curved road

r = \frac{D}{2} = \frac{1.8 k m}{2} = 900 m

$r = \dfrac{D}{2} = \dfrac{1.8 \ km}{2} = 900 \ m$

Speed

v = 30 m / s

$v = 30 \ m/s$

So, banking angle

tan θ = \frac{v^{2}}{r g}

$\tan \theta =\dfrac{v^2}{rg}$

∴

$\therefore$

tan θ = \frac{(30)^{2}}{900 \times 9.8} = 0.1

$\tan\theta =\dfrac{(30)^2}{900\times 9.8} = 0.1$

⟹ θ = t a n^{- 1} (0.1) = 6^{o}

$\implies \ \theta = tan^{-1}(0.1) = 6^o$

1

$1$ m long uniform beam is being balanced as shown in the given figure. What are the forces X and Y?

0%
Force X- $1$ N, Force Y- $0$ N
0%
Force X- $2$ N, Force Y- $1$ N
0%
Force X- $3$ N, Force Y- $3$ N
0%
Force X- $4$ N, Force Y- $7$ N

A particle is projected so as to just move along a vertical circle of radius

r

$r$ with the help of the massless string. The ratio of the tension in the string when the particle is at the lowest and highest point on the circle is?

0%
$1$
0%
Finite but large
0%
Zero
0%
Infinite

Explanation

T - m g cos θ = \frac{m V^{2}}{r}

$T-mg\cos\theta=\cfrac{mV^2}{r}$

At lowest point

A

$A$ :

T_{A} = \frac{m V_{A}^{2}}{r} + m g cos (0^{o})

$T_A=\cfrac{mV_A^2}{r}+mg\cos (0^o)$

⟹ T_{A} = \frac{m V_{A}^{2}}{r} + m g

$\implies T_A=\cfrac{mV_A^2}{r}+mg$

At highest point

B

$B$ :

T_{B} = 0

$T_B=0$

∴ \frac{T_{A}}{T_{B}} = \frac{(m V_{A}^{2} / r) + m g}{0}

$\therefore \cfrac{T_A}{T_B}=\cfrac{({mV_A^2/r})+mg}{0}$

⟹ \frac{T_{A}}{T_{B}} = \infty

$\implies \cfrac{T_A}{T_B}=\infty$

A solid sphere is placed on a smooth horizontal plane. A horizontal impluse

I

$I$ is applied at a distance

h

$h$ above the central line as shown in the figure. Soon after giving the impulse the sphere starts rolling.
The ratio

h / R

$h/R$ would be

0%
$\dfrac {1}{2}$
0%
$\dfrac {2}{5}$
0%
$\dfrac {1}{4}$
0%
$\dfrac {1}{5}$

Explanation

After application of impulse let velocity

= v

$=v$

I = m v \Rightarrow v = I / m

$I=mv\Rightarrow v=I/m$

Angular impulse about C.M

g m v = I ω \Rightarrow h m v = \frac{2}{5} m R^{2} ω \Rightarrow ω = \frac{5}{2} \frac{h v}{R^{2}}

$gmv=I\omega\\ \Rightarrow hmv=\cfrac{2}{5}mR^2\omega\Rightarrow\omega=\cfrac{5}{2}\cfrac{hv}{R^2}$

For pure rolling

v = r ω \Rightarrow ω = v / r \Rightarrow \frac{v}{R} = \frac{5}{2} \frac{h v}{R^{2}} \Rightarrow \frac{h}{R} = \frac{2}{5}

$v=r\omega\Rightarrow \omega=v/r\Rightarrow \cfrac{v}{R}=\cfrac{5}{2}\cfrac{hv}{R^2}\Rightarrow\cfrac{h}{R}=\cfrac{2}{5}$

A car sometimes overturns while taking a turn. When it overturns, it is

0%
The inner wheel which leaves the ground first
0%
The outer wheel which leaves the ground first
0%
Both the wheels leaves the ground simultaneously
0%
Either wheel which leaves the ground first

Explanation

While turning the maximum allowed speed is:

V_{m a x} = \sqrt{μ r g}

$V_{max}=\sqrt{\mu rg}$

where,

μ

$\mu$ is coefficient of friction,

r

$r$ is radius of the turn and

g

$g$ acceleration due to gravity.

For the inner wheel the radius of the turn is less than the outer wheel and therefore its maximum speed is also less than for the outer wheel. Thus the inner wheel leaves the ground first.

A railway track is banked for a speed v, by making the height of the outer rail h higher than that of the inner rail. The distance between the rails is d. The radius of curvature of the track is r. Then,

0%
$\dfrac{h}{d}=\dfrac{v^2}{rg}$
0%
$tan(sin^{-1}\dfrac{h}{d})=\dfrac{v^2}{rg}$
0%
$tan^{-1}(\dfrac{h}{d})=\dfrac{v^2}{rg}$
0%
$\dfrac{h}{r}=\dfrac{v^2}{dg}$

Explanation

N c o s θ = m g

$Ncos\theta = mg$

N s i n θ = m v^{2} / r

$N sin \theta = mv^2 / r$

t a n θ = v^{2} / r g

$tan \theta = v^2 / rg$

for train, AB= d and BB = h

s i n θ = h / d

$sin \theta = h/d$

θ = s i n^{- 1} (h / d)

$\theta = sin ^{-1}(h/d)$

t a n θ = t a n (s i n^{- 1} h / d) = v^{2} / r g

$tan \theta = tan (sin ^{-1 }h/d)= v^2 / rg$

So, option B is correct.

