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

If an object is in equilibrium, which of the following statements is not true?

0%
The speed of the object remains constant.
0%
The acceleration of the object is zero.
0%
The net force acting on the object is zero
0%
The object must be at rest.
0%
There are at least two forces acting on the object.

Explanation

Both answers (D) and (E) are not true: (D) is not true because the value of the velocity’s constant magnitude need not be zero, and (E) is not true because there may be no force acting on the object. An object in equilibrium has zero acceleration

$(\vec{a} = 0)$ , so both the magnitude and direction of the object’s velocity must be constant. Also, Newton’s second law states that the net force acting on an object in equilibrium is zero.

Answer the question:
The diagram shows a uniform beam being used as a balance. The beam is pivoted at its centre.
A

$1.0\,N$ weight is attached to one end of them. An empty pan weighing

$0.2\,N$ is attached to the other end of beam.
How many

$0.1\,N$ weights must be placed on the pan in order to balance the beam ?

0%
$5$
0%
$8$
0%
$10$
0%
$12$

Explanation

To balance the beam, net torque about pivot should be zero.

$\tau_{net}=\tau_p+\tau_w$

Let we need x number of weights.

$\tau_{net}=1\times l-(0.2+0.1x)l=0$

$\implies 1=0.2+0.1x$

$x=8$

so we need 8 weights of 0.1 N to balance beam.

A block is freely sliding down from a vertical height

$4\ m$ on a smooth inclined plane. The block reaches bottom of inclined plane then it describes vertical circle of radius

$1\ m$ along smooth track. The ratio of normal reactions on the block while it is crossing lowest point and highest point of vertical circle is:

0%
$6 : 1$
0%
$5 : 1$
0%
$3 : 1$
0%
$5 : 2$

Explanation

applying the law of conservation of energy at point A and B, we have

$mgh=\frac{1}{2}m{v_1}^2 \\\Rightarrow {v_1}=\sqrt{2gh}\\\Rightarrow {v_1}=\sqrt{8g}$
Now, net force is towards the center is the centripetal force, we have at point B

$\frac{m{v_1}^2}{r}=N_1-mg$

$\Rightarrow N_1=\frac{m{v_1}^2}{r}+mg =\frac{m(8g)}{1}+mg =9mg$ ..........

$(I)$
applying the law of conservation of energy at point B and C, we have

$\frac{1}{2}m{v_1}^2=mg(2r)+\frac{1}{2}m{v_2}^2 \\\Rightarrow \frac{1}{2}m(8g)=mg(2)+\frac{1}{2}m{v_2}^2\\\Rightarrow {v_2}=\sqrt{4g}$
Now, net force is towards the center is the centripetal force, we have at point C

$\frac{m{v_2}^2}{r}=N_2+mg$

$\Rightarrow N_2=\frac{m{v_2}^2}{r}-mg =\frac{m(4g)}{1}-mg =3mg$ ..........

$(II)$
So the ratio of normal reactions on the block while it is crossing lowest point, highest point of vertical circle is

$\frac{N_1}{N_2}=\frac{9mg}{3mg}=3:1$ ..from

$(I)$ and

$(II)$

A mass m starting from A reaches B on a smooth track. On reaching B, if it pushes the track with a force equal to

$x$ times its weight, then the correct relation is:

0%
$H=\dfrac{xR}{2}$
0%
$H=R$
0%
$H=\dfrac{\left ( x+5 \right )}{2}R$
0%
$H=\dfrac{\left ( x-1 \right )}{2}R$

Explanation

As per given condition:

$N_{B}=xmg$

At point B

$N_{B}-mg=\dfrac{mv_{B}^{2}}{R}.......(1)$

and,

$mg H=\dfrac{1}{2}mv_{B}^2....(2)$

From (1) and (2), we get,

$N_{B}=\dfrac{2mgH}{R}+mg$

or,

$x mg=mg \left ( \dfrac{2H}{R}+1 \right )$

or,

$\dfrac{R\left ( x-1 \right )}{2}=H.$

A vehicle is moving with a speed

$v$ on a curved smooth road of width

$b$ and radius

$R$ . For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is

0%
$\dfrac{v^{2}b}{Rg}$
0%
$\dfrac{vb}{Rg}$
0%
$\dfrac{vb^{2}}{Rg}$
0%
$\dfrac{vb}{R^{2}g}$

Explanation

tan

$\theta$

$=\dfrac{v^{2}}{Rg} \Rightarrow \dfrac{h}{b}=\dfrac{v^{2}}{Rg} \Rightarrow h=\dfrac{v^{2}b}{Rg}$ .

A small block of mass

$m$ is released from rest from point A inside a smooth hemisphere bowl of radius R, which is fixed on ground such that OA is horizontal. The ratio (x) of magnitude of centripetal force and normal reaction on the block at any point B varies with

$\theta$ as:

The horizontal component

$N_{x}$ and vertical component

$N_{y}$ of the reaction at end P are given by:

0%
$\mathrm{N}_{\mathrm{x}}$ $=$ $1300\mathrm{N}, \mathrm{N}_{\mathrm{y}}=960\mathrm{N}$
0%
$\mathrm{N}_{\mathrm{x}}=960\mathrm{N}, \mathrm{N}_{y}=140\mathrm{N}$
0%
$\mathrm{N}_{\mathrm{x}}=960\mathrm{N}, \mathrm{N}_{\mathrm{y}}=960\mathrm{N}$
0%
$\mathrm{N}_{\mathrm{x}}=1300\mathrm{N}, \mathrm{N}_{\mathrm{y}}=1300\mathrm{N}$

Explanation

For equilibrium,

$\sum F_{x}=0$ thus

$Tcos37^{0}-N_{x}=0$

$N_{x}=\displaystyle\frac{4}{5}T$

$=\displaystyle\frac{4}{5}(1200)$

$=960N$

$\sum F_{y}=0$ , thus

$N_{y}-Tsin37^{0}+500+80=0$
or

$N_{y}=(1200)\left( \displaystyle\frac{3}{5} \right )-500-80$

$=140N$

Two persons

$P$ and

$Q$ of the same height are carrying a uniform beam of length

$3 m$ . If

$Q$ is at one end, then the distance of

$P$ from the other end such that

$P$ ,

$Q$ receive loads in the ratio

$5 : 3$ is:

