MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 11

The work done by the force

$\vec{F}=A(y^{2}\hat{i}+2x^{2}\hat{j})$ , where

$A$ is a constant and

$x$ &

$y$ are in meters around the path shown is :

0%
$zero$
0%
$Ad$
0%
$Ad^{2}$
0%
$Ad^{3}$

Explanation

Variable force is apllied across the path

$OA$ to

$AB$ to

$BC$ to

$CO$

$W=\int \bar { F } .d\bar { r }$

$\\ =\int A(y^{ 2 }\bar { i } +2x^{ 2 }\bar { j } ).(dx\bar { i } +dy.\bar { j } )$

$\\ =\int A(y^{ 2 }dx+2x^{ 2 }dy)$

Now following the path of displacement along

$OA$ the

$y=0$

$\\ W_{ OA }=\int _{ x=0 }^{ x=d }{ A(0\quad +\quad 2 } x^{ 2 }.0)\quad =\quad 0+0\quad =0$

Similerly, for

$W_{ AB },\quad x=d$

$\\ W_{ AB }=A[0+2d^{ 2 }d]\quad =\quad 2Ad^{ 3 }$

Similarly, for

$W_{ BC },\ x=d,\quad y=d$

$\\ W_{ BC }=\int _{ x=d }^{ x=0 }{ A(d^{ 2 }dx\quad +\quad } 0)$

$\\ W_{ BC }=A[d^{ 2 }(-d)+0]=-Ad^{ 3 }$

$\\ W_{ CO }=A[0+0]$

$\\ W=0+2Ad^{ 3 }-Ad^{ 3 }+0=Ad^{ 3 }$

A ball of mass m moves towards a moving wall of infinite mass with a speed 'v' along the normal to the wall. The speed of the wall is 'u' toward the ball. The speed of the ball after elastic collision with wall is

0%
$u + v$ away from the wall
0%
$2u + v$ away from the wall
0%
$|u - v|$ away from the wall
0%
$|v- 2u |$ away from the wall

Explanation

In the frame of wall, the ball moves towards it with speed

$u + v$ and due to momentum conservation velocity after collision returns with velocity

$u + v$ in the direction in which wall is moving.

So from the ground frame, the velocity of ball after collision with wall is

$2u + v.$

A force

$\vec{F}=(3t\hat{i}+5\hat{j})$ N acts on a particle whose position vector varies as

$\vec{S}=(2t^{2}\hat{i}+5\hat{j})$ m, where

$t$ is time in seconds. The work done by this force from

$t=0$ to

$t=2s$ is:

0%
23 J
0%
32 J
0%
zero
0%
can't be obtained

Explanation

The work done by the force

$\vec{F}=(3t\hat{i}+5\hat{j})$ is

$W=\int \vec{F}.d\vec{s}=\int (3t \hat{i}+5\hat{j}).(4t dt \hat{i})$

$=\int_{0}^{2}12t^{2}dt=\dfrac{12[t^{3}]_{0}^{2}}{3}=\dfrac{12(8-0)}{3}=32 \ J$

A particle of mass

$m_0,$ travelling at speed

$v_0,$ strikes a stationary particle of mass

$2m_0.$ As a result, the particle of mass

$m_0$ is deflected through

$45^o$ and has a final speed of

$\dfrac{v_0}{\sqrt2}$ . Then the speed of the particle of mass

$2m_0$ after this collision is

0%
$\dfrac{v_0}{2}$
0%
$\dfrac{v_0}{2\sqrt2}$
0%
$\sqrt2v_0$
0%
$\dfrac{v_0}{\sqrt2}$

Explanation

A particle of mass

$m_0,$
Before collision

$mv_{0}=2mv cos\theta+\displaystyle \dfrac{mv_{0}}{2}..........(i)$

$\displaystyle \theta=\dfrac{mv_{0}}{2}-2mv sin \theta............(ii)$

By (i) & (ii),

$\theta=45^{0}$
Now again momentum conservation in y direction

$\displaystyle \dfrac{mv_{0}}{2}=2mv.\displaystyle \dfrac{1}{\sqrt2}\Rightarrow v=\displaystyle \dfrac{v_0}{2\sqrt2}$