"To every action, there is equal and opposite reaction". It is Newton's

0%
First Law
0%
Second Law
0%
Third Law of Motion
0%
None of these

A simple pendulum of length

L

$L$ carries a bob of mass

m

$m$ . When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal the net force on the bob is:

0%
$\sqrt { 10 } mg$
0%
$\sqrt { 5 } mg$
0%
$4mg$
0%
$1mg$

Explanation

V=minimum horizontal speed=

\sqrt{5 L g}

$\sqrt { 5Lg }$

using energy conservation,

\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} = m g L ∴ \frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} - m g L = \frac{1}{2} m 5 g L - m g L ∴ v^{2} = \frac{3 g L}{2} \times 2 = 3 g L T = \frac{m v^{2}}{L} = 3 m g ∴ f o r c e = \sqrt{{(3 m g)}^{2} + {(m g)}^{2}} = \sqrt{10} m g

$\frac { 1 }{ 2 } m{ v }^{ 2 }-\frac { 1 }{ 2 } m{ u }^{ 2 }=mgL\\ \therefore \quad \frac { 1 }{ 2 } m{ u }^{ 2 }=\frac { 1 }{ 2 } m{ v }^{ 2 }-mgL\\ =\frac { 1 }{ 2 } m{ 5gL }-mgL\\ \therefore \quad { v }^{ 2 }=\frac { 3gL }{ 2 } \times 2=3gL\\ T=\frac { m{ v }^{ 2 } }{ L } =3mg\\ \therefore \quad force=\quad \sqrt { { \left( 3mg \right) }^{ 2 }+{ \left( mg \right) }^{ 2 } } \quad \\ =\sqrt { 10 } mg$

The term inertia was first used by

0%
Newton
0%
Galileo
0%
Aristotle
0%
Kepler

Explanation

Galileo, a premier scientist in the seventeenth century, developed the concept of inertia. Galileo reasoned that moving objects eventually stop because of a force called friction. First thing it was discovered and not invented. It was discovered and stated and proved by Sir Isaac Newton.

Hence,

option

B

$B$ is correct answer.

A rigid ball of mass m strikes a rigid wall at

60^{0}

$60^0$ and gets reflected without loss of speed as shown in the figure. The value of impulse imparted by the wall on the ball will be:

0%
mV
0%
2 mV
0%
$\dfrac{mV}{2}$
0%
$\dfrac{mV}{3}$

Explanation

Given,

P_{i} = P_{f} = m V

$P_i= P_f= mV$
Change in momentum of the ball =

{¯ ¯¯ ¯ P}_{f} - {¯ ¯¯ ¯ P}_{i}

$\overline {P}_f-\overline{P}_i$
=

(- P_{f x}^i - P_{f y}^j) - (P_{i x}^i - P_{i y}^j)

$(-P_{fx}\hat{i}-P_{fy}\hat{j})-(P_{ix}\hat{i}-P_{iy}\hat{j})$

-^i (P_{f x} + P_{i x}) -^j (P_{f y} - P_{i y})

$-\hat{i}(P_{fx}+P_{ix})-\hat{j}(P_{fy}-P_{iy})$

- 2 P_{i x}^i = - m V^i

$-2P_{ix}\hat{i}= -mV\hat{i}$

[P_{f y} - P_{i y} = 0]

$[ P_{fy}-P_{iy}=0]$

Impulse imparted by the wall = change in the momentum of the ball = mV.

1078399_939483_ans_5354fdb4bd214733a7c5f48567da0a08.png

A girl presses her physics textbook against a rough vertical wall with her hand. The direction of the frictional force on the book exerted by the wall is:

0%
Downwards
0%
Upwards
0%
Out from the wall
0%
Into the wall

If a car is moving at a velocity

v

$v$ in a circular road with coefficient of friction

μ

$\mu$ and banking angle

θ

$\theta$ . Find the radius of the road such that friction force doesn't act on the car in the radial direction.

0%
$r=\dfrac{v^2}{g\tan \theta}$
0%
$r = \dfrac{\sqrt{v}}{g\tan \theta}$
0%
$r = \sqrt{\dfrac{v^2}{g\tan \theta}}$
0%
$r=\dfrac{v^2}{\sqrt{g}}$

Explanation

For the force of friction to be zero, the centripetal force has to be provided by the normal reaction force only. Which means

N s i n (θ) = \frac{m v^{2}}{r}

$N\ sin(\theta)=\dfrac{mv^2}{r}$ .....................

(1)

$(1)$

now the other component of the normal reaction

N c o s (θ)

$N\ cos(\theta)$ balances the weight of the car.

\Rightarrow N c o s (θ) = m g

$\Rightarrow N\ cos(\theta)=mg$ ....................