0%
$0.5\ m$
0%
$0.6\ m$
0%
$0.75\ m$
0%
$1\ m$

Explanation

$N_{P}+N_{Q}=Mg$
also

$\dfrac{N_{P}}{N_{Q}}=\dfrac{5}{3}$

$\Rightarrow N_{P}=\dfrac{5}{8}Mg; N_{Q}=\dfrac{3}{8}Mg$

Calculating torque about the end opposite to

$Q$

$N_{P}d_{1}+N_{Q}(3-d_{2})=Mg\dfrac{3}{2}$

$\dfrac{5}{8}Mgd_{1}+\dfrac{3}{8}Mg \times 3=Mg\dfrac{3}{2}$

$\therefore \dfrac{5}{8}Mgd_{1}+\dfrac{3}{8}Mg\times 3=Mg\dfrac{3}{2}$

$\dfrac{5}{8}Mgd_{1}=(\dfrac{12-9}{8})Mg$

$d_{1}=\dfrac{3}{5}=0.6\: m$

The tension in the string is:

0%
$600$ N
0%
$80$ N
0%
$1200$ N
0%
$800$ N

Explanation

For equilibrium about P

$\sum \tau _{p}=0$
or

$[(500)1+80(0.5)-(Tsin37^{0})(0.75)]=0$
(note that torque produced by

$Tcos30^{0}$ ,

$N_{x}$ and

$N_{y}$ about the point A is equal to zero),Thus

$500+40-T\left ( \displaystyle\frac{3}{5} \right )(0.75)=0$
or

$T=\displaystyle\frac{540(5)}{3(0.75)}=1200N$

The stick

$OA$ rotates about a horizontal axis through

$O$ with a constant counter clockwise velocity

$\omega =3$ rad/sec. As it passes through the position

$\theta =0^{0},$ a small mass

$m$ is placed upon it at a radial distance

$r=0.5\ m$ . If the mass is observed to be slipping when

$\theta = 37^{o}$ , then

$\mu$ is :

0%
$\dfrac{3}{16}$
0%
$\dfrac{9}{16}$
0%
$\dfrac{4}{9}$
0%
$\dfrac{5}{9}$

Explanation

As the mass is at the verge of slipping

$\therefore$ mg sin 37 -

$\mu$ mg cos 37

$=m\omega ^{2}r$

$6 - 8\mu = 4.5$

$\therefore \mu =\dfrac{3}{16}$

Two identical ladders are arranged as shown in the figure. Mass of each ladder is

$M$ and length

$L$ . The system is in equilibrium. Find direction and magnitude of friction force acting at A or B.

0%
$f=(\displaystyle \frac{M+m}{2})g\tan\theta$ horizontally outwards
0%
$f=(\displaystyle \frac{M+m}{2})g\cot\theta$ horizontally inwards
0%
$f=(\displaystyle \frac{M+m}{2})g\cos\theta$ horizontally outwards
0%
None of these.

Explanation

Drawing the FBD of both rods
As the rods are in equilibrium

$\Sigma F_{X}=0$

$\Sigma F_{y}=0$

$\tau_{\mathrm{n}\mathrm{e}\mathrm{t}}=0$
For right rod

$N_{2}+\displaystyle \frac{mg}{2}+Mg=N$ (1)

$N_{1}=f$ (2)
For left rod

$N_{2}+N=Mg+\displaystyle \frac{mg}{2}$ (3)

From

$(1)\mathrm{a}\mathrm{n}\mathrm{d}(3)N_{1}=f$

$= N_{2}=0 = N=Mg+\displaystyle \frac{mg}{2}$

Balancing torque about

$\mathrm{O}$ ,

$Mg\displaystyle \frac{L}{2}\cos\theta +fL \sin\theta =NL\cos\theta$

$\displaystyle \frac{Mg\cos\theta}{2}+f\sin\theta =(Mg+\displaystyle \frac{mg}{2})\cos\theta$

$f=(\displaystyle \frac{Mg}{2}+\frac{mg}{2})\cot\theta = f=(\displaystyle \frac{M+m}{2})g\cot\theta$

An elastic string carrying a body of mass

$m$ at one end extends by

$1.5\ cm$ . If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

0%
$3.0\ cm$
0%
$4.5\ cm$
0%
$1.5\ cm$
0%
$9.0\ cm$

Explanation

When the body is at rest, the extension is

$1.5\ cm$ while force acting on it is

$mg$
Force when it reaches bottom is

$F_b = \dfrac{mv^2}{r} + mg$
Now using energy balance between topmost point and bottom most point ,

$\dfrac{1}{2}mv^2=\dfrac{1}{2}m(\sqrt{rg})^2+mgh$ (h is the height difference between the top and bottom point ;

$\sqrt{gr}$ is critical velocity at topmost point )

$v^2 = rg + 2gh$
or,

$v^2 = rg + 4gr = 5 gr$
So,

$F_b = 6mg$
Using this in:

$\dfrac{e_2}{e_1} = \dfrac{F_2}{F_1}$ , we get:

$e_2 = 1.5 \times 6 = 9.0\: cm$

Which of the following statements for a rigid object undergoing pure translational motion are false?

0%
If an object receives an impulse its kinetic energy must change.
0%
An object's kinetic energy can change without the object changing momentum.
0%
An object can receive a net impulse without any work being done on it.
0%
A force may do work on an object without delivering any change in momentum.

A uniform of rod of length

$l$ is placed symmetrically on two walls as shown in the figure. The rod is in equilibrium. If

$N_1$ and

$N_2$ are the normal forces exerted by the walls on the rod, then :

0%
$N_1 > N_2$
0%
$N_1 < N_2$
0%
$N_1 = N_2$
0%
$N_1$ and $N_2$ would be in the vertical directions $N_1$ $N_2$

Explanation

At equilibrium,

$N_1+N_2= mg cos \ \theta$

Also

$\tau_A= 0 \implies N_1 (0) +mg \dfrac{l}{4} cos \theta - N_2 \dfrac{l}{2}$

$\dfrac{mg}{2} cos \theta = N_2$

Thus

$N_1= \dfrac{mg}{2} cos \theta$

Hence

$N_1=N_2$

An insect moves along a semi-circular transparent track of radius

$R$ , with a constant speed

$V_0$ . It starts from point O. A point source of light is fixed at the centre of the circle. What is the velocity of shadow of the insect on the surface at time

$t$ ?