A particle of mass

$m$ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is

$a$ . A force

$\vec{F}=y\hat{i}-x\hat{j}$ newton acts on the particle, where

$x, y$ denote the coordinates of position of the particle. The work done by this force in taking the particle from point A (

$a, 0$ ) to point B (

$0, a$ ) along the circular path is

0%
$\dfrac{\pi a^{2}}{4}J$
0%
$\dfrac{\pi a^{2}}{2}J$
0%
$\dfrac{\pi a^{2}}{3}J$
0%
$None\ of\ these$

Explanation

$\displaystyle W=\mathbf{\int F.dr}=\int (y\hat{i}-x\hat{j}).(dx\hat{i}+dy\hat{j})$

$=\displaystyle \int (ydx-xdy)$ .........(1)

$\because x^{2}+y^{2}=a^{2} \therefore xdx+y dy=0$

$\displaystyle \Rightarrow W=\int \left ( y\left ( \frac{-ydy}{x} \right )-xdy \right )=-\int \dfrac{x^{2}+y^{2}}{x}dy$

$\displaystyle =-\int_{0}^{a}\dfrac{a^{2}}{\sqrt{a^{2}-y^{2}}}dy=-\dfrac{\pi a^{2}}{2}$

Alternate Method :
It can be observed that the force is tangent to the curve at
each point and the magnitude is constant. The direction of
force is opposite to the direction of motion of the particle.

$\therefore$ work done = (force) (distance)

$=-\sqrt{x^{2}+y^{2}}=\dfrac{\pi a}{2}=-a\times \dfrac{\pi a}{2}=-\dfrac{\pi a^{2}}{2}J$

$W=-\dfrac{\pi a^{2}}{2}J$

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:

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

A force

$\vec F=(3t\hat i+5\hat j)$ N acts on a body due to which its displacement varies as

$\vec S=(2t^2\hat i-5\hat j)m$ . Work done by this force in

$2$ second is

0%
32 J
0%
24 J
0%
46 J
0%
20 J

Explanation

$\vec v=\dfrac {d\vec S}{dt}=(4t)\hat i$

$P=\vec F\cdot \vec v=12 t^2$

$\therefore W=\int_0^2Pdt=\int_0^2 (12)t^2dt=32J$ .

The world class Moses Mabhida football stadium situated in Durban has a symmetrical arc of length 350 m and a height of 106 m, as shown in the picture below on the left.
The picture on the right shows a funicular (Skycar), which takes tourists to the top of the arc. Suppose that the Skycar with tourists inside, starts from the base of the arc and travels a distance of 175 m along the arc to the viewing platform at the top. Assume that the work done by friction during the Skycars complete ascent is

$5.8\times 10^5J$ . If the combined mass of the Skycar and tourists is 5000 kg, then the work done by the motor that lifts the Skycar is approximately equal to,

0%
$4.6\times 10^6 J$
0%
$5.8\times 10^6 J$
0%
$8.0\times 10^6 J$
0%
$9.2\times 10^6 J$

One end of a light spring of spring constant

$k$ is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is

$+(\dfrac {1}{2})kx^2$ . The possible cases are:

0%
The spring was initially compressed by a distance $x$ and was finally in its natural length.
0%
It was initially stretched by a distance $x$ and finally was in its natural length.
0%
It was initially in its natural length and finally in a compressed position.
0%
It was initially in its natural length and finally in a stretched position.