(2)

$(2)$

Using this in

(1)

$(1)$ we get

m g t a n (θ) = \frac{m v^{2}}{r}

$mg\ tan(\theta)=\dfrac{mv^2}{r}$

\Rightarrow r = \frac{v^{2}}{g t a n (θ)}

$\Rightarrow r=\dfrac{v^2}{g\ tan(\theta)}$

A car is moving on a frictionless (in the radial direction) circular banked road, with a banking angle of

15^{o}

$15^o$ , find the velocity of the car.
Mass of car is

600 k g

$600\ kg$ , the radius of the circular road

= 10 m

$=10m$

Assume

g = 9.81 m / s^{2}

$g=9.81\ m/s^2$

0%
4.26 m/s
0%
5.12 m/s
0%
51.2 m/s
0%
42.6 m/s

Explanation

We know that the force acting on the car = mg

t a n (θ)

$tan(\theta)$

\Rightarrow \frac{m v^{2}}{r} = m g t a n (θ)

$\Rightarrow \dfrac{mv^2}{r}=mg\ tan(\theta)$

\Rightarrow v^{2} = r g t a n (θ)

$\Rightarrow v^2=rg\ tan(\theta)$

\Rightarrow v^{2} = 10 \times 9.81 \times 0.268

$\Rightarrow v^2=10\times 9.81 \times 0.268$

\Rightarrow v = 5.12 m / s

$\Rightarrow v=5.12\ m/s$

A ball of mass m strikes a rigid wall with speed u and rebounds with the same speed. The impulse imparted to the ball by the wall is:

0%
2mu
0%
mu
0%
Zero
0%
-2mu

Explanation

The situation is as shown in the figure.

p_{i n i t i a l} = m u, p_{f i n a l} = - m u

$p_{initial} = mu, p_{final}=-mu$
Impulse imparted to the ball = change in momentum
=

p_{i n i t i a l} = m u, p_{f i n a l} = - m u - m u = - 2 m u

$p_{initial} = mu, p_{final}=-mu-mu=-2mu$
1077603_938370_ans_791f37012be54ed2aef6ec71af1f9f9b.png

A particle is tied to one end of a light inextensible string and is moved in a vertical circle, the other end of the string is fixed at the centre. Then for a complete motion in a circle, which is correct.

(air resistance is negligible).

0%
Acceleration of the particle is directed towards the centre
0%
Total mechanical energy of the particle and earth remains constant
0%
Tension in the string remains constant
0%
Acceleration of the particle remains constant

A house is on built on the top of a hill with

45^{0}

$45^0$ slope. Due to the sliding of material and sand from top to the bottom of hill, the slope angle has been reduced.
If the coefficient of static friction between sand particles is 0.75, what is the final angle attained by hill? (

t a n^{- 1} 0.75 = 37^{0}

$tan^{-1}0.75 = 37^0$ )

0%
$8^0$
0%
$45^0$
0%
$37^0$
0%
$30^0$

Explanation

In equilibrium, when sliding stops

$N=mg\cos\theta…(1)$

And $f=mg\sin\theta=\mu N…(1)$

Dividing $(2)$ by $(1)\\ \Rightarrow \mu=\tan\theta=0.75=\cfrac{3}{4}\\ \Rightarrow \theta=37°$

1120613_987189_ans_0fdb2de124ae46f9a7c76bcabe2da021.png

A box of mass 8 kg is placed on a rough inclined plane of inclined

θ

$\theta$ . Its downward motion can be prevented by applying an upward pully F and it can be made to slide upwards by applying a force 2F. The coefficient of friction between the box and the inclined plane is

0%
$(tan\theta)/3$
0%
$tan\theta$
0%
$(tan\theta)/2$
0%
$2tan\theta$

Explanation

For downward direction

Force equation in perpendicular direction is given as

$N=mg\cos \theta$

${{f}^{'}}=\mu mg\cos \theta$

Force equation parallel to incline is

$f+{{f}^{'}}=mg\sin \theta$

$f+\mu mg\cos \theta =mg\sin \theta$

$f=mg\sin \theta -\mu mg\cos \theta ..............(1)$

For upward direction

Force equation in perpendicular direction is given as

$N=mg\cos \theta$

${{f}^{'}}=\mu N$

$f'=\mu mg\cos \theta$

Force equation parallel to incline is

$2f=mg\sin \theta +f'$

$2f=mg\sin \theta +\mu mg\cos \theta ............(2)$

From equation (1),equation (2) becomes

$2mg\sin \theta -2\mu mg\cos \theta =mg\sin \theta +\mu mg\cos \theta$

$mg\sin \theta =3\mu mg\cos \theta$

$\tan \theta =3\mu$

$\mu =\frac{\tan \theta }{3}$

1007159_985746_ans_08559547427b4eca8c36bc48bdadad7c.jpg

Two wires

A C

$AC$ and

B C

$BC$ are tied at

C

$C$ of a small sphere of mass

5 k g

$5\ kg$ , which revolves at a constant speed

v

$v$ in the horizontal plane with the speed

v

$v$ of radius

1.6 m

$1.6\ m$ . Find the minimum value of

v

$v$ .