0%
$V_{0}sin\left ( \dfrac{V_{0}t}{R} \right )$
0%
$V_{0}cosec^{2}\left ( \dfrac{V_{0}t}{R} \right )$
0%
$V_{0}sec^{2}\left ( \dfrac{R}{V_{0}t} \right )$
0%
$V_{0}sec^{2}\left ( \dfrac{V_{0}t}{R} \right )$

Explanation

$t$ time, angular displacement of the insect:

$\theta =\frac{arc}{radius}=\frac{V_{0}t}{R}$
Distance of the shadow from point O is

$x=R \tan \theta$

$x=R \tan (\dfrac{V_0t}{R})$

$\therefore$ Velocity

$V=\dfrac{dx}{dt}=(R)\left ( \dfrac{V_{0}}{R} \right ) \sec^{2}\left ( \dfrac{V_{0}t}{R} \right )=V _{0} \sec ^{2}\left ( \dfrac{V_{0}t}{R} \right )$

A uniform ladder of length

$5 m$ is placed against the wall as shown in the figure. If coefficient of friction

$\mu$ is the same for both the walls, what is the minimum value of

$\mu$ for it not to slip ?

0%
$\mu =\dfrac{1}{2}$
0%
$\mu =\dfrac{1}{4}$
0%
$\mu =\dfrac{1}{3}$
0%
$\mu =\dfrac{1}{5}$

Explanation

$\textbf{Hint}$ : Draw the FBD

$\textbf{Step 1}$ :

Given length of ladder is 5m. It is placed against the wall. Coefficient of friction is

$\mu$ .

Let mass of the ladder is

$m$ .

From the figure: for the vertical equilibrium

$N_1+\mu N_2=mg$

$(1)$

For horizontal equilibrium

$N_2=\mu N_1$

$(2)$

The torque about point A is given by:

$mg(2.5)cos \theta=N_2(4)+\mu N_2(3)$

$mg(2.5 \times \dfrac{3}{5})=(4+3 \mu)N_2$ , since

$cos \theta=\dfrac{3}{5}$

$mg(\dfrac{3}{2})=(4+3 \mu)N_2$

$(3)$

\textbf{Step 2}$$:

From equation

$(1)$ and

$(2)$

$N_2(\dfrac{1+\mu^2}{\mu})=mg \implies N_2= \dfrac{\mu mg}{1+\mu^2}$

So,

$mg \dfrac{3}{2}=(4+3 \mu) \dfrac{\mu mg}{1+\mu^2}$

$(3+3 \mu^2)=8 \mu+6 \mu^2$

$3 \mu^2+8 \mu-3=0$

$\mu=\dfrac{-8 \pm sqrt{64+36}}{6}=\dfrac{-8 \pm 10}{6}$

$\mu=\dfrac{1}{3}$

Thus option C is correct.

Due to an impulse, the change in momentum of a body is

$1.8 \: kg \: m \: s^{1}$ . If the duration of the impulse is 0.2 s, then what is the force produced in it?

0%
9 N
0%
8 N
0%
7 N
0%
6 N

A very long uniform helix is made of thin metal wire. The axis of helix is vertical. A small bead begins to slide down the fixed helix starting from rest. Considering friction between bead and wire of helix to be non-zero, which of the following statements is/ are true as long as bead moves on helix.

0%
The speed of bead keeps on increasing.
0%
The magnitude of frictional force on bead remains constant.
0%
The speed of bead first increases and then remains constant.
0%
The magnitude of frictional force increases and then remains constant.

A ladder of length l and mass m is placed against a smooth vertical wall, but the ground is not smooth. Coefficient of friction between the ground and the ladder is

$\mu$ . The angle

$\theta$ at which the ladder will stay in equilibrium is:

0%
$\theta=tan^{-1} (\mu)$
0%
$\theta=tan^{-1} (2\mu)$
0%
$\theta=tan^{-1} (\frac {\mu}{2})$
0%
none of these

Explanation

$mg=N_1$

$\mu N_1=N_2$ or

$N_2=\mu mg$
Taking moment about O, we get

$\mu \,m\,\,g\,\,l sin\theta-mg\dfrac {l}{2} cos\theta = 0$

$tan\theta=\dfrac {1}{2\mu}$ or

$\theta=tan^{-1} \left( \dfrac {1}{2\mu} \right)$

Two uniform boards, tied together with the help of a string, are balanced on a surface as shown in Fig. The coefficient of static friction between boards and surface is 0.The minimum value of

$\theta$ , for which this type of arrangement is possible is :

0%
$30^o$
0%
$45^o$
0%
$37^o$
0%
It is not possible to have this type of balanced arrangement.

Explanation

For the equilibrium of the entire structure,

$N_1+N_2=2Mg$

$Mg(\dfrac {1}{2} cos\theta)+N_2\times 2x-Mg(\dfrac {1}{2} cos\theta+2x)=0$

Given

$N_2=Mg, N_1=Mg$

For individual boards,

$T_1=f_1, N_1=Mg$

and

$T_1\times l sin \theta=Mg\dfrac {1}{2} cos\theta$

$f_1=T_1=\dfrac {Mg}{2 tan\theta}$

For safe equilibrium,

$f_1 < f_L=\mu_s Mg$

$\dfrac {Mg}{2 tan\theta}=\mu_s Mg\Rightarrow tan\theta > \dfrac {1}{2\mu_s}\Rightarrow \theta=45^o$

So the minimum value of

$\theta$ for this type of arrangement is

$45^o$

A vehicle is moving with a velocity

$v$ on a curved road of width

$b$ and radius of curvature

$R$ . For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edge of the road is

0%
$\dfrac{v^2b}{Rg}$
0%
$\dfrac{vb}{Rg}$
0%
$\dfrac{vb^2}{Rg}$
0%
$\dfrac{vb}{R^2g}$

Explanation

We know that for the vehicle banking
on a curved path, $tan\theta=\dfrac{v^{2}}{rg}$ .

Here, the vehicle moves with velocity
$v$ and radius of curvature $R$ and width is b.

So, here $\theta$ is small (refer figure) and thus, $sin\theta=\dfrac{h}{b}=tan\theta$ .

Now, we can equate $\dfrac{h}{b}=\dfrac{v^{2}}{Rg}$
$\Rightarrow h= \dfrac{bv^{2}}{Rg}$ , will be the required elevation between the outer and inner edges of the road.

