MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 7

A body is acted upon by force which is inversely proportional to the distance covered. The work done will be proportional to:

0%
$s$
0%
${s}^{2}$
0%
$\sqrt {s}$
0%
None of the above

Explanation

Given

$F\propto \dfrac{1}{s} \implies F=\dfrac{k}{s}$

$W=\int _{ { s }_{ 1 } }^{ s }{ \overrightarrow { F } .d } \overrightarrow { s } =\int _{ { s }_{ 1 } }^{ s }{ \dfrac { k }{ s } .d } s=k\ln { \left( \dfrac{s}{s_1}\right) }$

A particle moves along

$x-$ axis from

$x=0$ to

$x=5$ meter under the influence of a force

$F=7-2x+3x^{2}$ . The work done in the process is:

0%
$70$
0%
$135$
0%
$270$
0%
$35$

Explanation

Given,

$F=3x^2-2x+7$

Work done

$=\int { F\cdot ds }$

$=\int _{ 0 }^{ 5 }{ \left( { 3x }^{ 2 }-2x+7 \right) dx }$

$=\left[{ \frac { { 3x }^{ 3 } }{ 3 } -\frac { { 2x }^{ 2 } }{ 2 } +7x }\right]_{ 0 }^{ 5}$

$=(5)^3-(5)^2+7\times 5-0$

$=125-25+35$

$=135J$

A body of mass

$4\ kg$ moves under the action of a force

$\overrightarrow { F } =\left( \hat { 4i } +12{ t }^{ 2 }\hat { j } \right) N$ , where

$t$ is the time in second. The initial velocity of the particle is

$(2\hat { i } +\hat { j } +2\hat { k } )m{s}^{-1}$ . If the force is applied for

$1\ s$ , work done is:

0%
$4\ J$
0%
$8\ J$
0%
$12\ J$
0%
$16\ J$

Explanation

Given Mass

$m=4kg$

Force

$\overrightarrow { F } =\left( 4\hat { i } +12{ t }^{ 2 }\hat { j } \right) N$

Initial velocity

$\overrightarrow { v } =\left( 2\hat { i } +\hat { j } +2\hat { k } \right) m/s$

Magnitude of

$\overrightarrow { v } =\sqrt { { 2 }^{ 2 }+{ I }^{ 2 }+{ I }^{ 2 } } =\sqrt { 9 } \quad =3m/s$

Net force

$=\left| \left( 4i+\left| 2{ t }^{ 2 }\hat { j } \right| \right) N \right| \\ \Rightarrow ma=\sqrt { { 16 }^{ 2 }+144{ t }^{ 4 } } \\ \Rightarrow 4a=4\sqrt { 1+9{ t }^{ 4 } } \\ \Rightarrow a=\sqrt { 1+9{ t }^{ 4 } } \\ \Rightarrow \cfrac { dv }{ dt } =\sqrt { 1+9{ t }^{ 4 } } \\ \Rightarrow v=\sqrt { { 1 }^{ 2 }+{ \left( { 3t }^{ 2 } \right) }^{ 2 } } \\ at(t=1)\quad =4$

Work

$=$ Force

$\times$ displacement

$=12.6\times \cfrac { 1 }{ 3 } =4J$

A particle of mass

$2\ kg$ travels along a straight line with velocity

$v=a \sqrt {X}$ , where

$a$ is a constant. The work done by net force during the displacement of particle from

$x=0$ to

$x=4\ m$ is:

0%
$a^{2}$
0%
$2a^{2}$
0%
$4a^{2}$
0%
$\sqrt {2} a^{2}$

Explanation

Given,

$mass=2kg, velocity=a\sqrt{x}$

Since,

$v=a\sqrt{x}, v=a\sqrt4$

So,

$V^2=u^2+2as\Rightarrow v^2-u^2=2as\Rightarrow v^2=2as\rightarrow a=\dfrac{v^2}{2s}$ Since,

$u=0$

$a=\dfrac{2a^2}{2s}=\dfrac{4a^2}{4r^2}\Rightarrow a=\dfrac{a^2}{2}$

So, workdone

$=f\times ds=2\times\dfrac{a^2}{2}\times 4=4a^2$

A weightless thread can support tension upto

$30N$ . A particle of mass

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

$2m$ in a vertical plane. If

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

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

Explanation

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

978656_1025451_ans_24441a7ebe8c4447a51626af787a601d.png

A block suspended from a spring at natural length and is free to move vertically in the y-direction. Mark the CORRECT statement(s)

