MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 5

(i) What is the work done by the porter when he climbs up a height of 10 m

$\displaystyle (g\quad = 10\ {ms} ^{-2})$ ?

0%
$5kJ^2$
0%
$50kJ$
0%
$100kJ^2$
0%
$5kJ$

Explanation

The work done by the potter is defined as the product of the force and the displacement.

Work done = force

$\times$ displacement

$m \times g \times 10$ ( Since force = mass

$\times$ gravity)

$50 \times 10 \times 10$

$5KJ$

A ball moving with a velocity v strikes a wall moving toward the ball with a velocity u. An elastic impact lasts for t sec. Then the mean elastic force acting on the ball is

0%
$\displaystyle \frac { 2mv }{ t }$
0%
$\displaystyle \frac { 2m\left( \upsilon +u \right) }{ t }$
0%
$\displaystyle \frac { 2m\left( \upsilon +2u \right) }{ t }$
0%
$\displaystyle \frac { m\left( 2\upsilon +u \right) }{ t }$

Explanation

Relative speed of the ball

$\displaystyle =\left( \upsilon +u \right)$
Speed after rebouncing

$\displaystyle =-\left( \upsilon +u \right)$
Now,

$\displaystyle F=m\frac { \Delta \upsilon }{ \Delta t } =\frac { m }{ t } \left[ \left( \upsilon +u \right) \right] -[-\left[ \left( \upsilon +u \right) \right]]$

$\displaystyle =\frac { 2m }{ t } \left( \upsilon +u \right)$

Two bodies of equal weight are kept at heights of h and 1.5 h respectively. The ratio of their P.E. will be :

0%
3: 2
0%
2: 3
0%
1: 1
0%
none of these

Explanation

$\underline{\text{First body:}}$

Mass of the body =

$m$

Height at which the body is kept =

$h$

Potential energy of the body

$U_1=mgh$

$\underline{\text{Second body:}}$

Mass of the body =

$m$

Height at which the body is kept =

$1.5h$

Potential energy of the body

$U_2=1.5mgh$

The ratio of potential energy is

$U_1:U_2=1:1.5=\dfrac{2}{3}=2:3$

An object of mass

$1 kg$ has a P.E. of

$1 J$ relative to the ground when it is at a height of

0%
$0.1 m$
0%
$1 m$
0%
$9.8 m$
0%
$32 m$

Explanation

$P.E$ is given by

$=mgh$

where

$h$ is the height,

$m$ is mass and

$g$ is acceleration due to gravity

$1J=1\times 10\times h\\ \\$

$h=0.1m$

A body rolling down a hill has :

0%
K.E. only
0%
P.E. only
0%
neither K.E. nor P.E.
0%
both K.E. and P.E

A raised hammer possesses :

0%
K.E.
0%
P.E.
0%
electrical energy
0%
sound energy

The energy directly related to the speed of a moving body and its mass is:

0%
kinetic
0%
potential
0%
solar
0%
electric

A heavy object has ....... gravitational potential energy than a lighter one.

0%
less
0%
more
0%
same
0%
none of these

Explanation

Gravitational potential energy is given by

$PE=mgh$

where

$m$ is mass so heavy object means more mass and more mass means more potential energy.

The relationship between force and position is shown in the figure (in one dimensional case). Work done by the force in displacing a body from

$X=1\ cm\ to\$

$X=5\ cm\ is$ :

0%
700 ergs
0%
70 ergs
0%
60 ergs
0%
20 ergs

Explanation

Work is area under the curve.