A stationary ball weighing $0.25\ kg$ acquires a speed of $10 \ m/s$ when hit by a hockey stick. The impulse imparted to the ball is:

0%
$2.5\ N s$
0%
$2.0\ N s$
0%
$1.5\ N s$
0%
$0.5 \ N s$

Explanation

Given that the mass of the body is

$0.25 kg$ is at rest initially. when the body is hit by the stick it gains speed of

$10\ m/s$ .

means initial speed

$u = 0$ and after getting hit it's final speed

$v = 10 m/s$ .

we know that impulse is equal to change in momentum i.e.

$\text {Impulse } J =m(v-u)$

so here

$J = 0.25 ( 10 - 0 )$

$kg\ m/s$ or

$N$

hence The impulse imparted to the ball is

$2.5 N s$

The reaction at hinge 1, before hinge 2 breaks, is:

0%
$24 \ N$
0%
$12 \ N$
0%
$11 \ N$
0%
$10 \ N$

Explanation

As the rod is in equilibrium,

$\therefore \sum F_x=0$

$\sum F_y=0$

$\sum \tau=0$
Taking torque about hinge 2, we get

$N_1\times L=Mg\times \cfrac {L}{2}+F\times \cfrac {L}{4}$

$\therefore N_1=\cfrac {Mg}{2}+\cfrac {F}{2}=11 N$

A pendulum was kept horizontal and released. Find the acceleration of the pendulum when it makes an angle

$\theta$ with the vertical.

0%
$g\ \sqrt { 1+3\cos ^{ 2 }{ \theta } }$
0%
$g\ \sqrt { 1+3\sin ^{ 2 }{ \theta } }$
0%
$g\ \sin { \theta }$
0%
$2g\ \cos { \theta }$

Explanation

$\dfrac { { mv }^{ 2 } }{ 2 }$ =

$mgl\cos { \theta }$ or

$\dfrac { { v }^{ 2 } }{ l }$ =

$2g\ \cos { \theta }$
and

$\sqrt { { a }_{ t }^{ 2 }+{ a }_{ r }^{ 2 } } =\sqrt { { \left( g\sin { \theta } \right) }^{ 2 }+{ (2g\quad \cos { \theta } ) }^{ 2 } }$
=

$g \ \sqrt { 1+3\cos ^{ 2 }{ \theta } }$

What is the actual frictional force when the man has climbed

$1.0$ m along the ladder?

0%
$360 N$
0%
$171 N$
0%
$900 N$
0%
$740 N$

Explanation

At the point on the vertical wall there is normal force along horizontal direction and at the ground there will be normal force in vertical direction and friction force in horizontal direction.
Weight of Ladder is acting at 1.5m from ground contact point and is 160 N (downwards) in vertical direction and Weight of man is acting at (3/5)x1=0.6m from ground contact and is 740 N (downwards) in vertical direction

Net torque about ground is zero,

$740\times 0.6+160\times 1.5-{ N }_{ H }\times 4=0\\ { N }_{ H }=\quad 171\quad N$

By doing Force Balance along horizontal we get Friction force=Normal Horizontal= 171N

Just to check whether friction can be upto 171N or not.

${ N }_{ V }=740+160=900N\\ { F }_{ max }=900\times 0.4=360N$
As max value of friction can be upto 360N so our answer 171N is correct

A hemispherical bowl of raidus r is rotated about its axis of symmetry which is kept vertical. A small block is kept at a position where the radius makes an angle

$\theta$ with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl is

$\mu$ . The maximum speed for which the block will not slip

0%
$\displaystyle \left[\frac{g(\sin{\theta} -\mu \cos{\theta})}{r\sin{\theta}(\cos{\theta}+ \mu\sin{\theta})}\right]^{1/2}$
0%
$\displaystyle \left[\frac{g(\sin{\theta} +\mu \cos{\theta})}{r\sin{\theta}(\cos{\theta}+ \mu\sin{\theta})}\right]^{1/2}$
0%
$\displaystyle \left[\frac{g(\sin{\theta} +\mu \cos{\theta})}{r\sin{\theta}(\cos{\theta}- \mu\sin{\theta})}\right]^{1/2}$
0%
none of these

Explanation

When the bowl rotates, there will be a range of angular speed for which the block will not move.

When the angular speed of the bowl is minimum in this range, the block will have a tendency to roll down. So, the frictional force will act in the upward direction along the surface of the bowl.
When the angular speed of the bowl is maximum in this range, the block will have a tendency to roll up. So, the frictional force will act in the downward direction along the surface of the bowl.

We only need to consider the second case here.
Lets say the block moves in a horizontal circle with the centre at C as shown in diagram so that the radius is

$PC = OP \sin = r \sin \theta$
Its acceleration is therefore

$\omega^2r \sin\theta$ . Resolving the forces along PC and applying Newton's second law,

$N\sin\theta + \mu N\cos\theta = m\omega^2r \sin\theta$ --------(1)
As there is no vertical acceleration,

$N\cos\theta - \mu N\sin\theta = mg$

$\Rightarrow N = \dfrac{mg}{(cos\theta - sin\theta)}$ ---------(2)
Replacing the value of

$N$ from equation (2) in (1)

$\dfrac{mg( \sin\theta + \mu \cos\theta )} {(\cos\theta - \mu \sin\theta )} = m\omega^2r \sin\theta$

$\Rightarrow \omega = \left[\dfrac{g(\sin{\theta} +\mu \cos{\theta})}{r\sin{\theta}(\cos{\theta}- \mu\sin{\theta})}\right]^{1/2}$

A particle P of mass m attached to vertical axis by two strings AP and BP of length l each. The separation AB=l. The point p rotates around the axis with an angular velocity

$\omega$ . the tension in two strings are

${ T }_{ 1 }$ and

${ T }_{ 2 }$
taut only if

$\omega >\sqrt { \frac { 2g }{ l } }$

0%
${ T }_{ 1 }$ = ${ T }_{ 2 }$
0%
${ T }_{ 1 }$ + ${ T }_{ 2 }$ = $\cfrac{2}{\sqrt3}m\omega^2l$
0%
${ T }_{ 1 }$ - ${ T }_{ 2 }$ =2mg
0%
BP will remain

Explanation

From figure in equilibrium,

$T_1\cos60=mg+T_2\cos60$ here

$\theta=60$ for equilateral triangle.
or

$T_1(1/2)=mg+T_2(1/2) \Rightarrow T_1-T_2=2mg$
also,

$T_1\sin60+T_2\sin60=m\omega^2l$
or

$T_1+T_2=\frac{2}{\sqrt 3} m\omega^2l$

The spool shown in the figure is placed on a rough horizontal surface and has inner radius r and outer radius R. The angle