0%
Mass is released when y=0; the maximum value of y reached by the mass m is 2mg/k
0%
Mass is released when y=0; the maximum value of y reached by the mass m is mg/k
0%
Due to air resistance the mass settles down into an equilibrium position $y_{eq}$ , the mechanical energy loss is $\dfrac{1}{2} (m^2g^g/k)$
0%
Due to air resistance the mass settles down into an equilibrium position $y_{eq}$ , the mechanical energy loss is $(m^2g^g/k)$

A particle moves under the effect of a force

$F=Cx$ from

$x=0$ to

$x={x}_{1}$ . The work done in the process is

0%
$C{x}_{1}^{2}$
0%
$\cfrac{1}{2} C{x}_{1}^{2}$
0%
$C{x}_{1}$
0%
Zero

Explanation

Here F = Cx

Work done = F.d Where d is displacement

$w = \int^{x_2}_{x_1} F.dx$

$w = \int^{x_1}_{0} Fdx$

$w = \int^{x_1}_{0} cx \space dx$

$w = \dfrac{1}{2} cx^2_1$

Here (B) is correct answer

A force acts on a

$3\ g$ 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 second. The work done during the first

$4$ second is:

0%
$490\ mJ$
0%
$450\ mJ$
0%
$528\ mJ$
0%
$530\ mJ$

Explanation

Given,

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

$dx=d\left( 3t-4{{t}^{2}}+{{t}^{3}} \right)=\left( 3-8t+3{{t}^{2}} \right)dt$

Acceleration

$a(x)=\dfrac{{{d}^{2}}x}{dt}=\dfrac{{{d}^{2}}x}{dt}=\dfrac{d\left( 3-8t+3{{t}^{2}} \right)}{dt}=-8+6t$

Work done = $dw=ma.dx=m\left( 6t-8 \right)\left( 3-8t+3{{t}^{2}} \right)dt$

$\int_{0}^{W}{dw}=m\int_{0}^{4}{\left( 6t-8 \right)\left( 3-8t+3{{t}^{2}} \right)dt}$

$W=m\times 176=0.003\times 176=528\times {{10}^{-3}}\,J$

$W=528\times {{10}^{-3}} J$

A stone of mass

$1kg$ is tied to the end of a string of

$1m$ length. It is whirled in a vertical circle. If the velocity of the stone at the top be

$4m/s$ . What is the tension in the string (at that instant)?

0%
$6N$
0%
$16N$
0%
$5N$
0%
$10N$

Explanation

$T+mg=\dfrac { { mv }^{ 2 } }{ r } \\ T=\dfrac { { mv }^{ 2 } }{ r } -mg\\ \dfrac { 1\times { 4 }^{ 2 } }{ 1 } -1(10)\\ =16-10\\ =6N$

Under the action of a force, a

$2\ kg$ body moves such that its position

$x$ as a function of time

$t$ is given by

$x=\dfrac{t^{3}}{3}$ , where

$x$ is in meter and

$t$ in second. The work done by the force in first two seconds is:

0%
$1600\ J$
0%
$160\ J$
0%
$16\ J$
0%
$\dfrac{16}{9}\ J$

Explanation

Given,

$x=\dfrac{t^3}{3}$

Differentiate it wrt t

$\dfrac{dx}{dt}=\dfrac{3}{3}t^2$

$\dfrac{dx}{dt}=t^2$

$v=t^2$

Workdone (W)= Change in K.E

$W=K.E_f-K.E_i$

$K.E_i=0$

$K.E_f=\dfrac{1}{2}mv^2$

$K.E_f=\dfrac{1}{2}\times 2\times 4^2$

$K.E_f=16J$

$W=16J$

A boy is swinging in a swing. If he stands the time period will

0%
First decreases, then increase
0%
Decrease
0%
Increase
0%
Remain same

Explanation

As the child stand up the effective length of pendulum decreases due to the reason that center of gravity rises up.

According to

$T=2\pi\sqrt{\left(\dfrac{l}{g}\right)}$

Thus, the time peroid decreases

A particle of mass

$m$ initially moving with speed

$v$ . A force acts on the particle

$f=kx$ where

$x$ is the distance travelled by the particle and

$k$ is constant. Find the speed of the particle when the work done by the force equals

$W$ .