0%
$4\ m{s}^{-1}$
0%
$2\ m{s}^{-1}$
0%
$2.5\ m{s}^{-1}$
0%
None of these

Explanation

From

F B D

$FBD$

T_{1} cos 30^{0} + T_{2} cos 45^{0} = m g

$T_1\cos 30^0+T_2\cos 45^0=mg$

T_{1} sin 30^{0} + T_{2} sin 45^{0} = \frac{m v^{2}}{r}

$T_1\sin 30^0+T_2\sin 45^0=\dfrac{mv^2}{r}$

T_{1} sin 30^{0} - T_{1} cos 30^{0} = \frac{m v^{2}}{R} - m g

$T_1\sin 30^0-T_1\cos 30^0=\dfrac{mv^2}{R}-mg$

∵ {sin 45^{0} = cos 45^{0}}

$\because\{\sin45^0=\cos 45^0\}$

T_{1} = \frac{\frac{m v^{2}}{R} - m g}{[\frac{1}{2} - \frac{\sqrt{3}}{2}]}

$T_1=\dfrac{\dfrac{mv^2}{R}-mg}{\left[\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\right]}$

as it a string tension, So it can't be negative

T_{1} > 0

$T_1 > 0$ i.e

\frac{m v^{2}}{R} - m g \geq 0

$\dfrac{mv^2}{R}-mg \ge 0$

\frac{v_{2}}{r} - g \geq 0

$\dfrac{v_2}{r}-g \ge 0$

v \geq \sqrt{g R}

$v\ge \sqrt{gR}$

V_{m i n} \sqrt{g r} = \sqrt{10 \times 1.6} = \sqrt{16} = 4 m s^{- 1}

$V_{min}\sqrt{gr}=\sqrt{10\times 1.6}=\sqrt{16}=4ms^{-1}$

Hence (A) is the correct options.

1389702_987320_ans_07a0cf95b71045ae8a6a3848b5ba15ea.png

A particle of mass

m

$m$ slides on a frictionless surface ABCD, starting from rest as shown in Figure. The part BCD is a circular arc. If it loses contact at point P, the maximum height attained by the particle from point C is :

0%
$R[2 +\dfrac{1}{2\sqrt{2}}]$
0%
$R[2 +\dfrac{1}{2\sqrt{2}}]R$
0%
$3R$
0%
None of these

Explanation

m g \times 2 R - m g \frac{R}{\sqrt{2}} = \frac{1}{2} m v^{2}

$mg \times 2R - mg \dfrac{R}{\sqrt{2}} = \dfrac{1}{2} mv^2$ __(1)

V = \sqrt{4 g R - \sqrt{2} g R}

$V = \sqrt{4g R - \sqrt{2} g R}$

h_{1} = \frac{V^{2} {sin}^{2} θ}{2 g}

$h_1 = \dfrac{V^2 \sin^2 \theta}{2g}$

= \frac{V^{2} {sin}^{2} 45}{2 g}

$=\dfrac{V^2 \sin^2 45}{2g}$

H_{m a x} = R + \frac{R}{\sqrt{2}} + \frac{(4 g R - \sqrt{2} g R)}{2 g \times \sqrt{2}}

$H_{max} = R + \dfrac{R}{\sqrt{2}} + \dfrac{(4g R - \sqrt{2} g R)}{2g \times \sqrt{2}}$

H_{m a x} = R + \frac{R}{\sqrt{2}} + \frac{(4 g R - \sqrt{2} g R)}{2 g} \times \frac{1}{\sqrt{2}}

$H_{max} = R + \dfrac{R}{\sqrt{2}} + \dfrac{(4g R - \sqrt{2} g R)}{2g} \times \dfrac{1}{\sqrt{2}}$

= R [2 + \frac{1}{2 \sqrt{2}}]

$= R \left[2 + \dfrac{1}{2 \sqrt{2}} \right]$

1451517_984147_ans_1509b921e03549ebabee211f7e4225e2.png

A block of mass m rests on a rough inclined plane. The coefficient of friction between the surface and the block if

μ

$\mu$ . At what angle of inclination

θ

$\theta$ of the plane to the horizontal will the block just start to slide down the plane?

0%
$\theta=tan^{-1}\mu$
0%
$\theta=cos^{-1}\mu$
0%
$\theta=sin^{-1}\mu$
0%
$\theta=sec^{-1}\mu$

Explanation

The various forces acting on the block are as shown in the figure.
From figure

m g s i n θ = f

$mgsin\theta=f$ ............(i)

m g c o s θ = N

$mgcos\theta =N$ .........(ii)
Divide (i) by (ii), we get

t a n θ = \frac{f}{N} = \frac{μ N}{N}

$tan\theta=\dfrac{f}{N}=\dfrac{\mu N}{N}$ or

θ = t a n^{- 1} (μ)

$\theta = tan^{-1}(\mu)$
1078449_939742_ans_394d270640204646b20550447e693358.png

The velocity of a body moving in a vertical circle of radius r is

\sqrt{7 g r}

$\sqrt{7gr}$ at the lowest point of the circle. What is the ratio of maximum and minimum tension?