A spring of natural length

$l$ is compressed vertically downward against the floor so that its compressed length becomes

$\dfrac {l}{2}$ . On releasing, the spring attains its natural length. If

$k$ is the stiffness constant of spring, then the work done by the spring on the floor is:

0%
Zero
0%
$\dfrac {1}{2}kl^2$
0%
$\dfrac {1}{2}l(\dfrac {1}{2})^2$
0%
$kl^2$

A block of mass

$10kg$ is released on a fixed wedge inside a cart which is moved with constant velocity

$10 \ ms^{-1}$ towards right. There is no relative motion between block and cart. Then work done by normal reaction on block in two seconds from ground frame will be

$(g=10 \ ms^{-2})$ :

0%
$1320 J$
0%
$960 J$
0%
$1200 J$
0%
$240 J$

Explanation

Constant velocity means net force

$=0$
Using Lami's theorem in the figure,
We have,

$\displaystyle \dfrac {N}{sin (180^o- 53^o)}=\frac {100}{sin 90^o}$

$\therefore N=100 sin 53^o=80 N$

Now,

$W_N=NS cos 53^o$

$=(80)(20)(0.6)$

$=960 J$

A particle of mass m, moving with velocity v collides a stationary particles of mass 2m. As a result of collision, the particle of mass m deviates by

$45^o$ and has final speed of

$\dfrac {v}{2}$ . For this situation mark out the correct statement (s).

0%
The angle of divergence between particles after collision is $\dfrac {\pi}{2}$
0%
The angle of divergence between particles after collision is less than $\dfrac {\pi}{2}$
0%
Collision is elastic
0%
Collision is inelastic

Explanation

Let velocity of mass 2m is

$\vec{{v}'}$ after collision.;
From conservation of momentum in the xy plane,

$\vec{p_i}=\vec{p_f}$ ;

$mv\hat{i}=\dfrac{mv}{2}(\cos 45^0\hat{i}+\sin 45^0\hat{j})+2m\vec{{v}'}=\dfrac{mv}{2\sqrt2}(\hat{i}+\hat{j})+2m\vec{{v}'}$ ;

$\therefore 2\vec{{v}'}=(\dfrac{2\sqrt2-1}{2\sqrt2})v\hat{i}-\dfrac{v}{2\sqrt2}\hat{j}$

$\Rightarrow \vec{{v}'} =0.32v\hat{i}-0.18v\hat{j}$ ;

$\therefore \tan \theta =\dfrac{0.18}{0.32}=0.56$ ;

$\therefore\theta =29.35^0$ ;

$\left | \vec{v}' \right |= \sqrt{(.32)^2+(.18)^2}v=0.37v$ ;

Since

$( \theta +45^0) < 90^0$ , the angle of divergence between particles after collision is less than

$90^0$ ;

${KE}_i=\dfrac{1}{2}mv^2$ ;

${KE}_f=\dfrac{1}{2}m(\dfrac{v}{2})^2 + \dfrac{1}{2}2m(0.37v)^2$ ;

$\therefore {KE}_f < {KE}_i$ ;

$\therefore$ collision is inelastic.

A spring of spring constant

$5\times 10^3$ N/m is stretched initially by

$5$ cm from the unstretched position. The work required to further stretch the spring by another

$5$ cm is:

0%
6.25 N-m
0%
12.50 N-m
0%
18.75 N-m
0%
25.00 N-m

Explanation

Energy stored in a spring when stretched by

$x$ is

$\dfrac {1}{2}kx^2$

The amount of work required is the change in energy of the spring.

$W=U_f-U_i=\dfrac {1}{2}k({x_2}^2-{x_1}^2)$

$\Rightarrow W=0.5(5000)(0.1^2-0.05^2)=18.75 \ N-m$
Option C.

The force exerted by a compression device is given by

$F(x) = kx (x-l)$ for

$0 \leq x \leq l$ , where

$l$ is the maximum possible compression,

$x$ is the compression and

$k$ is the constant. Work done to compress the device by a distance

$d$ will be maximum when

0%
$d = \dfrac {l}{4}$
0%
$d = \dfrac {l}{\sqrt{2}}$
0%
$d = \dfrac {l}{2}$
0%
$d = l$

Explanation

For W to be maximum ;

$\dfrac{dW}{dx} = 0$ ;

i.e.

$F(x) = 0 \Rightarrow x= l, x = 0$

clearly for

$d = l$ , the work done is maximum.

Alternate Solution :
External force and displacement are in the same direction

$\therefore$ Work will be positive continuously so it will be maximum when displacement is maximum.