$\theta$ between the applied force and the horizontal can be varied. The critical angle

$(\theta)$ for which the spool does not roll and remains stationary is given by :

0%
$\theta=cos^{-1}\left (\cfrac {r}{R}\right )$
0%
$\theta=cos^{-1}\left (\cfrac {2r}{R}\right )$
0%
$\theta=cos^{-1}\sqrt {\cfrac {r}{R}}$
0%
$\theta=sin^{-1}\left (\cfrac {r}{R}\right )$

Explanation

If the spool is not to translate,

$F cos \theta =f$ ....(i)
If the spool is not to rotate,

$fR=Fr$ .....(ii)
or

$\frac {fR}{r}cos\theta=f$
or

$cos\theta=\frac {r}{R}\Rightarrow \theta=cos^{-1}\left (\frac {r}{R}\right )$
Also the line of action of F should pass through the bottom most point.

A body is in equilibrium under the influence of a number of forces. Each force has a different line of action. The minimum number of forces required is :

0%
2, if their lines of action pass through the centre of mass of the body.
0%
3, if their lines of action are not parallel.
0%
3, if their lines of action are parallel.
0%
4, if their lines of action are parallel and all the forces have the same magnitude.

A massless spool of inner radius r, outer radius R is placed against vertical wall and tilted split floor as shown. A light inextensible thread is tightly wound around the spool through which a mass m is hanging. There exists no friction at point A, while the coefficient of friction between spool and point B is

$\mu$ . The angle between two surfaces is

$\theta$ .

0%
the magnitude of force on the spool at B in order to maintain equilibrium is $mg\sqrt {(\cfrac {r}{R})^2+(1-\cfrac {r}{R})^2\cfrac {1}{tan^2\theta}}$
0%
the magnitude of force on the spool at B in order to maintain equilibrium is $mg \left (1-\cfrac {r}{R}\right )\cfrac {1}{tan\theta}$
0%
the maintain value of $\mu$ for the system to remain in equilibrium is $\cfrac {cot\theta}{(R/r)-1}$
0%
the maintain value of $\mu$ for the system to remain in equilibrium is $\cfrac {tan\theta}{(R/r)-1}$

Explanation

$mgr=fR$ ....(i)

$N_1 sin\theta+f=mg$ ....(ii)

$N_1 cos\theta=N_2$ ....(iii)

From (i),

$f=\dfrac {mgr}{R}$

From (ii) and (iii),

$N_2=\dfrac {mg-f}{tan\theta}=\dfrac {mg}{tan\theta}\left [1-\dfrac {r}{R}\right ]$

Net force at B:

$F_B=\sqrt {f^2+N_2^2}$

$F_B=mg\sqrt {(\cfrac {r}{R})^2+(1-\cfrac {r}{R})^2\cfrac {1}{tan^2\theta}}$

For minimum value of

$\mu: f\leq \mu N_2$

$\Rightarrow \dfrac {mgr}{R}\leq \dfrac {\mu mg}{tan\theta}\left [1-\dfrac {r}{R}\right ]\Rightarrow \mu \geq \dfrac {tan\theta}{R/r-1}$

A small spherical ball is suspended through a string of length l. The whole arrangement is placed in a vehicle which is moving with velocity v. Now, suddenly the vehicle stops and ball starts moving along a circular path. If tension in the string at the highest point is twice the weight of the ball then (Assume that the ball completes the vertical circle)

0%
$\displaystyle v= \sqrt{5gl}$
0%
$\displaystyle v= \sqrt{7gl}$
0%
velocity of the ball at highest point is $\displaystyle \sqrt{gl}$
0%
velocity of the ball at the highest point is $\displaystyle \sqrt{3gl}$

Explanation

When the vehicle stops suddenly then the ball will have velocity

$v$ at point A due to inertia of motion. Let its velocity at B be

$v'$ .

Work-energy theorem:

$-mg(2l) = \dfrac{1}{2}m(v')^2-\dfrac{1}{2}mv^2$ ..............(1)

Also, circular motion equation:

$\dfrac{m(v')^2}{l} = T + mg$

As given

$T= 2mg \implies \dfrac{m(v')^2}{l} = 3mg$

Thus

$v' = \sqrt{3gl}$

From (1),

$-2mgl = \dfrac{1}{2}\times 3mgl - \dfrac{1}{2}mv^2$

$\implies v= \sqrt{7gl}$

A stone is tied to a string of length,

$\ell$ , and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed,

$u$ . The magnitude of the change in velocity as it reaches a position where the string is horizontal (

$g$ being acceleration due to gravity) is

0%
$\sqrt{2g\ell}$
0%
$\sqrt{2(u^2-g\ell)}$
0%
$\sqrt{u^2-g\ell}$
0%
$u-\sqrt{u^2-2g\ell}$

Explanation

Hint:

The principle of energy conservation states that energy is neither created nor destroyed.
It may transform from one type to another.

Step 1 : Find the string when it is horizontal.

So, by conservation of energy we have,

$\Rightarrow$

$\frac{1}{2} m u^{2}=\frac{1}{2} m v^{2}+m g h$

$\Rightarrow \frac{1}{2} m u^{2}-\frac{1}{2} m v^{2}=m g h$

$\Rightarrow \frac{1}{2}\left(u^{2}-v^{2}\right)=g l$

$\Rightarrow v^{2}-u^{2}=-2 g l$

Step 2 : Evaluate by doing addition on both sides

Now, adding

$2 u^{2}$ on both sides, we get:

$v^{2}-u^{2}+2 u^{2}=2 u^{2}-2 g l$

$\Rightarrow v^{2}+u^{2}=2\left(u^{2}-g l\right)$

$\sqrt{v^{2}+u^{2}}=\sqrt{2\left(u^{2}-g l\right)}$

Therefore, the change in magnitude of velocity will be

$\sqrt{2\left(u^{2}-g l\right)}$

Therfore the correct option is 'B' is $\sqrt{2\left(u^{2}-g \ell\right)}$

A ball attached to one end of a string swings in a vertical plane such that its acceleration at point A (extreme position) is equal to its acceleration at point B (mean position). The angle

$\displaystyle \theta$ is

0%
$\displaystyle \cos ^{-1}\left ( \frac{2}{5} \right )$
0%
$\displaystyle \cos ^{-1}\left ( \frac{4}{5} \right )$
0%
$\displaystyle \cos ^{-1}\left ( \frac{3}{5} \right )$
0%
None of these

Explanation

Let acceleration at both points be 'a'.