0%
$\sqrt { \frac { k }{ m } +{ v }^{ 2 } }$
0%
$\sqrt { \frac { 2W }{ m } +{ v }^{ 2 } }$
0%
$\sqrt { \frac { 2W }{ k } +{ v }^{ 2 } }$
0%
$\sqrt { \frac { W }{ 2m } +{ v }^{ 2 } }$

Explanation

Given,

Final velocity, ${{v}_{1}}$

Change in kinetic Energy = work done by force

$\dfrac{1}{2}mv_{1}^{2}-\dfrac{1}{2}m{{v}^{2}}=W$

${{v}_{1}}=\sqrt{\dfrac{2W}{m}+{{v}^{2}}}$

Hence, final velocity is $\sqrt{\dfrac{2W}{m}+{{v}^{2}}}$

A wheel rotating at an angular speed of 20 rad

${ s }^{ -1 }$ , is brought to rest by a constant torque in 4s.If the M.I is 0.2 kg

${ m }^{ 2 }$ , the work done in first 2s is:

0%
$50\ J$
0%
$30\ J$
0%
$20\ J$
0%
$10\ J$

Explanation

Angular acceleration

$\alpha=\dfrac{20}{4}=5$

Torque=

$I\alpha=5\times 0.2=1$

Angular displacement in first 2 secs=

$20\times 2+\dfrac{1}{2}5\times 2^2=50$

Work done=Torque×angular displacement

$=1×50=50J$

Under the action of a force, a

$2lg$ body moves such that it position

$x$ as a function of time is given by

$x=\cfrac{{t}^{3}}{3}$ , where

$x$ is in metre and

$t$ in seconds. The work done by the force in the first two seconds is

0%
$1600J$
0%
$160J$
0%
$16J$
0%
$1.6J$

Explanation

Given : Mass m

$=2kg$

$x(t)=\cfrac { { t }^{ 3 } }{ 3 }$

Time

$t=2s$

Differentiating position

$x(t)= velocity v$

$=\cfrac { d }{ dt } \left( \cfrac { { t }^{ 3 } }{ 3 } \right) =\cfrac { 1 }{ 3 } .3{ t }^{ 2 }\quad ={ t }^{ 2 }$

$t=2s\Rightarrow v={ t }^{ 2 }\quad =4m/s$

Kinetic energy

$=\cfrac { 1 }{ 2 } m{ v }^{ 2 }\quad =16J$

A particle is moving in a vertical circle the tension in the string when passing through two position at angle

${30}^{o}$ and

${60}^{o}$ from vertical lowest position are

${T}_{1}$ and

${T}_{2}$ respectively then-

0%
${T}_{1}={T}_{2}$
0%
${T}_{1}> {T}_{2}$
0%
${T}_{1}< {T}_{2}$
0%
${T}_{1}\ge {T}_{2}$

Explanation

$T-mgcos\theta =\dfrac { m{ v }^{ 2 } }{ r } \\ T=mgcos\theta +\dfrac { m{ v }^{ 2 } }{ r } \\ for\quad \theta =30\\ { T }_{ 1 }=mgcos30+\dfrac { m{ v }^{ 2 } }{ r } \\ for\quad \theta =60\\ { T }_{ 2 }=mgcos60+\dfrac { m{ v }^{ 2 } }{ r } \\ { T }_{ 1 }>{ T }_{ 2 }$

A force of

$(4{ x }^{ 2 }+3x)N$ acts on a particle which displaces it from

$x=2m$ to

$x=3m$ . The work done by the force is:

0%
$32.8 J$
0%
$3.28 J$
0%
$0.328 J$
0%
$zero$

Explanation

$W=\int Fdx=\int (4x^2+3x)dx=\dfrac{4x^3}{3}+\dfrac{3x^2}{2}$

$W=\dfrac{4\times 3^3}{3}+\dfrac{3\times 3^2}{2}-\dfrac{4\times 2^3}{3}-\dfrac{3\times 2^2}{2}=32.8J$

A force

$F=(3{x}^{2}+2x-7)N$ acts on a

$2kg$ body as a result of which the body gets displaced from

$x=0$ to

$x=5m$ . The work done by the force will be:

0%
$5J$
0%
$70J$
0%
$115J$
0%
$270J$

Explanation

Given,

$=3x^2+2x-7, mass=2kg$

So the work-done

$W=\int_{0}^{5}(3x^2=2x-7) dx=[x^3+x^2-7x]_{0}^{5}=[5^3+5^2-7\times5=115J$

A stone of mass

$0.2kg$ is tied to one end of a thread of length

$0.1m$ whirled in a vertical circle. When the stone is at the lowest point of circle, tension in thread is

$52N$ , then velocity of the stone will be:

0%
$4m/s$
0%
$5m/s$
0%
$6m/s$
0%
$7m/s$

Explanation

$T-mg=\dfrac { { mv }^{ 2 } }{ r } \\ T=\dfrac { { mv }^{ 2 } }{ r } +mg\\ 52=0.2(10)\quad +\dfrac { 0.2({ v }^{ 2 }) }{ 0.1 } \\ 2{ v }^{ 2 }=52-2=50\\ \qquad v=5m/sec$

Two springs

$A$ and

$B(k_{A}=2 k_{B})$ are stretched by a applying forces of equal magnitudes at the four ends. If the energy stored in

$A$ is

$E$ , that in

$B$ is

0%
$E/2$
0%
$2E$
0%
$E$
0%
$E/4$ .

Force constant of spring one is

$K$ and another spring is

$2K$ . When both spring are stretched through same distance, then the work done

0%
${W}_{2}=2{W}_{1}^{2}$
0%
${W}_{2}=2{W}_{1}$
0%
${W}_{2}={W}_{1}$
0%
${W}_{2}=0.5{W}_{1}$

Explanation

Work done = -

$\Delta (Potential energy)$

$-\frac{1}{2}kx^{2}$

Both springs are stretched by the same length , let it be

$l$ .

Then ,

$W_{1}$ =

$-\dfrac{1}{2}kl^{2}$

$W_{2}$ =

$-\dfrac{1}{2}2kl^{2}$

Therefore ,

$W_{2}$ =

$2$

$W_{1}$

A string of length

$L$ and force constant

$K$ is stretched to obtain extension

$l$ . It is further stretched to obtain extension

${l}_{1}$ .The work done in second stretching is

0%
$\cfrac{1}{2}k{l}_{1}(2l+{l}_{1})$
0%
$\cfrac{1}{2}K{l}_{1}^{2}$
0%
$\cfrac { 1 }{ 2 } K\left( { l }^{ 2 }-{ l }_{ 1 }^{ 2 } \right)$
0%
$\cfrac { 1 }{ 2 } K\left( { 2{ l }_{ 1 }^{ 2 }-l }^{ 2 } \right)$

Explanation

Work done= Final total energy-Initial Total energy

Let us assume difference in first and second extention is

$l-l_1$

Workdone

$=\dfrac{Kl^2}{2}-\dfrac{Kl_1^2}{2}$

$=\dfrac{1}{2}K(l^2-l_1^2)$

A force of

$(4x^{2}+3x)\ N$ acts on a particle which displaces it from

$x=2m$ to

$x=3m$ . Te work done by the force is:

0%
$32.8\ J$
0%
$3.281\ J$
0%
$0.328\ J$
0%
$zero$

Explanation

$W=F dx=\int (4x^2+3x)dx=\dfrac{4x^3}{3}+\dfrac{3x^2}{2}=32.8J$

A boy is swinging on a swing such that his lowest and highest positions are at heights of

$2m$ and

$4.5m$ respectively. His velocity at the lowest position is:

0%
$2.5{ms}^{-1}$
0%
$7{ms}^{-1}$
0%
$14{ms}^{-1}$
0%
$20{ms}^{-1}$

Explanation

The height of the boy falls with the swing,

$h=4.5-2=2.5m$

Therefore,

The velocity at the lowest position

$=\sqrt{2gh}$

$v=\sqrt{2\times9.8\times2.5}=\sqrt{49}$

$v=7 m/s$

A position dependent force F =

$3x^{2} - 2x + 7$ acts on a body of mass 7 kg and displace it from

$x$ = 10 m to

$x$ = 5 m. The work done on the body is

$x'$ joule. If both

$F$ and

$x'$ are measured in SI units, the value of

$x'$ is :

0%
-835
0%
235
0%
335
0%
935

Explanation

Given that

$F= 3x^2-2x+7$

Mass of body =

$7\ kg$

Work done by force,

$W=\int_{10}^5 F\cdot dx=\int_{10}^5(3x^2-2x+7)dx= [x^3- x^2+7x]_{10}^5= -835\ J$