A spring of force constant

$k$ is cut in two parts at its one-third length. When both the parts have same elongation, the work done in the two parts will be (Spring constant of a spring is inversely proportional to length of spring)

0%
Equal to both
0%
Greater for the longer part
0%
Greater for the shorter part
0%
Data insufficient

Explanation

$k\alpha \dfrac {1}{l}$
l of shorter part is less, therefore value of k is more

$W=\dfrac {1}{2}kx^2$

$\therefore W_{\text {shorter part}}$ will be more.

The relationship between the force

$F$ and position

$x$ of a body is as shown in figure. The work done by force

$F$ , in displacing the body from

$x=1 \ m$ to

$x=5 \ m$ will be

0%
30 J
0%
15 J
0%
25 J
0%
20 J

Explanation

Hint: Work done is the area under F-x graph.

Correct Option: B

$\textbf{Step1: Calculation of work done}$

As work done = area enclosed by F-x graph

Work done in displacing the body from x=1 m to x=5 m will be:

W = Area of $ABNM$ + Area of $CDEN$ - Area of $EFGH$ + Area of $HIJ$

(Area of $EFGH$ is taken negative as the force is in negative direction)

$\Rightarrow W = \left( {1 \times 10} \right) + \left( {1 \times 5} \right) - \left( {1 \times 5} \right) + \left( {\dfrac{1}{2} \times 10 \times 1} \right) = 15J$

So, the work done in displacing the body from x=1 m to x=5 m will be $15J$ .

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

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

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

A mass

${ m }_{ 1 }$ with initial speed

${ v }_{ 0 }$ in the positive

$x$ -direction collides with a mass

${ m }_{ 2 }=2{ m }_{ 1 }$ which is initially at rest at the origin, as shown in figure. After the collision

${ m }_{ 1 }$ moves off with speed

${ v }_{ 1 }={ v }_{ 0 }/2$ in the negative

$y$ - direction, and

${ m }_{ 2 }$ moves off with speed

${ v }_{ 2 }$ at angle

$\theta$ . Find the velocity (magnitude and direction) of the centre of mass after the collision :

0%
${ v }_{ 0 }/3$
0%
${ v }_{ 0 }/2$
0%
${ v }_{ 0 }/5$
0%
${ v }_{ 0 }/6$

Explanation

Since there is no external force acting on the 2 mass system, the momentum of the system and the velocity of the centre of mass remains constant.
Therefore,

$V_{cm}$ is the same before and after the collision, i.e;

$V_{cm,x} = \dfrac {m_1v_1+m_2v_2}{m_1 +m_2}$ , as

$m_2=2m_1$ &

$V_{cm,y}=0.$

$V_{cm,x}= \dfrac { m_1v_o+2m_1 \times 0}{m_1+2m_1} = \dfrac{v_o}{3} .$

A body of mass

$m$ is hauled from the Earth's surface by applying a force

$\vec{F}$ varying with the height of ascent

$y$ as

$\vec{F}=2(ay-1)mg$ , where

$a$ is a positive constant. Find the work performed by this force

$W$ and the increment in the body's potential energy

$\Delta U$ in the gravitational field of the Earth over the first half of the ascent.

0%
$\displaystyle W=\frac{3mg}{4a}$ , $\displaystyle\Delta U=\frac{mg}{2a}$
0%
$\displaystyle W=\frac{3mg}{a}$ , $\displaystyle\Delta U=\frac{mg}{2a}$
0%
$\displaystyle W=\frac{3mg}{4a}$ , $\displaystyle\Delta U=\frac{mg}{a}$
0%
$\displaystyle W=\frac{3mg}{4a}$ , $\displaystyle\Delta U=\frac{mg}{3a}$

Explanation

Force acting on the body of mass m=

$\vec { F } =2\left( a\times y-1 \right) \times m\times g$

Body will be ascenting until

$\vec { F } =0\ \Rightarrow 2\left( a\times y-1 \right) \times m\times g=0\ \Rightarrow a\times y-1=0\ \Rightarrow a\times y=1\ \Rightarrow y =\dfrac { 1 }{ a }$