At point A,

$mg\sin { \theta } = ma$

At point B,

$\dfrac { m{ v }^{ 2 } }{ r } = ma$

By energy conservation,

$\dfrac { m{ v }^{ 2 } }{ 2 } = mgr\times (1-\cos { \theta } )$

Solving, to get

$\cos { \theta } = -1 \ or \dfrac { 3 }{ 5 }$

=> θ =

$\cos ^{ -1 }{ \dfrac { 3 }{ 5 } }$

A ball tied to the end of the string swings in a vertical circle under the influence of gravity.

0%
When the string makes an angle $\displaystyle 90^{\circ}$ with the vertical, the tangential acceleration is zero and radial acceleration is somewhere between minimum and maximum.
0%
When the string makes an angle $\displaystyle 90^{\circ}$ with the vertical, the tangential acceleration is maximum and radial acceleration is somewhere between maximum and minimum.
0%
At no place in circular motion, tangential acceleration is equal to radial acceleration.
0%
When radial acceleration has its maximum value, the tangential acceleration is zero.

Explanation

$A\ and \ B: \left| { a }_{ tangential } \right| = g$ which is maximum and radial acceleration is between\ maximum and minimum when string makes ${ 90 }^{ 0 }$ .

$\Rightarrow$ A is not correct, B is correct. C is not correct as tangential acceleration and radial acceleration can become equal multiple times in a circular motion.D is correct because radial acceleration becomes maximum at the bottom most point where tangential acceleration is zero.

just before the sphere comes in contact with the peg.

0%
$\displaystyle \frac{mg}{2}$
0%
$\displaystyle {mg}$
0%
$\displaystyle \frac{3mg}{2}$
0%
$\displaystyle \frac{5mg}{2}$

Explanation

Given :

$r = 0.8$ m

$u = 0$ m/s

Before coming in contact with the peg, the sphere rotates in a circle of radius

$r =0.8$ m

Let the velocity of the sphere when it reaches the point C be

$v$ .

From figure,

$PC = r sin 30^o = \dfrac{r}{2}$ m

Work-energy theorem :

$W = \Delta K.E$

$\therefore$

$mg (PC) =\dfrac{1}{2} mv^2 - \dfrac{1}{2}mu^2$

$mg \times \dfrac{r}{2} = \dfrac{mv^2}{2} - 0$

$\implies \dfrac{mv^2}{r} =mg$

Using:

$ma_r = \dfrac{mv^2}{r} = T - mgsin30^o$

$\therefore$

$mg = T - \dfrac{mg}{2}$

$\implies T = \dfrac{3mg}{2}$

if AB is a massless rod,

0%
$\displaystyle \frac{L}{2}$
0%
$\displaystyle \frac{3L}{2}$
0%
$\displaystyle \frac{5L}{2}$
0%
$\displaystyle \frac{7L}{2}$

Explanation

Initial velocity of the rod

$u = \sqrt{3Lg}$

Let the maximum height attained to be

$H$ . At the starting point, its potential energy is zero while at the highest point, its kinetic is zero.

Applying conservation of energy :

$P.E_i + K.E_ i =P.E_f + K.E_f$

$\therefore$

$0 + \dfrac{1}{2}mu^2 = mgH + 0$

$\implies$

$H = \dfrac{u^2}{2g} = \dfrac{3Lg}{2g} = \dfrac{3L}{2}$

A spool of mass

$M$ and radius

$2R$ lies on an inclined plane as shown in figure. A light thread is wound around the connecting tube of the spool and its free end carries a weight of mass

$m$ . The value of

$m$ so that system is in equilibrium is

0%
$2M\sin { \alpha }$
0%
$M\sin { \alpha }$
0%
$2M\tan { \alpha }$
0%
$M\tan { \alpha }$

Explanation

frictional force will act downwards so as to balance the torque of tension force.

For rotational equilibrium,

$\tau_o = 0 \implies f(2R) = TR$

For translational equilibrium:

$Mg sin \alpha + f = T$

Thus

$Mg sin \alpha + \dfrac{T}{2} = T \implies Mg sin \alpha = \dfrac{T}{2}$

Also,

$T= mg$

$Mg sin \alpha = \dfrac{mg}{2}$

$\implies m = 2M sin \alpha$

A particle is given an initial speed

$u$ inside a smooth spherical shell of radius

$R$ so that it is just able to complete the circle. Acceleration of the particle, when its velocity is vertical, is

0%
$\displaystyle g\sqrt{10}$
0%
$\displaystyle g$
0%
$\displaystyle g\sqrt{2}$
0%
$\displaystyle g\sqrt{6}$

Explanation

Let, the velocity at highest point be v then

$m{ v }^{ 2 }/r = mg \dfrac { m{ u }^{ 2 } }{ 2 } = 2\times mgR + \dfrac { mv^{ 2 } }{ 2 }$

$\Rightarrow u = \sqrt { 5Rg } \dfrac { mx^{ 2 } }{ 2 } = \dfrac { m{ u }^{ 2 } }{ 2 } - mgR$

Net acceleration at the point, when velocity is vertical is:

$a = \sqrt { { \left( \dfrac { mx^{ 2 } }{ 2 } \right) }^{ 2 }+ { g }^{ 2 } } = \sqrt { 10\times g }$

$=g\sqrt{10}$

A simple pendulum is released from rest with the string in horizontal position. The vertical component of the velocity of the bob becomes maximum, when the string makes an angle

$\displaystyle \theta$ with the vertical. The angle

$\displaystyle \theta$ is equal to

0%
$\displaystyle \frac{\pi }{4}$
0%
$\displaystyle \cos ^{-1}\left ( \frac{1}{\sqrt{3}} \right )$
0%
$\displaystyle \sin ^{-1}\left ( \frac{1}{\sqrt{3}} \right )$
0%
$\displaystyle \frac{\pi }{3}$

Explanation

At angle theta , l = length of rod v = pendulum velocity u = vertical velocity

By energy conservation,

$mgl\cos { \theta } = \dfrac { m{ v }^{ 2 } }{ 2 } = \dfrac { m{ u }^{ 2 } }{ 2\sin ^{ 2 }{ \theta } }$

Differentiating wrt

$\theta$ , and putting

$\dfrac { du }{ d\theta } =0$ ;

we get,

$\theta = \cos ^{ -1 }{ \left( \dfrac { 1 }{ \sqrt { 3 } } \right) } \\$

A particle is moving in a circular path in the vertical plane. It is attached at one end of a string of length

$l$ whose other end is fixed. The velocity at lowest point is

$u$ . The tension in the string is

$\displaystyle \vec{T}$ and acceleration of the particle is

$\displaystyle \vec{a}$ at any position. Then

$\displaystyle \vec{T}.\vec{a}$ is zero at highest point if

0%
$\displaystyle u> \sqrt{5gl}$
0%
$\displaystyle u= \sqrt{5gl}$
0%
Both (a) and (b) correct
0%
Both (a) and (b) are wrong

Explanation

Tension is

$0$ in case of

$B$ .