A particle moves along the

$x-$ axis from

$x=0$ to

$x=5m$ under the influence of a force given by

$F=$

$7 - 2x + 3{x^2}N$ . The work done in the process is

0%
$107\ J$
0%
$270\ J$
0%
$100\ J$
0%
$135\ J$

Explanation

$W=\int_{0}^{5}Fdx$

$=\int_{0}^{5}(7-2x+3x^2)dx$

$=35-25+125$

$=135J$

A body mass of

$6kg$ is under a force which causes displacement in it given by

$= \dfrac{{{t^2}}}{4}$ metres where

$t$ is time.The work done by the force in

$2$ seconds is

0%
$12J$
0%
$9J$
0%
$6J$
0%
$3J$

A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of

$0.1 \ weber / m^2$ . the moment of work done in rotating it through

$180^\circ$ from its equilibrium position will be

0%
0.1 J
0%
0.2 J
0%
0.4 J
0%
0.8 J

Explanation

Magnetic moment of current carrying loop

$= M = iAN = i\left( {\pi {r^2}} \right)N$

$W=MB \left( {\cos {\theta _1} - \cos {\theta _2}} \right) = MB\left( {\cos {0^0} - \cos {{180}^0}} \right) = 2MB$

$=0.1 J$

$\therefore$ Option

$A$ is correct.

A particle of mass

$2 kg$ travels along straight line with velocity

$v=a\sqrt{x}$ , where

$x$ is constant. The work done by net force during the displacement of particle from

$x=0$ to

$x=4m$ is

0%
$a^2$
0%
$2a^2$
0%
$4a^2$
0%
$\sqrt{2}a^2$

Explanation

$v=a\sqrt{x}\quad \Rightarrow \cfrac{dx}{dt}=ax^{1/2}\quad \Rightarrow \int x^{1/2} dx=\int a dt$

$2\sqrt{x}=at\quad \Rightarrow x=\cfrac{a^2t^2}{4}$

$\Rightarrow v=\cfrac{a^2t}{2}$

$\Rightarrow a^{\prime}=\cfrac{a^2}{2}$

$F=ma^{\prime}=\cfrac{ma^2}{2}$

$Work = Force\times Displacement$

$\Rightarrow Work =\cfrac{ma^2}{2}\times 4=2ma^2$
As,

$m=2$

$\therefore Work = 4a^2$

A sphere, a cube and a thin circular plate; all are of the same material and same mass and all of them are initially heated to same high temperature. Then:

0%
plate will cool fastest and cube the slowest
0%
sphere will cool fastest and cube the slowest
0%
plate will cool fastest and sphere the slowest
0%
cube will cool fastest and plate the slowest

Explanation

1. The surface area of sphere < surface area of the cube for the given same mass and same density.

2. The rate of cooling

$\propto$ area of contact with surroundings.

$\therefore$ the plate will cool faster than the sphere.

In the figure, a 4.0 kg ball is on the end of a 1.6 m rope that is fixed at O. The ball is held at point A, with the rope horizontal is given an initial downward velocity. The ball moves theough three qyarters of a circle with no friction and arrives at B, with the rope barely under tension. Thee initial velocity of the ball, at point A, is closest to

0%
4.0 m/s.
0%
5.6 m/s.
0%
6.2 m/s
0%
6.8 m/s

Explanation

The motion of the ball is always perpendicular to the tension in the Rope. Therefore, the tension does no work on the system and the total mechanical energy of the ball is conserved.

$E_A = E_B$

$mgR + \dfrac{1}{2} mv_A^2 = mg(2R) + \dfrac{1}{2} mv_B^2$

$v_A^2 = 2gR +v_B^2$

Newton's second law for the ball at point B

$mg + T = \dfrac{mv_B^2}{R}$

$T \approx 0$

$v_B = gR$

$v_A = 3gR$

= 3(9.8)(1.6)

= 6.9 m/s

Since the velocity is near to 6.8m/s.

Hence (D) option is correct answer

One end of a string of length

$1.0m$ is tied to a body of mass

$0.5kg$ . It is whirled in a vertical circle with angular velocity

$4 rad/s$ . The tension in the string when body is at the lower most point of its motion is equal to [Take

$g=10m/s^2$ ].

0%
$3N$
0%
$5N$
0%
$8N$
0%
$13N$

Explanation

$T - mg = \cfrac{mv^2}{r}$

or,

$T - mg = m \omega ^2 r$

$\therefore T = mg + m\omega^2 r$

$= (0.5 \times 10) + (0.5 \times 4^2 \times 1)$

$= 5+8 = 13N$

980092_1075445_ans_2ef8b06d2afb41ff81c6ea7d6737324d.png

The velocity

$(v)$ of a particle of mass

$m$ moving along x-axis is given by

$v = \alpha \sqrt {x}$ , where

$\alpha$ is a constant. Find work done by force acting on particle during its motion from

$x = 0$ to

$x = 2m$ .