So, the body will ascent until the height of y=

$\dfrac { 1 }{ a }$

Now, the work done by the force

$\vec {F}$ for half of the ascent height

$=\int _{ 0 }^{ \dfrac { 1 }{ 2a } }{ \vec { F } } dy\ =\int _{ 0 }^{ \dfrac { 1 }{ 2a } }{ 2\left( a\times y-1 \right) } \times m\times g\times dy\ =\dfrac { 3mg }{ 4a }$

Now, the potential energy stored in the body due to gravitational field force for half of ascent height

$=\dfrac{ mg }{ 2a }$

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

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

One end of a light spring of spring constant

$k$ is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is

$+\left(\displaystyle\frac{1}{2}\right){kx^2}$ .The possible cases are:

0%
The spring was initally compressed by a distance $x$ and was finally in its nautral length
0%
It was initially streteched by a distance $x$ and finally was in its natural length
0%
It was initially in its natural length and finally in compressed position
0%
It was initially in its natural length and finally in a stretched position

A spring of constant

$k$ is fixed to a wall. A boy stretches this spring by distance

$x$ and in the mean time the compartment moves by a distance

$s$ . The work done by boy with reference to earth is

0%
$\displaystyle \dfrac{1}{2} kx^2$
0%
$\displaystyle \dfrac{1}{2} (kx)(s +x)$
0%
$\displaystyle \dfrac{1}{2} kxs$
0%
$\displaystyle \dfrac{1}{2} kx(s + x + s)$

Explanation

Work done by boy to stretch the spring

$\displaystyle = \left ( \frac{1}{2} kx \right ) (S + x)$
Work done by man on the floor

$\displaystyle = - \left ( \frac{1}{2} kx \right ) (S)$

$\therefore$ Total work done

$= \displaystyle \frac{1}{2} kx^2$

One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is

$1/2 \:k\:x^2$ . The possible cases are

0%
the spring was initially compressed by a distance x, was finally in its natural length
0%
it was initially stretched by a distance x and was finally in its natural length
0%
it was initially in its natural length and finally in a compressed position
0%
It was initially in its natural length and finally in the stretched position

Explanation

Stored elastic potential energy of spring

$=\dfrac{1}{2}kx^2$ where

$x$ is compression or elongation of spring from its natural length. In this position the spring can do work on the block tied to it, which is equal to

$\dfrac{1}{2}kx^2$ , so both option (a) & b are correct.

Two blocks A and B of masses m and 2m respectively placed on a smooth floor are connected by a spring. A third body C of mass m moves with velocity

$v_0$ along the line joining A and B and collides elastically with A. At a certain instant of time after collision it is found that the instantaneous velocities of A and B are same then :

0%
the common velocity of A and B at time t $_0$ is v/3.
0%
the spring constant is k $= \displaystyle \frac{3mv_0^2}{2x_0^2}$ .
0%
the spring constant is k $= \displaystyle \frac{2mv_0^2}{3x_0^2}$ .
0%
none of these

Explanation

Since the masses of A and C are same and the collision is elastic, their velocities are exchanged after the collision. So, now the system is (A+B) and initially B is at rest wheras, A has velocity

${ v }_{ 0 }$ . After some time they have same velocities, so conserving momentum

${ mv }_{ 0 }=(m+2m)v$

$v =\dfrac { { v }_{ 0 } }{ 3 }$

A brick of mass

$1.8$ kg is kept on a spring of spring constant

$K = 490 N m^{-1}$ . The spring is compressed so that after the release brick rises to

$3.6$ m. Find the compression in the spring. (Take

$g= 10 \ m/s^2$ )

0%
0.21 m
0%
0.322 m
0%
0.414 m
0%
0.514 m

Explanation

As the brick raises, it gains potential energy, which is the work done by the spring on the brick.