0%
$4: 1$
0%
$\sqrt 7:1$
0%
$3: 1$
0%
$2: 1$

Explanation

Since

V_{b} > \sqrt{5 g r} \to

$V_{b} >\sqrt{5 gr}\rightarrow$ It will complete vertical circular

↓

$\downarrow$

velocity at bottom

Thus

T_{b o t t o m} - T_{t o p} = 6 m g

$T_{bottom}-T_{top}=6\ mg$

T_{m a x} - T_{m i n} = 6 m g . . . . . . (1)

$T_{max}-T_{min}=6\ mg ...... (1)$

Now,

\Rightarrow T_{m a x} - m g = m (a_{c})

$\Rightarrow T_{max}-mg=m(a_{c})$

T_{m a x} - m g = \frac{m v^{2}}{r}

$T_{max}-mg =\dfrac{mv^{2}}{r}$

T_{m a x} = \frac{m}{r} (7 g r) + m g

$T_{max}=\dfrac{m}{r}(7 gr)+mg$

T_{m a x} = 8 m g

$T_{max}=8\ mg$

From

(1)

$(1)$

T_{m i n} = 2 m g

$T_{min}=2\ mg$

Thus

\frac{T_{m a x}}{T_{m i n}} = 4 : 1

$\dfrac{T_{max}}{T_{min}}=4:1$

1441537_950200_ans_f7a138257f11432b8308f2de8be44e5d.png

Fig A.43 show three block on a rough surface under the influence of a force

P

$P$ of the same magnitude in all the three cases.
Co0efficient of friction is the same between each block and the ground. What possible relation holds between the magnitude of normal
reaction and friction forces? (Assume that the block do not overturn about edge.) Here,

f_{A}

${ f }_{ A }$ ,

f_{B}

${ f }_{ B }$ and

f_{C}

${ f }_{ C }$ are frictional forces and

N_{A}

${ N }_{ A }$ ,

N_{B}

${ N }_{ B }$ and

N_{C}

${ N }_{ C }$ are reactions.

0%
${ N }_{ A }\ >\ { N }_{ C }\ >\ { N }_{ B }$
0%
${ f }_{ A }\ >\ { f }_{ C }\ >\ { f }_{ B }$
0%
${ f }_{ C }\ >\ { f }_{ A }\ =\ { f }_{ B }$
0%
${ N }_{ C }\ >\ { N }_{ A }\ =\ { N }_{ B }$

Explanation

N_{A} = m g + P sin θ

$N_A=mg+P\sin \theta$

N_{B} = m g - P sin θ

$N_B=mg-P\sin \theta$

N_{C} = m g

$N_C=mg$
Clearly,

N_{A} > N_{C} > N_{B}

$N_A > N_C > N_B$
Now to find relation between frictions:
(i) Let sliding does not occur:

f_{A} = P cos θ, f_{b} = P cos θ, f_{C} = P

$f_A=P\cos \theta, f_b =P\cos \theta , f_C=P$

\Rightarrow f_{C} > f_{A} = f_{B}

$\Rightarrow f_C > f_A =f_B$
(ii) Let sliding occur at all contact surfaces :

f_{A} = μ N_{A}, f_{B} = μ N_{B}, f_{C} = μ N_{C}

$f_A=\mu N_A, f_B=\mu N_B, f_C=\mu N_C$

\Rightarrow f_{A} > f_{C} > f_{B}

$\Rightarrow f_A > f_C > f_B$

For the conditions of the equilibrium of the body, i.e. the rigid body only the external forces defines the equilibrium. Because the internal forces cancels out so not to be considered.

0%
The first part of the statement is false and other part is true
0%
The first part of the statement is false and other part is false too
0%
The first part of the statement is true and other part is false
0%
The first part of the statement is true and other part is true too

A man revolves a stone of mass m tied to the end of a string in a vertical circle of radius R, The net force at the lowest and height points of the circle directed vertical downwards are
Here

T_{1}, T_{2}

$T_1, T_2$ and

v_{1}, v_{2}

$v_1, v_2$ denote the tension in the string and the speed of the stone at the lowest and highest points, respectively.

0%
Lowest point: $mg - T_1$ Highest point : $mg + T_2$
0%
Lowest point: $mg + T_1$ Highest point : $mg - T_2$
0%
Lowest point: $mg + T_1 -\frac{mv^2}{R}$ Highest point: $mg - T_2 +\frac{mv^2}{r}$
0%
Lowest point: $mg - T_1 -\frac{mv^2}{R}$ Highest point: $mg + T +\frac{mv^2}{r}$

A box of mass m is made to move from A to B as shown in figure. Where will the normal force be larger:

0%
A
0%
B
0%
Same at both A and B
0%
None of the above

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"

0%
Motion; Rest; External Force
0%
Rest; Motion; External Force
0%
Rest; Motion; Internal Force
0%
None of these

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

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

Explanation

Since momentum

p = m v

$p=mv$

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

Which of the following is correct?

0%
The application of the conditions of the equilibrium of the body is valid only in the 2D
0%
The application of the conditions of the equilibrium of the body is valid only in the 3D
0%
The application of the conditions of the equilibrium of the body is valid only in the 1D
0%
The application of the conditions of the equilibrium of the body is valid throughout

A particle of mass 1 kg is suspended by means of a string of length L=2 m. The string makes

6 / π

$6/\pi$ rps around a vertical axis through the fixed end. The tension in the string is

0%
72 N
0%
36 N
0%
288 N
0%
10 N

Explanation

The tension is the centripetal force =

m w^{2} R

$mw^2R$

The angular velocity is w=

(6 / π) (2 π) = 12

$(6/\pi)(2 \pi) =12$ rads/s

Substituting we get, Tension =

1 \times 144 \times 2 = 288 N

$1 \times 144 \times 2 = 288 N$

Thus the correct options is (c)