0%
$m\alpha^{2}$
0%
$m\alpha$
0%
$\dfrac {m\alpha}{2}$
0%
None of these

Explanation

$v=\alpha\sqrt{x}$

$\cfrac{dx}{dt}=\alpha\sqrt{x}$

$\Rightarrow \int_{0}^{x}{\cfrac{dx}{\sqrt{x}}}=\int_{0}^{t}{\alpha dt}$

$x=\cfrac{\alpha^2t^2}{4}$

$v=\cfrac{\alpha^2t}{2}$

$a=\cfrac{\alpha^2}{2}$

$F=ma=\cfrac{m\alpha^2}{2}$

$W=F\cdot \Delta x=\cfrac{m\alpha^2}{2}\times 2=m\alpha^2$

A small sphere of mass m suspended by a thread is first taken aside so that the thread forms the right angle with the vertical and then released, then:

The total acceleration of the sphere and the thread tension as a function of

$\theta$ , the angle of deflection of the thread from the vertical will be

0%
$g\sqrt{1+3cos^2} \theta, T=3 mg\,cos\,\theta$
0%
$g \, cos\,\theta, T=3 mg \, cos\,\theta$
0%
$g\sqrt{1+3sin^2} \theta, T=5 mg\,cos\,\theta$
0%
$g \, sin\,\theta, T=5 mg \, cos\,\theta$

Which of the following bodies has the largest kinetic energy?

0%
Mass $3M$ and speed $V$
0%
Mass $3M$ and speed $2V$
0%
Mass $2M$ and speed $3V$
0%
Mass $M$ and speed $4V$

Explanation

We know that, the kinetic energy is

$K.E=\dfrac{1}{2}m{{v}^{2}}....(I)$

Checking all options given -

A) $m=3M$ & $v=v$

$K.E=\dfrac{1}{2}m{{v}^{2}}$

$K.E=\dfrac{1}{2}\times 3M\times {{v}^{2}}$

$K.E=\dfrac{3M{{v}^{2}}}{2}$

B) $m=3M$ & $v=2v$

$K.E=\dfrac{1}{2}m{{v}^{2}}$

$K.E=\dfrac{1}{2}\times 3M\times 4{{v}^{2}}$

$K.E=6M{{v}^{2}}$

C) $m=2M$ & $v=3v$

$K.E=\dfrac{1}{2}m{{v}^{2}}$

$K.E=\dfrac{1}{2}\times 2M\times 9{{v}^{2}}$

$K.E=9M{{v}^{2}}$

D) $m=M$ & $v=4v$

$K.E=\dfrac{1}{2}\times M\times 16{{v}^{2}}$

$K.E=8M{{v}^{2}}$

is $9M{{v}^{2}}$ when $m=2M$ and $v=3v$ .

The angle

$\theta$ between the thread and the vertical at the moment when the total acceleration vector of the sphere is directed horizontally will be

0%
$cos\, \theta =\dfrac{1}{\sqrt{3}}$
0%
$cos\, \theta =\dfrac{1}{3}$
0%
$sin\, \theta =\dfrac{1}{\sqrt{3}}$
0%
$sin\, \theta =\dfrac{1}{\sqrt{2}}$

A force

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

$\vec { S }=(2{t^2}\hat { i } -5\hat { j })$ . Find the work done by this force in initial

$2s$ .

0%
$23\ J$
0%
$32\ J$
0%
$Zero$
0%
$48\,J$

Explanation

Given,

Force, $F=\left( 3t\hat{i}+5\hat{j} \right)N$

Displacement, $s=\left( 2{{t}^{2}}\hat{i}-5\hat{j} \right)\,m$

$ds=\left( 4t\hat{i}+0\hat{J} \right)\,dt$

$dw=F.ds$

$\int_{0}^{w}{dw}=\int_{0}^{2}{\left( 3t\hat{i}+5\hat{j} \right)}\left( 4t\hat{i}+0\hat{j} \right)\,dt$

$W=\left. 12{{t}^{2}} \right|_{0}^{2}=48\,J$

Net work done is $48\,J$

A body of mass

$m$ starts moving from rest along x-axis so that its velocity varies as

$v = a{s}^{1/2}$ where

$a$ is a constant and

$s$ is the distance covered by the body. The total work done by all the forces acting on the body in the first

$t$ seconds after the start of the motion is

0%
$\dfrac {ma^{4}t^{2}}{4\sqrt {2}}$
0%
$8ma^{4}t^{2}$
0%
$4ma^{4}t^{2}$
0%
$\dfrac {1}{8}ma^{4}t^{2}$