Let the spring be compressed by

$x$ , then the work done is

$\dfrac {1}{2}kx^2$

$\therefore \dfrac {1}{2}kx^2=mgh$

$\Rightarrow x=(\dfrac {2mgh}{k})^{1/2}=\sqrt {\dfrac {2\times 1.8\times 10\times 3.6}{490}}=0.514m$

Option D.

A force

$F = - K (y \widehat i + x \widehat j)$ (where

$K$ is positive constant) acts on a particle moving in the

$x-y$ plane. Starting from the origin, the particle is taken along the

$x$ -axis to the point

$(a, 0)$ and then parallel to

$y$ -axis to the point

$(0, a)$ . The total work done by the force

$F$ on the particle is:

0%
$- 2 Ka^2$
0%
$2 Ka^2$
0%
$- Ka^2$
0%
$Ka^2$

Explanation

Given

$\vec F = -K(y \vec i+x \vec j)$
Consider the small work done by the force

$F$ in displacing the particle through area

$dS$ as

$dW = \vec F.\vec {dS}$

where,

$\vec {dS} = dx \hat{i} + dy \hat j$

$\therefore dW = ( -K(y \hat i+x \hat j)).( dx \hat i + dy \hat j)$

$\therefore dW = -K(y dx + x dy) = - Kd(xy)$ .....................(1)

Since, first the particle is taken along the position x-axis to the point (a, 0) and then parallel to y-axis to the point (0, a) hence, integrating for whole area we get

$\displaystyle W = \int_{x=0}^{x=a} \int_{y=0}^{y=a} dW$

$\displaystyle W = \int_{x=0}^{x=a} \int_{y=0}^{y=a} (- Kd(xy))$ .............from(1)

$\displaystyle W = -K(\left | x \right |_{0}^{a})(\left | y \right |_{0}^{a})$

$W = -Ka^2$

The work done by the spring in case (i):

0%
$\displaystyle \frac{kl^2}{2} - \frac{rl^3}{3}$
0%
$\displaystyle \frac{kl^2}{2} + \frac{rl^2}{3}$
0%
$\displaystyle \frac{kl^3}{6} - \frac{rl^2}{8}$
0%
$\displaystyle \frac{kl^3}{8} - \frac{rl^2}{6}$

Explanation

Given,

$F=$

$kx-rx^{2}$
workdone by spring in case(i)

$=$

$\int_{0}^{l}Fdx$

$=-$ [

$\dfrac{kl^{2}}{2} -\dfrac{rl^{3}}{3}$ ] {since,

$F$ and displacement are in opposite direction }

A force

$F = - K (y \widehat i + x \widehat j)$ , (where

$K$ is a +ve constant) acts on a particle moving in xy plane starting from origin, the particle is taken along the positive x-axis to the point (

$a, 0$ ) and then parallel to y axis to the point (

$a, a$ ). The total work done by force

$F$ on the particle is

0%
$- 2 Ka^2$
0%
$2 Ka^2$
0%
$- Ka^2$
0%
$Ka^2$

Explanation

Work was done

$\int Fdt$

When particle moves from

$0$ to

$a$ along x-axis

Then

$W_1 = \displaystyle \int_0^a - (Ky \widehat i) dx = 0$

$\because\,\, y=0$

Now, when the particle moves parallel to the y-axis

$W_2 = \displaystyle \int_0^a - (Ka \widehat j) dy$

$\because\,\, x=a$

$\therefore W_2= - Ka \displaystyle \int_0^a dy = - Ka^2$

$\therefore W = W_1 + W_2 = - Ka^2$

The work done by the tension T in the above process is

0%
$Zero$
0%
$T(L-L\cos \theta )$
0%
$-TL$
0%
$-TL\,\sin \theta$

Explanation

The instantaneous displacement is directed normal to

$T$ all the time.

Therefore

$W_T=0$ .

Two springs have their force constant as

$k_1$ and

$k_2\:(k_1 > k_2)$ . When they are stretched by the same force

0%
no work is done in case of both the springs.
0%
equal work is done in case of both the springs
0%
more work is done in case of second spring
0%
more work is done in case of first spring.