A 0.25kg ball attached to a 1.5 m rope moves with a constant speed of 15 m/s around a vertical circle. Calculate the tension force on the rope at the middle of the circle:

0%
37.5 N
0%
137.5 N
0%
2.5 N
0%
25 N

Explanation

The centripetal force acts along the direction of the tension and hence

T = m v^{2} / R = 0.25 \times 15^{2} / 1.5 = 37.5 N

$T=mv^2/R = 0.25 \times 15^2/1.5=37.5 N$

A mass attached to a string that is itself attached to the ceiling swings back and forth. If the bob is observed to be moving upward at a given instance, as shown to the right, which arrow best depicts the direction of the net force acting on the bob at that instant

0%
A
0%
B
0%
C
0%
D

Explanation

Two forces act on the bob. (i) is the tension in the string and the (ii) mg sin

θ

$\theta$ , which will be tangential to the path. The resultant of both the forces will be along vector C

Thus the correct option is (c)

Two blocks of same mass (

4

$4$ kg) are placed according to diagram. Initial velocities of bodies are

4

$4$ m/s and

2

$2$ m/s and the string is taut. Find the impulse on

4

$4$ kg when the string again becomes taut.

0%
$24$ N-s
0%
$6$ N-s
0%
$4$ N-s
0%
$2$ N-s

Explanation

s = 4 t = 2 t + 5 t^{2}

$s=4t=2t+5t^2$

5 t^{2} = 2 t

$5t^2=2t$

t = 0.4

$t=0.4$

v = 2 + g t = 6

$v=2+gt=6$ m/s

4 \times 4 \times 4 \times 6 = 8

$4\times 4\times 4\times 6=8$ v

v = 5

$v=5$

J = 4 (v - 4) = 4

$J=4(v-4)=4$ .

Momentum of an object changes from

5 k g m / s

$5\ kg\ m/s$ to

15 k g m / s

$15\ kg\ m/s$ in

2

$2$ seconds. What is the force applied on the object?

0%
$10\ N$
0%
$5\ N$
0%
$20\ N$
0%
$40\ N$

Explanation

F = \frac{Δ p}{Δ t} = \frac{(p 2 - p 1)}{t} = \frac{(15 - 5)}{2} = 5 N

$F = \dfrac{\Delta p}{\Delta t} = \dfrac{(p2 - p1)}{t} = \dfrac{(15-5)}{2} = 5\ N$

A football player kicks a

0.25 k g

$0.25kg$ ball and imparts it a velocity of

10 m / s

$10m/s$ . The contact between foot and the ball is only

\frac{1}{50}

$\cfrac{1}{50}$ th of a second. The kicking force is

0%
$250N$
0%
$125N$
0%
$0N$
0%
$3.78N$

Explanation

Force =

\frac{△ P}{△ t}

$\dfrac{\triangle P}{\triangle t}$ (Newtons 2nd law)

F = \frac{m △ V}{△ f} = \frac{m (V_{f} - V_{i})}{△ f}

$F=\dfrac{m\triangle V}{\triangle f}=\dfrac{m(V_f-V_i)}{\triangle f}$

Force,

m = 0.25 k g

$m=0.25kg$

and

V_{f} = 10 m / s, V_{i} = 0 m / s

$V_f=10m/s,V_i=0m/s$

∴ F = \frac{(0.25) (10 - 0)}{\frac{1}{50}}

$\therefore F=\dfrac{(0.25)(10-0)}{\dfrac{1}{50}}$

F = 0.25 (500)

$F=0.25(500)$

F = 25 \times 5

$F=25\times 5$

F = 125 N

$\boxed{F=125N}$

A weightless thread can bear tension up to

30 N

$30N$ . A stone of mass

500 g

$500\ g$ is tied to it and revolved in a circular path of radius

2 m

$2m$ in a vertical plane. If

g = 10 {m s}^{- 2}

$g=10{ms}^{-2}$ , then the maximum angular velocity of the stone will be:

0%
$2\ rad\ {s}^{-1}$
0%
$5\ rad\ {s}^{-1}$
0%
$8\ rad\ {s}^{-1}$
0%
$16\ rad\ {s}^{-1}$

Explanation

${ T }_{ max }=\cfrac { m{ v }_{ max }^{ 2 } }{ r } +mg$

where ${ T }_{ max }$ is maximum tension

$m$ is mass

$x$ is radius

and $g$ is acceleration due to gravity

${ T }_{ max }=m{ { w }_{ max }^{ 2 } }r+mg$

where $w=\cfrac { v }{ r }$ is angular velocity

$\cfrac { { T }_{ max } }{ m } =m{ { w }_{ max }^{ 2 } }r+g$

$\cfrac { 30 }{ 0.5 } ={ { w }_{ max }^{ 2 } }2+10\\ \Rightarrow { w }_{ max }^{ 2 }=25\\ \Rightarrow { w }_{ max }=5 rad/s$

Banking of roads is done due to

0%
provide enough friction for circular motion of the vehicle
0%
provide necessary centripetal force required for circular motion of the vehicle
0%
provide enough radius of curvature for circular motion of the vehicle
0%
provide enough area for navigating in the circular motion of the vehicle

