MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 13

A chain is held on a frictionless table with

$L/4$ hanging over. Knowing total mass of the chain is

$M$ and total length is

$L$ , the minimum work required to pull hanging part back too the table is:

0%
$\cfrac{MgL}{16}$
0%
$\cfrac{MgL}{8}$
0%
$\cfrac{MgL}{32}$
0%
$\cfrac{MgL}{24}$

The time period of a simple harmonic pendulum in a stationary train is

$T$ . The time period of a mass attached to a spring is also

$T$ . The train accelerates at the rate

$5 ms^{-2}$ . If the new time periods of the pendulum and spring be

$T_p$ and

$T_s$ respectively, then

0%
$T_p = T_s$
0%
$T_p > T_s$
0%
$T_p < T_s$
0%
Cannot be predicted

A small solid sphere of mass

$m$ is released from a point

$A$ at a height

$h$ above the bottom of a rough track as shown in the figure. If the sphere rolls down the track without slipping, its rotational kinetic energy when it comes to the bottom of track is

0%
$mgh$
0%
$\cfrac{10}{7}mgh$
0%
$\cfrac{5}{7}mgh$
0%
$\cfrac{2}{7}mgh$

A worker lifts a

$20.0 kg$ bucket of concrete from the ground up to the top of a

$20.0 m$ tall building. What is the amount of work that the worker did in lifting the bucket?

0%
$3.92 kJ$
0%
$400 J$
0%
$560 kJ$
0%
$4.08 J$

Explanation

Given:

The mass of the bucket $m=20\ kg$

The height of the building $h=20\ m$

$\text{Work done by the worker} = \text{Potential energy of bucket at the top of the building}$

$\therefore W = mgh$

$\Rightarrow W = 20 \times 9.8 \times 20$

$\Rightarrow W = 3920\;{\rm{J}}$

Thus, the required work done is $3.92\;k{\rm{J}}$

A block attached to a spring, pulled by a constant horizontal force, is kept on a smooth surface as shown in the figure. Initially, the spring is in the natural state. Then the maximum positive work that the applied force

$F$ can do is: (Given that spring does not break)

0%
$\cfrac{{F}^{2}}{k}$
0%
$\cfrac{{2F}^{2}}{k}$
0%
$\infty$
0%
$\cfrac{{7F}^{2}}{k}$

If initial charge on all the capacitors were zero, work done by the battery in the circuit shown is

0%
$0.2 mJ$
0%
$200 mJ$
0%
$0.4 mJ$
0%
$400 mJ$

A bob of mass

$m$ is attached at one end of a string of length

$\iota$ . Other end of the string is fixed point

$O$ . Bob is rotating in circular path of radius

$\iota$ in horizontal plane about

$O$ with constant speed

$v$ , as shown in the figure. The average force exerted by string on the bob during its:

0%
half revolution will be $\dfrac{mv^{2}}{\pi \iota }$
0%
half revolution will be $\dfrac{2mv^{2}}{\pi \iota }$
0%
one fourth revolution will be $\dfrac{\sqrt{2mv^{2}}}{\pi \iota }$
0%
one revolution will be zero

A force

$\vec {F} = -k(x\hat {i} + y\hat {j})$ , 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 positive x-axis to the point

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

$(a, a)$ .

0%
Work done by the force in moving particle along x-axis is $-\dfrac {1}{2}ka^{2}$
0%
Work done by the force in moving particle along x-axis is $-ka^{2}$
0%
Work done by the force in moving particle along y-axis is $-\dfrac {1}{2}ka^{2}$
0%
Total work done by the force for overall motion is $-ka^{2}$

A solid sphere of mass

$4kg$ and radius

$1m$ is rotating about the given axis

$xx'$ with angular velocity

$10rad/s$ shown in figure. The

${K.E}_{rotation}$ is given by

0%
$180J$
0%
$200J$
0%
$240J$
0%
$280J$

The work done in extending a spring by

$x_0$ is

$W_0$ . Find the work done in further extension

$x_0$ .