$W=W_1+W_2+W_3+W_4$

$W_1=area\ under\ A_1BCM_2$

$W_2=area\ under\ M_2DEF_3$

$W_3=area\ under\ F_3GHI_4$

$W_4=area\ under\ I_4JKL5$

$W_1=10\times 1=10ergs$

$W_2=20\times 1=20ergs$

$W_3=-20\times 1=-20ergs$

$W_4=10\times 1=10ergs$

$W=W_1+W_2+W_3+W_4=10+20-20+10=20\ ergs$

The mass of an object

$P$ is double the mass of

$Q$ . If both move with the same velocity, then the ratio of K.E. of

$P$ to

$Q$ is

0%
1 : 2
0%
2 : 1
0%
1 : 4
0%
4 : 1

Explanation

$M_P =2 \times M_Q$

$V_P = V_Q$

$K.E_P: K.E_Q = ?$

$\implies \dfrac{1}{2}M_PV_P^2 : \dfrac{1}{2}M_QV_Q^2$

Since

$V_P$ =

$V_Q$

$\implies M_P : M_Q$

$\implies 2 \times M_Q : M_Q$

$\implies 2 : 1$

Which one of the following types of energy is possessed by a body when placed at a certain height?

0%
Kinetic energy
0%
Potential energy
0%
Heat energy
0%
Sound energy
0%
None of these

Explanation

When a body is placed at a certain height, the body possesses potential energy.

$P.E. = mgH$

$m$ ; the mass of the object

$g$ ; acceleration due to gravity

$H$ ; height gained by object

Find the potential energy if a body is placed at a certain height of 5 m from the earth.

0%
196 J
0%
200 J
0%
150 J
0%
300 J
0%
None of these

The water stored in a reservoir possesses :

0%
Kinetic energy
0%
Muscular energy
0%
Potential energy
0%
Magnetic energy

Kinetic energy of a body depends on its:

0%
Position
0%
Velocity
0%
Shape
0%
Colour

Explanation

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

So the kinetic energy of a body depends upon its mass and velocity.

When an aeroplane takes off from the ground :

0%
Kinetic energy increases and Potential energy decreases
0%
Potential energy increases and Kinetic energy remains constant
0%
Both Kinetic energy and Potential energy remain constant
0%
Both Kinetic energy and Potential energy increase

Explanation

Kinetic energy is due to the motion of the object and potential energy is due to relative elevation.

So as the plane takes off height increases and also its speed increase, so both kinetic and potential energy increases.

$\text{Kinetic energy} = \dfrac{1}{2}mv^2$

$\text{Potential energy} = mgH$

Which of the following does not have potential energy ?

0%
An inflated balloon
0%
Water in a flowing river
0%
A fruit on the tree
0%
A spinning top

A stone tied to a string is rotated in a vertical plane. If mass of the stone is

$m$ , the length of the string is

$r$ and the linear speed of the stone is

$v$ ,When the stone is at its lowest point, then the tension in the string will be :

0%
$\displaystyle \frac { m{ v }^{ 2 } }{ r } +mg$
0%
$\displaystyle \frac { m{ v }^{ 2 } }{ r } -mg$
0%
$\displaystyle \frac { mv }{ r }$
0%
$\displaystyle mg$

Explanation

At the lowest point, as shown in the figure both mg and centrifugal force

$\dfrac { m{ v }^{ 2 } }{ r } \\$ will act in the same direction so,

$T=mg+\dfrac { m{ v }^{ 2 } }{ r } \\$

A crane pulls up a car of mass

$500kg$ to a vertical height of

$4m$ . So, work done by the crane is:

0%
$19.6J$
0%
$19.6kJ$
0%
$19600kJ$
0%
$4900J$

Explanation

To raise the car, the crane has to do work against the force of gravity.
Therefore, the force required to lift the car,

$F=mg=500\times{9.8}N=4900N$
Displacement,

$S=$ vertical height raised

$=4m$ .