$u$ is more than this, then Tension is never zero.

$u$ is less than this, then the particle would never be able to complete the circular loop.

$m{ v }^{ 2 }/r\quad =\quad mg\\ \dfrac { m{ u }^{ 2 } }{ 2 } =\quad 2\times mgR\quad +\quad \dfrac { mv^{ 2 } }{ 2 } \\ \Rightarrow \quad u\quad =\quad \sqrt { 5Rg }$

A particle of mass

$m$ is suspended by a string of length

$l$ from a fixed rigid support. A sufficient horizontal velocity

$\displaystyle v_{0}= \sqrt{3gl}$ is imparted to it suddenly. Calculate the angle made by the string with the vertical when the acceleration of the particle is inclined to the string by

$\displaystyle 45^{\circ}.$

0%
$\displaystyle \theta = \frac{\pi }{2}$
0%
$\displaystyle \theta = \frac{\pi }{3}$
0%
$\displaystyle \theta = \frac{\pi }{4}$
0%
$\displaystyle \theta = \pi$

Explanation

Given :

$u = \sqrt{3gl}$

Let the velocity of the bob be

$v$ when it reaches the point C.

From figure,

$AB = l - lcos\theta = l(1-cos\theta)$

Using work-energy theorem :

$W = \Delta K.E$

$\therefore$

$-mg l (1- cos\theta) = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$

$-2 gl (1 - cos\theta) =v^2 - (3gl)$

$\implies v^2 = g l+ 2g lcos\theta$

Using

$a_r =\dfrac{v^2}{l} = g + 2g cos\theta$

Also tangential acceleration

$a_t =g sin\theta$

As the acceleration makes an angle

$45^o$ with the string, thus

$tan45^o = \dfrac{a_t}{a_r}$

$\implies$

$a_r = a_t$

$\therefore$

$g + 2g cos\theta = g sin\theta$

$\implies$

$sin\theta - 2cos\theta -1 = 0$ where

$\theta =\dfrac{\pi}{2}$ satisfies the equation.

What is the maximum speed at which a railway carriage can move without toppling over along a curve of radius

$R=200m$ if the distance from the centre of gravity of the carriage to the level of the rails is

$h=1.0m$ , the distance between the rails is

$l=2.0m$ and the rails are laid horizontally? (Take

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

0%
$\displaystyle 11.18ms^{-1}$
0%
$\displaystyle 22.36ms^{-1}$
0%
$\displaystyle 44.72ms^{-1}$
0%
$\displaystyle 74ms^{-1}$

Explanation

In railway carriage's frame, the carriage is in translational as well as rotational equilibrium.

Rotational equilibrium : Torque about point O is zero i.e

$\tau_{o} = 0$

$\therefore$

$f \times 1 - N \times 1 = 0$

$\implies f = N$

Translational equilibrium :

$\therefore$

$mg = N$

$\implies mg = f$ ..........(1)

Also

$f = ma_r$ .........(2)

From (1) and (2), we get

$ma_r = mg$

$\therefore$

$\dfrac{mv^2}{r} = mg$

$\implies v = \sqrt{rg}$

$\therefore$

$v = \sqrt{200 \times 10} =44.72$

$m/s$

just after it comes in contact with the peg.

0%
$\displaystyle \frac{mg}{2}$
0%
$\displaystyle mg$
0%
$\displaystyle \frac{3mg}{2}$
0%
$\displaystyle \frac{5mg}{2}$

Explanation

Given :

$r = 0.8$ m

$u = 0$ m/s

$OA =0.4$ m

$\therefore$

$AM = r' =0.8 - 0.4 =0.4$ m

After coming in contact with the peg, the sphere rotates in a circle of radius

$r' =0.4$ m

Let the velocity of the sphere when it reaches the point v be

$v$ .

From figure,

$PC = r sin 30^o =0.4$ m

Work-energy theorem :

$W = \Delta K.E$

$\therefore$

$mg (PC) =\dfrac{1}{2} mv^2 - \dfrac{1}{2}mu^2$

$mg \times 0.4 = \dfrac{mv^2}{2} - 0$

$\implies mv^2 =0.8mg$

Using:

$ma_r = \dfrac{mv^2}{r'} = T' - mgsin30^o$

$\therefore$

$\dfrac{0.8mg}{0.4} = T' - \dfrac{mg}{2}$

$\implies T' = \dfrac{5mg}{2}$

A car is travelling along a circular curve that has a radius of 50 m. If its speed is 16 m/s and is increasing uniformly at

$\displaystyle 8 m/s^{2},$ determine the magnitude of its acceleration at this instant.

0%
$9.5\ m/s^2$
0%
$9.8\ m/s^2$
0%
$8.5\ m/s^2$
0%
$7.5\ m/s^2$

Explanation

$\displaystyle a= \sqrt{{a_{t}}^{2}+{a_{n}}^{2}}= \sqrt{\left ( \frac{dv}{dt} \right )^{2}+\left ( \frac{v^{2}}{R} \right )^{2}}= \sqrt{\left ( 8 \right )^{2}+\left ( \frac{256}{50} \right )^{2}}= 9.5m/s^{2}$

A thin circular wire of radius R rotates about its vertical diameter with an angular frequency

$\displaystyle \omega .$ Show that a small bead on the wire remains at its lowermost point

$\displaystyle \omega \leq \sqrt{g\:l\:R}.$ What is angle made by the radius vector joining the centre to the bead with the vertical downward direction for

$\displaystyle \omega = \sqrt{2g\:l\:R}?$ Neglect friction.