Explanation

From Hooke's law

$F\propto x\Rightarrow F=kx$ , where k is spring constant

Since force is same in Stretching for both spring so

$F=k_1x_1 = k_2x_2\Rightarrow x_1<x_2 \:$ because

$\: k_1>k_2$

so work done in case of first spring is

$W_1 =\dfrac{1}{2} k_1x_1^2$ and work done in case of second spring is

$\displaystyle W_2 =\dfrac{1}{2}k_2x_2^2 \:so \:\frac{W_1}{W_2}=\frac{x_1}{x_2}\Rightarrow W_1<W_2$

It means that more work is done in case of second spring (work done on spring is equal to stored elastic potential energy of the spring)

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

Calculate the work done on the tool by

$\vec{F}$ if this displacement is along the straight line

$y =x$ that connects these two points.

0%
$2.50 J$
0%
$500 J$
0%
$50.6 J$
0%
$2 J$

Explanation

as ,

$dW=f.ds$

$W=\int { f.ds }$

and force is only along the y-axis, work-done along the x-axis is zero

$\displaystyle W=\int_{0,0}^{3,3} { -\alpha x{ y }^{ 2 }.dy } \cos 180^{\circ}$ and

$x= y$

$\displaystyle W=\int_{0}^{3} { \alpha { y }^{ 3 }.dy }$

$\displaystyle W=\alpha \int _{ 0 }^{ 3 }{ { y }^{ 3 }dy }$

$\displaystyle W=2.5*\{ \frac { { 3 }^{ 4 } }{ 4 } -0\}$

$W=50.6 J$

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 force

$\displaystyle F= -\frac{k}{x^{2}}\left ( x\neq 0 \right )$ acts on a particle in x-direction. Find the work done by this force in displacing the particle from.

$\displaystyle x= +a\:$ to

$\:x= +2a.$ Here,

$k$ is a positive constant.

0%
$\displaystyle -\frac{k}{2a}$
0%
$\displaystyle \frac{k}{2a}$
0%
$\displaystyle -\frac{k}{a}$
0%
$\displaystyle +\frac{k}{a}$

Explanation

$\displaystyle W= \int F\:dx= \int_{+a}^{+2a}\left ( \frac{-k}{x^{2}} \right )dx= \left [ \frac{k}{x} \right ]_{+a}^{+2a}$

$\displaystyle = -\frac{k}{2a}$
Note: It is important to note that work comes out to be negative which is quite obvious as the force acting on the particle is in negative x-direction

$\displaystyle \left ( F= -\frac{k}{x^{2}} \right )$ while displacement is along positive x-direction. (from x=a to x=2a)

An object is displaced from position vector

$\displaystyle \vec{r}_{1}= \left ( 2\hat{i}+3\hat{j} \right )m\:$ to

$\:\vec{r}_{2}= \left ( 4\hat{i}+6\hat{j} \right )$ m under a force

$\displaystyle \vec{F}= \left ( 3x^{2}\hat{i}+2y\hat{j} \right )N.$ Find the work done by this force.

0%
$83 J$
0%
$41.5 J$
0%
$166 J$
0%
$164 J$

Explanation

$\displaystyle W= \int_{\vec{r}_{1}}^{\vec{r}_{2}}\vec{F}\cdot \vec{dr}$

$\displaystyle = \int_{\vec{r}_{1}}^{\vec{r}_{2}}\left ( 3x^{2}\hat{i}+2y\hat{j} \right )\cdot \left ( dx\hat{i}+dy\hat{j}+dz\hat{k} \right )$

$\displaystyle = \int_{\vec{r}_{1}}^{\vec{r}_{2}}\left ( 3x^{2}dx+2ydy \right )= \left [ x^{3}+y^{2} \right ]_{\left ( 2,3 \right )}^{\left ( 4,6 \right )}$

$=83 J$

Which of the following statements is correct regarding the work done

$\vec{F}$ along these two paths.