0%
$W=2W_0$
0%
$W=3W_0$
0%
$W=W_0$
0%
$W=\frac{1}{3} W_0$

A gun is mounted on a trolley free to move on horizontal tracks. Mass of gun and trolley

$25$ kg. Gun fires a two shells of mass

$5$ kg each other. Velocity of shells with roll to gun is

$60$ m/s . In above question, velocity of gun with respect to ground after firing second shell is-

0%
$12$ m/s
0%
$18$ m/s
0%
$50$ m/s
0%
$60$ m/s

Two spring of spring constant k and 3k are stretched separately by same force. The ratio of the potential energy stored in them respectively will be

0%
3 : 1
0%
9 : 1
0%
1 : 3
0%
1 : 9

The P.E of a certain spring a, when stretched from natural through a distance

$0.3m$ , is

$10J$ . The amount of work in joule that must be done on this spring to stretch it through an additional distance

$0.15m$ will be

0%
$10J$
0%
$20J$
0%
$7.5J$
0%
$12.5J$

The potential energy of a particle varies with position

$x$ according to the reaction

$U(x) = 2x^{4} - 27x$ the point

$x = 3/2$ is point of

0%
Unstable equilibrium
0%
Stable equilibrium
0%
Neutral equilibrium
0%
None of these

Explanation

Hint: Use the formula $F=-\dfrac{dU}{dx}$

A vertical spring of force constant

$100 N/m$ is attached with a hanging mass of

$10$ kg. Now an external force is applied on the mass so that the spring is stretched by the additional

$2$ m, The work done by the force F is (

$g= 10 m/s^2$ )

0%
$200$ J
0%
$400$ J
0%
$450$ J
0%
$600$ J

A ball of mass

$10kg$ moving with velocity

$20m/s$ collides elastically with wall and rebound with the same speed.Then the magnitude of change in momentum of the ball will be

0%
Zero
0%
$300kg$ $m/s$
0%
$400kg$ $m/s$
0%
$100kg$ $m/s$

Select the incorrect option:-
(

$\tau = torque, \,v = velocity, \,\omega = angular \,velocity, \, a_c = centripetal \, acceleration$ )

0%
$\vec \tau \times \vec \tau = 0$
0%
$\vec v . \vec \omega = 0$
0%
$\vec \omega \times \vec v= 0$
0%
$\vec a_c \times \vec v= 0$

$540$ calories of heat convert

$1\ cubic\ centimeter$ of water at

$100^{o}\ C$ into

$1671\ cubic\ centimeter$ of steam at

$100^{o}\ C$ at a pressure of one atmosphere. Then the work done against the atmospheric pressure is nearly:

0%
$540\ cal$
0%
$40\ cal$
0%
$zero\ cal$
0%
$500\ cal$

A bucket full of water weighs

$5$ kg, it is pulled from a well

$20m$ deep. There is a small hole In the bucket through which water leaks at a constant rate of

$0.2 k g m ^ { - 1 }.$ The total work done in pulling the bucket up from the well is

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

0%
$600J$
0%
$400J$
0%
$100J$
0%
$500J$

$AB$ is a long frictionless horizontal surface. One end of an ideal spring of spring constant

$K$ is attached to a block of mass

$m$ which is being moved left with constant velocity

$v _ { 1 }$ and another end is free. Another block of mass

$2\mathrm { m }$ is given a velocity

$3\mathrm { v }$ towards the spring. The magnitude of work done by the external agent in moving

$m$ with constant velocity

$v$ in a long time is

$\beta$ times

$m v ^ { 2 }$ . Find the value of

$\beta .$

0%
$-5\ mv^2$
0%
$-8\ mv^2$
0%
$-3\ mv^2$
0%
None of these

The average kinetic energy of an idea gas per molecule at

$25^{o}C$ , will be

0%
$6.1\times10^{-20}J$
0%
$6.1\times10^{-21}J$
0%
$6.1\times10^{-22}J$
0%
$6.1\times10^{-23}J$

A smooth track in the form of a quarter circle of radius

$6 m$ lies in the vertical plane. A particle moves from

$P_1$ to

$P_2$ under the action of forces

$\vec F_1, \vec F_2$ and

$\vec F_3$ . Force

$\vec F_1$ is always toward

$P_2$ and is always

$20 N$ in magnitude. Force

$\vec F_2$ always acts horizontally and is always

$30 N$ in magnitude. Force

$\vec F_3$ always acts tangentially to the track and is of magnitude

$15 N$ . Select the correct alternative(s)

0%
work done by $\vec F_1$ is $120 J$
0%
work done by $\vec F_2$ is $180 J$
0%
work done by $\vec F_3$ is 45 $\pi$
0%
$\vec F_1$ is conservative in nature

The mass of object X is

$M_1$ and that of object Y is

$M_2$ . Keeping their kinetic energy constant, if the velocity of object Y is doubled the velocity of object X, what will be the relation between their masses?

0%
$M_1=2M_2$
0%
$4M_1=M_2$
0%
$M_1=4M_2$
0%
$2M_1=M_2$

Explanation

Mass of object X =

$M_1$ , Velocity of object X =

$v_1$

Mass of object Y =

$M_2$ , Velocity of object Y =

$v_2$

Also,

$v_2$ = 2

$v_1$

K.E is constant.