0%
$\displaystyle 30^{\circ}$
0%
$\displaystyle 45^{\circ}$
0%
$\displaystyle 60^{\circ}$
0%
$\displaystyle 90^{\circ}$

Explanation

Let the angle be

$\theta$ .

balancing forces in vertical direction,

$Ncos\theta=mg - (1)$

balancing forces in horizontal direction,

$Nsin\theta=m\omega^2Rsin\theta$

$\implies N=m\omega^2R=2mg$

substituting in (1),

$2mgcos\theta=mg\implies cos\theta=0.5\implies \theta=60^0$

The bob of the pendulum shown in figure describes an arc of circle in a vertical plane. If the tension in the cord is

$2.5\ times$ the weight of the bob for the position shown. Find the velocity and the acceleration of the bob in that position.

0%
$\displaystyle 16.75ms^{-1}, 5.66ms^{-2}$
0%
$\displaystyle 5.66ms^{-1}, 16.75ms^{-2}$
0%
$\displaystyle 2.88ms^{-1}, 16.75ms^{-2}$
0%
$\displaystyle 5.66ms^{-1}, 8.34ms^{-2}$

Explanation

Let the mass of the bob be

$m$ and the velocity of the at point B be

$v$

Given :

$r =2$ m

$T =2.5 mg$

Using circular motion equation at B :

$\dfrac{mv^2}{r} = T - mg cos30^o$

$\therefore$

$\dfrac{mv^2}{2} = 2.5 mg - mg \times 0.866$

$v^2 = 3.27 g =3.27 \times 9.8 = 32$

$\implies v =5.66$ m/s

$\therefore$ Radial acceleration

$a_r = \dfrac{v^2}{r} = \dfrac{32}{2} =16$

$m/s^2$

Also tangential acceleration

$a_t = g sin 30^o =9.8 \times 0.5 = 4.9$

$m/s^2$

Total acceleration

$a = \sqrt{a_r^2 + a_t^2} = \sqrt{16^2 + (4.9)^2} = 16.75$

$m/s^2$

The simple

$2 kg$ pendulum is released from rest in the horizontal position. As it reaches the bottom position, the cord wraps around the smooth fixed pin at

$B$ and continues in the smaller are in the vertical plane. Calculate the magnitude of the force

$R$ supported by the pin at

$B$ when the pendulum passes the position

$\displaystyle \theta = 30^{\circ}.\left ( g= 9.8m/s^{2} \right )$

0%
$15 N$
0%
$30 N$
0%
$45 N$
0%
$60 N$

Explanation

using energy conservation

$\Rightarrow ({ K.E.) }_{ E }+({ P.E.) }_{ E }=({ K.E.) }_{ C }+({ P.E.) }_{ C }$

$\Rightarrow 0+2g\times 800\times { 10 }^{ -3 }=\frac { 1 }{ 2 } \times 2\times { { v }_{ 1 }^{ 2 } }+0$

$\Rightarrow v_{ 1 }=4m/s$

again using energy conservation at point

$C$ and

$D$

$\Rightarrow ({ K.E.) }_{ C }+({ P.E.) }_{ C }=({ K.E.) }_{ D }+({ P.E.) }_{ D }$

$\Rightarrow \frac { 1 }{ 2 } \times 2\times { { v }_{ 1 }^{ 2 } }+0=\frac { 1 }{ 2 } \times 2\times { { v }_{ 2 }^{ 2 }+ }2g \times 400(1-\cos { 30° } )\times { 10 }^{ -3 }$

$\Rightarrow { v }_{ 2 }^{ 2 }=8+4\sqrt { 3 }$

From the diagram, at point

$D$

$\Rightarrow T=mg\cos { \theta } +\frac { m{ v }_{ 2 }^{ 2 } }{ R } \\ \Rightarrow T=2\times 10\times \cos { 30° } +\frac { 2\times (8+\sqrt { 3 } ) }{ 400\times { 10 }^{ -3 } }$

$\Rightarrow T=90.2N$

A car is travelling along a circular curve that has a radius of

$50$ m. If its speed is

$16$ m/s and is incrcasing uniformly at

$\displaystyle 8m/s^{2}.$ Determine the magnitude of its acceleration at this instant.

0%
$\displaystyle 7.5 ms^{-2}$
0%
$\displaystyle 8.5 ms^{-2}$
0%
$\displaystyle 9.5 ms^{-2}$
0%
$\displaystyle 9.8 ms^{-2}$

Explanation

Given :

$r =50$ m

$a_t =8$

$m/s^2$

$v =16$

$m/s$

Centripetal acceleration

$a_r = \dfrac{v^2}{r} = \dfrac{(16)^2}{50} =5.12$

$m/s^2$

$\therefore$ Total acceleration

$a =\sqrt{a_r^2 + a_t^2} = \sqrt{(5.12)^2 + 8^2} =\sqrt{90.2} = 9.5$

$m/s^2$

A uniform rod of length L is placed symmetrically on two walls as shown in the figure. The rod is in equilibrium. If

$N_1$ and

$N_2$ are the normal forces exerted by the walls on the rod then

0%
$N_1$ < $N_2$
0%
$N_1$ > $N_2$
0%
$N_1$ = $N_2$
0%
$N_1$ and $N_2$ would be in the vertical directions

Explanation

$N_{1} \times \dfrac{l}{4}=N_{2} \times \dfrac{l}{4}$

$N_{1}=N_{2}$

(by torque conservation)

Correct option is (C).

The mass of the bob of a simple pendulum of length

$L$ is

$m$ . If the bob is left from its horizontal position then the speedof the bob and the tension in the thread in the lowest position of the bob will be respectively:

0%
$\;\sqrt{2gL}\;and\;3\;mg$
0%
$\;3\;mg\;and \sqrt{2gL}$
0%
$\;2\;mg\;and\;\sqrt{2gL}$
0%
$\;2\;gL\;and\;3\;mg$

Explanation

Velocity of the bob at point A is zero, thus its kinetic energy at A is zero.

Let the velocity of the bob at point B be

$v$ and tension in the string be

$T$ .

Using work-energy theorem from A to B :

$W = \Delta K.E$

$\therefore$

$mgL = \dfrac{1}{2} mv^2 - 0$

$\implies v =\sqrt{ 2gL}$

Circular motion equation at B :

$\dfrac{mv^2}{L} = T - mg$

$\therefore$

$T = \dfrac{mv^2}{L} + mg = \dfrac{m (2gL)}{L} + mg = 3mg$

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

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.