The figure shows the position -time (x- t) graph of one-dimensional motion of a body of mass

0.4

$0.4$ kg. The magnitude of each impulse is :

0%
$0.4$ Ns
0%
$0.8$ Ns
0%
$1.6$ Ns
0%
$0.2$ Ns

Explanation

0.8 N s

$0.8Ns$

Impulse= change in momentum

= final momentum- initial momentum

Initial velocity,

u = \frac{2 - 0}{2 - 0} = 1 m / s^{- 1}

$u=\dfrac{2-0}{2-0}=1m/s^{-1}$

Final velocity,

v = \frac{0 - 2}{4 - 2} = - 1 m / s^{- 1}

$v=\dfrac{0-2}{4-2}=-1m/s^{-1}$

Initial momentum,

P_{i} = m u = 0.4 \times 1 = 0.4 N s

$P_i= mu=0.4\times1=0.4Ns$

Final momentum,

P_{f} = m v = 0.4 \times (- 1) = - 0.4 N s

$P_f=mv= 0.4\times (-1)=-0.4Ns$

Impulse

= P_{f} - P_{i}

$=P_f-P_i$

= - 0.4 - 0.4 = - 0.8 N s

$=-0.4-0.4=-0.8Ns$

The magnitude of each impulse is

0.8 N s

$0.8Ns$

Impulse of the force exerted by A and B during the collision, is equal to

0%
$\left( \sqrt { 3 } mi+3mj \right) kg-m/s$
0%
$\left( \cfrac { \sqrt { 3 } }{ 2 } mi-\sqrt { 3 } mj \right) kg-m/s$
0%
$\left( 3mi-\sqrt { 3 } mj \right) kg-m/s$
0%
$\left( 2\sqrt { 3 } mi+3mj \right) kg-m/s\quad$

Explanation

{\to J}_{A o n B} = m {\to v}_{B_{f}} - {\to v}_{B_{i}} = m [4 cos 30^{o} (cos 30^i - sin 30^{o} j) - 0] = (3 m i - \sqrt{3} m j) k g - m / s

${ \overrightarrow { J } }_{ Aon B }=m{ \overrightarrow { v } }_{ { B }_{ f } }-{ \overrightarrow { v } }_{ { B }_{ i } }=m\left[ 4\cos { { 30 }^{ o } } \left( \cos { 30\hat { i } } -\sin { { 30 }^{ o }j } \right) -0 \right] =\left( 3mi-\sqrt { 3 } mj \right) kg-m/s$

Though friction can provide necessary centripetal force, the banking of roads is considered as advantageous in highways:

0%
True
0%
False

A block of mass

m

$m$ is on an inclined plane of angle of angle

θ

$\theta$ . The coefficient of friction between the block and the plane is

μ

$\mu$ and

tan θ > μ

$\tan { \theta } >\mu$ . The block is held stationary by applying a force

P

$P$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As

P

$P$ is varied from

P_{1} = m g (sin θ - μ cos θ)

${ P }_{ 1 }=mg\left( \sin { \theta } -\mu \cos { \theta } \right)$ to

P_{2} = m g (sin θ - μ cos θ)

${ P }_{ 2 }=mg\left( \sin { \theta } -\mu \cos { \theta } \right)$ , the frictional force

f

$f$ versus

P

$P$ graph will look like

A vehicle is at rest on a banked road with angle of banking

θ

$\theta$ . The normal reaction of the vehicle is

N_{1}

${N}_{1}$ . When the vehicle takes a turn on the same road the normal reaction is

N_{2}

${N}_{2}$ . Then

N_{1} / N_{2}

${N}_{1}/{N}_{2}$ is equal to:

0%
$1$
0%
$\sin ^{ 2 }{ \theta }$
0%
$\cos ^{ 2 }{ \theta }$
0%
$0.5 \sin (2\theta)$

Explanation

N_{1} = m g cos θ

${N_1} = mg\cos \theta$

- - - - - - (1)

$------(1)$

When it takes a turn on the same road

(\frac{m v^{2}}{r})

$\left( {\dfrac{{m{v^2}}}{r}} \right)$ is the centripetal force acting on it

.

$.$

Balancing the force

m g sin θ = \frac{m v^{2}}{r} cos θ

$mg\sin{\theta} = \dfrac{{m{v^2}}}{r}\cos \theta$

;

$;$

\frac{v^{2}}{r} = g tan θ

$\dfrac{{{v^2}}}{r} = g\tan \theta$

N_{2} = m g cos θ + \frac{m v^{2}}{r} sin θ = m g cos θ + m g tan θ sin θ

${N_2} = mg\cos \theta + \dfrac{{m{v^2}}}{r}\sin \theta \, = mg\cos \theta + mg\tan \theta \,\sin \theta$

- - - - - - - - - - - (2)

$-----------(2)$

\frac{N_{1}}{N_{2}} = \frac{m g cos θ}{m g cos θ + m g sin θ tan θ} = \frac{cos θ}{cos θ \frac{{sin}^{2} θ}{cos θ} = \frac{cos θ}{\frac{{sin}^{2} θ + {cos}^{2} θ}{cos θ}}} = {cos}^{2} θ