K.E $_1$ = K.E $_2$

$\dfrac{1}{2}$ $M_1$ $v_1^2$ = $\dfrac{1}{2}$ $M_2$ $v_2^2$

$\dfrac{1}{2}$

$M_1$

$v_1^2$ =

$\dfrac{1}{2}$

$M_2$

$(2v_1)^2$

$\dfrac{1}{2}$

$M_1$

$v_1^2$ =

$\dfrac{1}{2}$

$M_2$

$4\ v_1^2$

$M_1$ = 4

$M_2$

The spring is compressed by a distance

$a$ and released. The block again comes to rest when the spring is elongated by a distance

$b.$ During this process:

0%
work done by the spring on the block = $\frac{1}{2} k(a - b)^2$
0%
work done by the spring on the block = $\frac{1}{2} k(a^2 + b^2)$
0%
co-efficient of friction = $\frac{k(a - b)}{2mg}$
0%
co-efficient of friction = $\frac{k(a + b)}{2mg}$

A vertical spring of force constant

$100$ N/m is attached with a hanging mass of

$10$ kg. Now an external force is applied on the mass so that the spring is stretched by additional

$2$ m. The work done by the force F is?

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

0%
$200$ J
0%
$400$ J
0%
$450$ J
0%
$600$ J

Explanation

Given that,

Spring constant $k=100$

Mass $m=10\,kg$

Stretching $x=2\,m$

We know that,

The work done is

$W=\dfrac{1}{2}k{{x}^{2}}$

$W=\dfrac{1}{2}\times 100\times 4$

$W=200\,J$

Hence, the work done is $200\ J$

In a simple hydraulic press, the cross-sectional area of the two cylinders is

$5\times 10^{-4}m^2$ and

$10^{-2}m^2$ , respectively. A force of

$20$ N is applied at the small plunger. How much work is done by the operator, if the smaller plunger moves down

$0.1$ m?

0%
40 J
0%
2 J
0%
20 J
0%
200 J

A block is attached with a spring and is moving towards a fixed wall with speed V as shown in figure. As the spring reaches the wall, it starts compressing. the work done by the spring on the wall during the process of compression is

0%
$\frac{1}{2} mv^2$
0%
$mv^2$
0%
kmv
0%
zero

A ball moving with a velocity of 6 m/s strikes an identical stationary ball. After collision each ball moves at an angle of

$30^{\circ}$ with the original line of motion. What are the speeds of the balls after the collision?

0%
$\frac{\sqrt{3}}{2}m/sec$
0%
3 m/sec
0%
2 $\sqrt{3}$ m/sec
0%
$\sqrt{3}$ m/sec

In an elastic collision which of the following is correct?

0%
Linear momentum is conserved
0%
Total energy is conserved
0%
Kinetic energy is conserved
0%
Both (1) and (2)

A practical of mass

$m$ is moving in

$X-Y$ plane under the action of a central force

$\overrightarrow { F }$ in such a manner that at any instant

$t$ , the components of the velocity of the particle are-

${ v }_{ x }=4\cos { t,{ v }_{ y }=4sint}$ Then the amount of the work done by the force in a time interval

$t=0$ to

$t$ is-

0%
$zero$
0%
$4Ft$
0%
$\frac { 4F }{ t }$
0%
None of the above

The collision of two balls of equal mass takes place at the origin of co-ordinates. Before collision, the components of velocities are (

$y_{x}=50 cm/s$ ,

$v_{v}=0)$ and

$(v_{x}=-40 cm/s)$ and velocity components (

$v_{x}$ and

$v_{y}$ respectively) of the second ball are:

0%
$30\ and\ 10 cm/s$
0%
$10\ and\ 30 cm/s$
0%
$5\ and\ 15 cm/s$
0%
$15\ and\ 5 cm/s$

A uniform rod AB of length land mass m is lying on a smooth table. A small particle of mass m strike the rod with a velocity

$V_{0}$ at point C a distance X from the centre O. The particle comes to rest after collision. The value of x, so that point A of the rod remains stationary just after collision is:

0%
$L/3$
0%
$L/6$
0%
$L/4$
0%
$L/12$

Two identical balls are released from positions as shown. They collide elastically on horizontal surface. Ratio of heights attained by A & B after collision is( All surface are smooth,neglect energy loss at M & N)

0%
1:4
0%
2:1
0%
4:13
0%
2:5

A car weighing 1 ton is moving twice as fast as another car weighing 2 ton. The kinetic energy of the one-ton car is