0%
Work done on x-axis is zero
0%
Work done on x-axis is less than on y-axis
0%
Work done on x-axis is more than on y-axis but not zero
0%
Data insufficient

Explanation

since force is applied in -ve y-axis and angle between x & y axis is

${ 90 }^{ o }$
and

$dW=\vec{F}\cdot \vec{ds}$

$W=0$ on x-axis
Alternatively, since

$y=0$ along the path from

$x=0$ ( origin) to

$x=3$ ,

$\vec{F}=\alpha xy^2 \hat{j}=0$

$\Rightarrow W=\vec{F}\cdot \vec{ds}=0$
Similarly,

$W=0$ along y-axis.

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,

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

The work done by the varying force in changing the angular displacement from 0 to

$\theta$ is

0%
$Wh$
0%
$FL\,\sin \theta$
0%
$Fh$
0%
$\dfrac{1}{2}FL\,\sin \theta$

Explanation

Because the process is quasi static, work is equal to increase of potential energy.

$Work =mgh= Wh$ .

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 cutting tool under microprocessor control has several forces acting on it. One force is

$\vec{F}=-\alpha xy^2\hat{j}$ , a force in the negative y-direction whose magnitude depend on the position of the tool. The constant is

$\alpha = 2.50\ N/m^3$ . Consider the displacement of the tool from the origin to the point

$x = 3.00 \,m, y = 3.00 \,m$ . Calculate the work done on the tool by

$\vec{F}$ if the tool is first moved out along the x-axis to the point

$x = 3.00m, \:y= 0m$ and then moved parallel to the y-axis to

$x = 3.00m, y = 3.00 \,m$ .

0%
$67.5 J$
0%
$85 J$
0%
$102 J$
0%
$7.5 J$

Explanation

as ,

$dW=f.ds$

$W=\int { \vec{F}.\vec{ds} }$
and force is only along y-axis ,work done along x- axis is zero
work done in moving from

$(3,0)$ to

$(3,3)$ is

$\displaystyle W=\int_0^3 { -\alpha x{ y }^{ 2 }.dy } \cos 180^{\circ}$ and

$x= 3$

$\displaystyle W=\int_0^3 { \alpha \times 3 \times { y }^{ 2 }.dy }$

$\displaystyle W=2.5 \times 3 \times \int _{ 0 }^{ 3 }{ { y }^{ 2 }dy }$

$\displaystyle W=7.5\times\{ \dfrac { { 3 }^{ 3 } }{ 3 } -0\}$

$W=67.5 \ J$

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.

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),

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

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

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 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 :

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

$\therefore$

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

$\implies$

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

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

$D$

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

$\Rightarrow T=90.2N$

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 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 :

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

$\therefore$

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

The sphere at A is given a downward velocity

$\displaystyle v_{0}$ of magnitude

$5 m/s$ and swings in a vertical plane at the end of a rope of length

$l=2 m$ attached to a support at

$O$ . Determine the angle

$\displaystyle \theta$ at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.

0%
$\displaystyle \sin ^{-1}\left ( \frac{1}{4} \right )$
0%
$\displaystyle \sin ^{-1}\left ( \frac{1}{3} \right )$
0%
$\displaystyle \sin ^{-1}\left ( \frac{1}{2} \right )$
0%
$\displaystyle \sin ^{-1}\left ( \frac{3}{4} \right )$

Explanation

Given :

$v_o =5 m/s$

$l =2$ m

$T = 2mg$

From figure,

$PM = l sin\theta$

Circular motion equation at P :

$\dfrac{mv^2}{l} = T - mgsin\theta$

$mv^2 =l( 2mg - mgsin\theta) = 2 (2 m \times 10 - m(10) sin\theta)$

$\implies$

$mv^2 = m (40 - 20 sin\theta)$

Work-energy theorem :

$W = \Delta K.E$

$\therefore$

$mg (PM) = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_o^2$

$2mg (l sin\theta) =m(40 - 20sin\theta) - mv_o^2$

$2 m (10) (2 sin\theta) = m (40 - 20 sin\theta) - m(25)$

$\implies sin\theta = \dfrac{1}{4}$

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 11 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 11 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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