$\dfrac{{{N_1}}}{{{N_2}}} = \dfrac{{mg\cos \theta }}{{mg\cos \theta + mg\,\sin \theta \,\tan \theta \,}}\, = \dfrac{{\cos \theta }}{{\cos \theta \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} = \dfrac{{\cos \theta }}{{\dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta }}}}}} = {\cos ^2}\theta$

Hence

,

$,$

Option

(c)

$(c)$ is correct

1148235_1022199_ans_0c52fbba97b6404c8f4a8da2769a66ca.png

A cyclist leans with the horizontal at an angle of

30^{o}

$30^{o}$ , while negotiating round a circular road of radius

20 \sqrt{3}

$20\sqrt{3}$ meters. Speed of the cycle should be :

0%
$7\sqrt{3}\ m/s$
0%
$14\ m/s$
0%
$7\sqrt{6}\ m/s$
0%
$14\sqrt{3}\ m/s$

Explanation

Balancing the forces:

N sin 30 ° = m g ⟶ (1)

$N\sin { 30° } =mg\longrightarrow (1)$

and

N cos 30 ° = \frac{m v^{2}}{R} ⟶ (2)

$N\cos { 30 } °=\cfrac { m{ v }^{ 2 } }{ R } \longrightarrow (2)$

dividing

(2)

$(2)$ by

(1)

$(1)$

tan 30 ° = \frac{g R}{v^{2}} \Rightarrow v^{2} = \frac{g R}{tan 30 °} \Rightarrow v^{2} = 10 \times 20 \sqrt{3} \times \sqrt{3} = 200 \times 3 = 600 v = \sqrt{600} = 24.5 m / s = 14 \sqrt{3} m / s

$\tan { 30° } =\cfrac { gR }{ { v }^{ 2 } } \\ \Rightarrow { v }^{ 2 }=\cfrac { gR }{ \tan { 30° } } \\ \Rightarrow { v }^{ 2 }=10\times 20\sqrt { 3 } \times \sqrt { 3 } \\ \quad \quad =200\times 3=600\\ v=\sqrt { 600 } =24.5\quad m/s\\ \quad \quad \quad \quad =14\sqrt { 3 } m/s$

1036209_1024495_ans_eda1b8d4c8a14742adbd466fed760239.png

A weightless thread can support tension upto

30 N

$30N$ . A particle of mass

0.5 k g

$0.5kg$ is tied to it and is revolved in a circle of radius

2 m

$2m$ in a vertical plane. If

g = 10 m / s^{2}

$g=10m/{s}^{2}$ , then the maximum angular velocity of the stone will be

0%
$5rad/s$
0%
$\sqrt{30}rad/s$
0%
$\sqrt {60}rad/s$
0%
$10rad/s$

Explanation

T - m g = m ω^{2} r T_{m a x} = m g + m ω^{2} r ω^{2} = (\frac{T - m g}{m r}) = (\frac{30 - 5}{0.5 (2)}) = 5 r a d / s e c

$T-mg=m\omega ^{ 2 }r\\ { T }_{ max }=mg+m\omega ^{ 2 }r\\ { \omega }^{ 2 }=\left( \frac { T-mg }{ mr } \right) \\ =\left( \frac { 30-5 }{ 0.5(2) } \right) \\ =5rad/sec$
978656_1025451_ans_24441a7ebe8c4447a51626af787a601d.png

978656_1025451_ans_24441a7ebe8c4447a51626af787a601d.png

A car is moving in a circular horizontal track of radius

10 m

$10m$ with a constant speed of

10 m / s

$10m/s$ . A plumb bob is suspended from the roof of the car by a light rigid rod of length

1 m

$1m$ . The angle made by the rod with track is

0%
zero
0%
${30}^{o}$
0%
${45}^{o}$
0%
${60}^{o}$

Explanation

T cos

θ

$\theta$ = mg..........(1)

T Sin

θ

$\theta$ =

\frac{m v^{2}}{r}

$\dfrac {mv^2}r$ .....(2)

E q u^{n} (2) / (1)

$Equ^n\ (2)/(1)$

t a n θ = \frac{v^{2}}{r g} = \frac{100}{100} = 1

$tan \theta = \dfrac{v^2}{rg} = \dfrac{100}{100} =1$
I used

g = 10

$g=10$ and we get

θ = 45^{0}

$\theta =45^0$

959429_1022490_ans_93dee9f629c5480e8eaeaa2f9230d572.PNG

A road is

8 m

$8\ m$ wide. Its radius of curvature is

40 m

$40\ m$ . The outer edge is above the lower edge by distance of

1.2 m

$1.2\ m$ . The most suited velocity on the road is nearly:

0%
$5.7\ ms^{-1}$
0%
$8\ ms^{-1}$
0%
$36.1\ ms^{-1}$
0%
$9.7\ ms^{-1}$

Explanation

We know,

\frac{v^{2}}{r g} = \frac{h}{l}

$\frac{v^2}{rg}=\frac{h}{l}$

v = \sqrt{\frac{40 \times 1.2 \times 10}{8}}

$v=\sqrt{\frac{40\times1.2\times10}{8}}$

v = 7.74 \approx 8 m / s

$v=7.74\approx8 m/s$

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 7 - MCQExams.com

0:0:2

Practice Class 11 Engineering Physics Quiz Questions and Answers

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