0%
less than that of the two-ton car is
0%
some as that of the two-ton car is
0%
more than that of the two-ton car is
0%
impossible to compare with that of the two-ton car unless the height of each

Explanation

$K_1=\dfrac{1}{2}(1)(2v)^2=2v^2$

$K_2=\dfrac{1}{2}(2)(v)^2=v^2$

$\implies K_1 > K_2$

A particle of mass m starts moving from origin along x-axis and its velocity varies with position (x) as

$v$ =

$k\sqrt{x}$ . The work done by force acting on it during first 't' seconds is

0%
$\frac{mk^4t^2}{4}$
0%
$\frac{mk^2t}{2}$
0%
$\frac{mk^4t^2}{8}$
0%
$\frac{mk^2t^2}{4}$

Explanation

$\textbf{Step 1 - Calculate velocity in terms of t.}$

Given,

$v = k\sqrt {x}$

$v = \dfrac {dx}{dt} = k\sqrt {x}$

$kdt = x^{-\dfrac {1}{2}} dx$

Integrating both side

$\int kdt = \int x^{-\dfrac {1}{2}} dx$

$kt = \dfrac {x^{\dfrac {1}{2}}}{\dfrac {1}{2}} = 2\sqrt {x}$

$\Rightarrow x = \dfrac {k^{2}t^{2}}{4}$

$\Rightarrow v = k\sqrt {x} = k \sqrt {\dfrac {k^{2}t^{2}}{4}}$

$\Rightarrow v = \dfrac {k^{2}t}{2}$

$\textbf{Step 2 - By Work-Energy theorem}$

$W = \triangle k \Rightarrow W = k_{f} - k_{i}$

$\Rightarrow W = \dfrac {1}{2} mv_{f}^{2} - \dfrac {1}{2}mv_{i}^{2}$

$= \dfrac {1}{2} m\left (\dfrac {k^{2}t}{2}\right )^{2} - 0 (at\ x_{i} = 0, v_{i} = 0)$

$\Rightarrow W = \dfrac {mk^{4} t^{2}}{8}$

Correct option : C

A block of mass M initially at height 'h' on to a spring of force constant k. the maximum compression in the spring is x. then

0%
$mgh = \frac {1}{2} k x^2$
0%
$mgh(h+x)= \frac {1}{2} k(x+h)^2$
0%
$mgh = \frac {1}{2} k(x+h)^2$
0%
$mg(h+x)= \frac {1}{2} k(x)^2$

A stone is projected from a horizontal plane. It attains maximum height

$H$ and strikes a stationary smooth wall and falls on the ground vertically below the maximum height. Assume the collision to be elastic the height of the point on the wall where ball will strike is:

0%
$H/2$
0%
$H/4$
0%
$3H/4$
0%
none of these

N identical balls are placed on a smooth horizontal surface. Another ball of same mass collides elastically with velocity

$u$ with first ball of N balls. A process of collision is thus started in which first ball collides with second ball and the second ball with the third ball and so on. The coefficient of resulting for each collision is

$e$ . Find speed of Nth ball :

0%
$(1+e)^{N}u$
0%
$u(1+e)^{N-1}$
0%
$\cfrac{u(1+e)^{N-1}}{2^{N-1}}$
0%
$u^{N}(1+e)^{N-1}$

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then the energy stored in the wire is

0%
0.1 J
0%
0.2 J
0%
10 J
0%
20 J

Explanation

Work done in stretching wire

$=\dfrac {1}{2}\dfrac {YAL^{2}}{L}=\dfrac {1}{2}F.l$

Where

$L=$ length of wire

$l=$ increase in length

We know that

$E=\dfrac {1}{2}\times F \times AL$

$=\dfrac {1}{2}\times 200 \times 10^{-3}$

$=0.1J$

Hence $(A)$ option is correct.

Under the action of variable force, at any instant the displacement is

0%
assumed to be infinitesimally small so that the force is assumed to be constant.
0%
constant whatever be the magnitude of force.
0%
half the magnitude of force.
0%
assumed to be infinitesimally small so that the force is assumed to be perpendicular to the displacement.