A body of mass

$2kg$ makes an elastic collision with another body at rest and comes to rest .The mass of the second body which collides with the first body is

0%
$2 kg$
0%
$1.2 kg$
0%
$3 kg$
0%
$1 kg$

A force

$\vec{F}=x\hat{i}+2y\hat{j}$ is applied on a particle. Find out work done by

$F$ to move the particle from point

$A$ to

$B$

0%
$-3.5\ J$
0%
$-2.5\ J$
0%
$-4.5\ J$
0%
$-4\ J$

Explanation

It is given that,

$F=x\hat{i}+2y\hat{j}$

$Let,\,\,dx=dx\hat{i}+dy\hat{j}$

$dw=\int{F.dx}$

$=\int{(x\hat{i}+2y\hat{j}})(dx\hat{i}+dy\hat{j})$

$=\int\limits_{1}^{0}{xdx}+2\int\limits_{2}^{1}{ydy}$

$={{\left. \dfrac{{{x}^{2}}}{2} \right|}_{1}}^{0}+{{\left. {{y}^{2}} \right|}_{2}}^{1}$

$=\dfrac{-1}{2}+(1-4)$

$=-0.5-3$

$=-3.5\,\,J$

Work done from

$d=0\ m$ to

$d=4\ m$

0%
$12.5\ J$
0%
$15\ J$
0%
$17.5\ J$
0%
$20\ J$

Force acting on a particle moving in a straight line varies with the velocities of the particle as

$F=K.V$ . Where

$K$ is constant. The work done by this force in time

$t$ is

0%
$KVt$
0%
$K^{2}V^{2}t^{2}$
0%
$K^{2}Vt$
0%
$KV^{2}t$

Explanation

Given that,

Force $F=KV$

Velocity= $V$

We know that,

$F=Ma$

Now,

$Ma=KV$

$M\left( \dfrac{dV}{dt} \right)=KV$

$MdV=KVdt$

Multiply by v in both side

$MVdV=K{{V}^{2}}dt$

On integrating both side

$\int\limits_{{{v}_{1}}}^{{{v}_{2}}}{MVdV=\int\limits_{0}^{t}{K{{V}^{2}}dt}}$

$M\left[ \dfrac{V_{2}^{2}}{2}-\dfrac{V_{1}^{2}}{2} \right]=K{{V}^{2}}t$

$\dfrac{M}{2}\left[ V_{2}^{2}-V_{1}^{2} \right]=K{{V}^{2}}t$

Now, the change in K.E

Now, from work energy theorem

Change in energy = work done

$K.E=K{{V}^{2}}t$

Hence, the work done is $K{{V}^{2}}t$

The distance

$(x)$ converted by a body of

$2\ kg$ under the action of a force is related to time

$t$ as

$x=t^{2}/4$ . What is the work done by the force in first

$2$ seconds?

0%
$4\ J$
0%
$2\ J$
0%
$1\ J$
0%
$0.5\ J$

Explanation

Given that,

Mass $m=2\,kg$

Distance $x=\dfrac{{{t}^{2}}}{4}$

Now, on differentiate

$\dfrac{dx}{dt}=\dfrac{1}{2}t$

Now, velocity and acceleration is

$v=\dfrac{dx}{dt}=\dfrac{t}{2}$

$a=\dfrac{dv}{dt}=\dfrac{1}{2}$

Now, the force is

$F=ma$

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

$F=1\,N$

Now, the work done is

$dW=F\centerdot dx$

$\int{dW}=\int\limits_{0}^{2}{1}\times \dfrac{t}{2}dt$

$W=\left[ \dfrac{{{t}^{2}}}{4} \right]_{0}^{2}$

$W=\left[ \dfrac{4}{4}-0 \right]$

$W=1\,J$

Hence, the work done is $1\ J$

A body of mass travels in a straight line with a velocity

$v=kx^{3/2}$ where

$k$ is a constant. The work done in displacing the body from

$x=0$ to

$x$ is proportional to:

0%
$x^{1/2}$
0%
$x^{2}$
0%
$x^{3}$
0%
$x^{5/2}$

Explanation

Given,

Velocity, $\dfrac{dx}{dt}=v=k{{x}^{3/2}}$

Acceleration,

$a=\dfrac{dv}{dt}=\dfrac{d\left( k{{x}^{3/2}} \right)}{dt}=\dfrac{1}{2}k{{x}^{1/2}}\dfrac{dx}{dt}=\dfrac{1}{2}k{{x}^{1/2}}v$

$a=\dfrac{1}{2}k{{x}^{1/2}}\left( k{{x}^{3/2}} \right)=\dfrac{1}{2}{{k}^{2}}{{x}^{2}}$

$work\,done\,=\,force\times displacemet\,=Mass\times acceleration\times displacement$

$\Rightarrow dW=m.a\,.dx=m.\dfrac{1}{2}{{k}^{2}}{{x}^{2}}dx$

On integrating

$W=m\dfrac{1}{2}{{k}^{2}}\left( \dfrac{{{x}^{3}}}{3} \right)$

Hence, Work done is proportional to ${{x}^{3}}$

A particle moves from origin to position

$\vec {r}_{1}=3\hat {i}+2\hat {j}-6\hat {k}$ under the action of force

$4\hat {i}+\hat {j}+3\hat {k}N$ . the work done will be

0%
$10\ J$
0%
$5\ J$
0%
$2\ J$
0%
$-4\ J$

Under influence of a force

$\overrightarrow { F } \left( x \right) =\left( 3{ x }^{ 2 }-2x+5 \right) \hat { i } N$ there is displacement of a particle from

$x=0$ and

$x=5m$ on

$x-$ axis. So work done is

$J$ .

0%
$100$
0%
$150$
0%
$125$
0%
$120$

A particle moves under the effect of a force

$F=c\ x$ from

$x=0$ to

$x=x_{1}$ . The work done in the process is

0%
$\dfrac{c{ { x }_{ 1 } }^{ 2 }}{2}$
0%
$c{ { x }_{ 1 } }^{ 2 }$
0%
$c{ { x }_{ 1 } }^{ 3 }$
0%
$Zero$

An inclined track ends in a circular loop of diameter

$D$ . From what height on the track a particles should be released so that it completes that loop in the vertical plane?

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

A stone, tied at the end of a string

$80$ cm long, is whirled in a horizontal circle with a constant speed. If the stone makes

$14$ revolutions in

$25$ sec, what is the magnitude of acceleration of the stone.

0%
$680$ cm $/s^2$
0%
$720$ cm $/s^2$
0%
$860$ cm $/s^2$
0%
$990$ cm $/s^2$

Explanation

Given length of string = 80 cm and angular velocity

$\omega$ = 2

$\pi$ f=

$\dfrac { 88 }{ 25 }$

And we know centripetal acceleration =

$r{ \omega }^{ 2 }$

$80\times { (\frac { 88 }{ 25 } ) }^{ 2 }$

=991.232 cm/s

=990 cm/s

A force acts on a

$3$ g 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%
$2.88$ J
0%
$450$ mJ
0%
$490$ mJ
0%
$530$ mJ

Explanation

We have,

mass,

$m=3\ g=0.003\ kg$

$x=3t-4t^2+t^3$

Now,

$v=\dfrac{dx}{dt}=3-8t+3t^2\Rightarrow dx=(3-8t+3t^2)dt$

$\Rightarrow a=\dfrac{dv}{dt}=0-8+6t$

Now,

$dw=Fdx$

$\Rightarrow dw=(ma)dx$

$\Rightarrow dw=(0.003)(-8+6t)(3-8t+3t^2)dt$

$\Rightarrow dw=(0.003)(18t^3-72t^2+82t-24)dt$

$\Rightarrow w=(0.003)\displaystyle \int_{0}^{4}{(18t^3-72t^2+82t-24)dt}$

$\Rightarrow w=0.003\times 176=0.528\ J$

$\Rightarrow w=530\ mJ$

$4\ kg$ particle moves along the

$X-$ axis. Its position

$x$ varies with time according to

$x\left( t \right) = t + 2{t^3}$ , where

$X$ is in

$m$ and

$t$ is in seconds. Compute kinetic energy of the particle in time t

0%
$KE=72t^2+24t+2$
0%
0
0%
20t
0%
40t

Explanation

Kinetic energy is given by,

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

$v=\dfrac{dx}{dx}$

$=\dfrac{d}{dt}(t+2t^3)$

$v=1+6t^2$

$\Rightarrow v^2=(1+6t^2)^2=1+36t^4+12t^2$

$KE=\dfrac{1}{2}\times 4\times (1+36t^4+12t^2)$

$=2(1+36t^4+12t^2)$

$\therefore KE=72t^2+24t+2$

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 7 - 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 7 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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