A mass is performing circular motion in a vertical plane centered at

$O$ . The average velocity of the particle is in creased, then at which point the string will break:

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

The kinetic energy of a particle moving along a circle of radius

$R$ depends on the distance covered s as

$T={ KS }^{ 2 }$ where K is a constant. Find the force acting on the particle as a function of

$S$ -

0%
$\dfrac { 2K }{ S } \sqrt { 1+(\frac { S }{ R } ) } ^{ 2 }$
0%
$2KS\sqrt { 1+(\frac { R }{ S } ) } ^{ 2 }$
0%
$2KS\sqrt { 1+(\frac { S }{ R } ) } ^{ 2 }$
0%
$\dfrac { 2S }{ K } \sqrt { 1+(\frac { S }{ R } ) } ^{ 2 }$

A body of mass 3 kg is under a constant force, Which causes a displacement s in meter in it, given by the relation

$s = \frac { 1 }{ 3 } t^2$ , Where t is in second. Work done by the force in 2 s is.

0%
$\frac { 5 }{ 19 } J$
0%
$\frac { 3 }{ 8 } J$
0%
$\frac { 8 }{ 3 } J$
0%
$\frac { 19 }{ 5 } J$

A uniform chain of length

$2$ m is kept on a table such that a length of

$60$ cm hangs freely from the edge of the table. The total mass of the chain is

$4$ kg. What is the work done in pulling the entire chain on the table?

0%
$7.2$ J
0%
$3.6$ J
0%
$120$ J
0%
$1200$ J

Explanation

$\begin{array}{l} Fraction\, of\, length\, of\, the\, chain\, hanging\, from\, the\, table- \\ =\frac { 1 }{ n } =\dfrac { { 60\, cm } }{ { 200\, cm } } =\dfrac { 3 }{ { 10 } } \\ n=\frac { { 10 } }{ 3 } \\ Work\, done\, in\, pulling\, the\, chain\, on\, the\, table- \\ W=\dfrac { { mgl } }{ { 2{ n^{ 2 } } } } =\frac { { 4\times 10\times 2 } }{ { 2\times { { \left( { \dfrac { { 10 } }{ 3 } } \right) }^{ 2 } } } } =3.6J \end{array}$

Hence,

option

$(B)$ is correct answer.

A point starts from rest and moves along a circular path with a constant tangential acceleration. After one rotation, the ratio of its radial acceleration to its tangential acceleration will be equal to:

0%
$1$
0%
$2\pi$
0%
$\cfrac{1}{2}\pi$
0%
$4\pi$

A block 'A' of mass 2m placed on another block 'B' of mass 4m which in turn is placed on a fixed table. The two blocks have the same length 4d and they are placed as shown in the figure. The coefficient of friction (both static and kinetic) between block 'B' and table is

$\mu$ . There is no friction between the two blocks. A small object of mass m moving horizontally along a line passing through the centre of mass (CM) of block B and perpendicular to its face with a speed v collides elastically with block B at a height d above the table. What is the minimum value of v required to make the block A topple?

0%
$\sqrt { 6\mu gd }$
0%
$\sqrt { 3\mu gd }$
0%
$\dfrac { 5 }{ 2 } \sqrt { 3\mu gd }$
0%
$\dfrac { 5 }{ 2 } \sqrt { 6\mu gd }$

A force acts on a

$30$ gm particle in such a way that the position of the particle as a function of time is given by

$x = 3t - { 4t }^{ 2 } + { t }^{ 3 }$ , where

$x$ is in meters and

$t$ is in seconds. The work done during the first

$4$ second is :-

0%
$5.28\ J$
0%
$450\ mJ$
0%
$490\ mJ$
0%
$530\ mJ$

A road is banked at an angle of

$37^{ 0 }.$ If a car moving at 54 km/hr does not experience any friction force while negotiating the curve, the radius of curve is :-

0%
$\dfrac { 67.5 }{ 4 } m$
0%
45 m
0%
54 m
0%
30 m

A bullet of mass m strikes a pendulum bob of mass M with velocity u. It passes through and emerges out with a velocity u/2 from bob. The length of the pendulum is l. What should be the minimum value bob u if the pendulum bob will swing through a complete circle?

0%
$\frac{{2M}}{m}\times \,\sqrt {5gl\,}$
0%
$\frac{M}{{2m}}\times \,\sqrt {5gl\,}$
0%
$\frac{{2M}}{m}\times \,\dfrac{1}{{\sqrt {5gl\,} }}$
0%
$\frac{M}{{2m}}\times \,\dfrac{1}{{\sqrt {5gl\,} }}$

A particle is whirled in a vertical circle of radius 1.0 m using a string with one end fixed. If the ratio of maximum and minimum tension in the string is

$\frac{5}{3}$ , the minimum velocity of the particle during circular motion is :

0%
$\sqrt {10}$ m/s
0%
$\sqrt {50}$ m/s
0%
$10$ m/s
0%
$10\sqrt 5$ m/s

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 13 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 13 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.