$\therefore$ Work done,

$W=F.S=4900\times{4}J=19600J=19.6kJ$

A boy of mass

$40\ kg$ runs up a flight of

$50$ steps, each

$10\ cm$ high in

$14\ s$ . So, work done by the boy is:

0%
$1960\ J$
0%
$19.6\ J$
0%
$980\ J$
0%
$9.8\ J$

Explanation

Mass of the boy,

$m = 40\ kg$

Number of steps =

$50$

Height of each step =

$10\ cm$

Force on the boy due to gravity,

$F=mg=40\times{9.8}N=392\ N$
While climbing up the steps, the boy does work against gravity.
Displacement in the vertical direction,

$s=(50\times{10})\ cm=500\ cm=5\ m$

Displacement is in the direction of force applied by the boy against gravity.
So, work done,

$W=F\times{s}=392\times{5}\ J=1960\ J$

A fruit hanging from the top branch of a tree possesses

0%
gravitational potential energy
0%
elastic potential energy
0%
kinetic energy
0%
sound energy

A body of mass

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

$x=\cfrac{{t}^{2}}{4}$ metre, where

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

$2s$ is

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

Explanation

The displacement of a body is given by

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

$\therefore$ Velocity,

$v = \dfrac{dx}{dt} = \dfrac{1}{4} \times 2t = \dfrac{t}{2}$

Initial velocity,

$u = v\bigg|_{t= 0 } = 0$

Final velocity,

$v = v\bigg|_{t= 2} = \dfrac{2}{2} = 1$

$m/s$

From work-energy theorem,

$W = \Delta K.E$

$W = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$

$W = \dfrac{1}{2} \times 6 \times 1^2 - 0$

$\Rightarrow$

$W = 3J$

In which form, the energy is stored in a fuel?

0%
Potential energy
0%
Thermal energy
0%
Kinetic energy
0%
Chemical energy

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

Explanation

Work done in stretching a string to obtain an extension

$l$ is

${W}_{1} = \dfrac{1}{2} Kl^2$

Similarly, work done in stretching a string to obtain extension

${l}_{1}$ is

${W}_{2} = \dfrac{1}{2} K{l}_{1}^{2}$

$\therefore$ Work done in second case

$= {W}_{2} - {W}_{1} = \dfrac{1}{2} K \left({l}_{1}^{2} - {l}^{2}\right)$

An object of mass

$40\ kg$ is raised to a height of

$5\ m$ above ground. If the object is allowed to fall down find its kinetic energy when it is half-way down.

0%
1960 J
0%
980 J
0%
990 J
0%
1000 J

Explanation

Gravitational Potential energy, $W = mgh$

$h=$ vertical displacement $=5\ m$

$m =$ Mass of object $= 40\ Kg$

$g =$ Acceleration due to gravity $= 9.8\ ms^{-2}$

$Kinetic$ $Energy$ = $Loss$ $in$ $PE$ = $40 \times 5 \times 9.8= 1960\ J$

For half way down $=\dfrac{1960}{2}= 980\ J$

At this point i.e. half way object has an equal amount of potential and kinetic energy.

Which one of the following energies cannot be possessed by a body at rest?

0%
Potential energy
0%
Kinetic energy
0%
Thermal energy
0%
Magnetic energy

If the string were to break at point II, what would be the path of the ball?

0%
0%
0%
0%
0%
None of these

Explanation

When the bob is swinging in vertical circle the centripetal force and the pseudo force varies. Centripetal force causes the tension in the string.

Let centripetal force be

$C$ , pseudo force be

$P$ and weight of bob be

$W$ .

At position 2,

$C = W$

If the string were to break at this position, the only external force acting on the bob is

$W$ (downward). Hence the bob would move vertically down.

Therefore option D is correct.

Which of the following statements is true?

0%
The tension in the string is greater at points III than at point I
0%
The tension in the string is greater at points I than at point III
0%
The tension at point I is equal to the tension at point III
0%
The tension in the string is greatest at point II
0%
The tension in the string is same at points I, II and III

Explanation

When the bob is swinging in vertical circle the centripetal force and the pseudo force varies. Centripetal force causes the tension in the string.

Let centripetal force be

$C$ , pseudo force be

$P$ and weight of bob be

$W$ .

At position 1,

$C = P - W$

At position 2,

$C = W$

At position 3,

$C = P + W$

As C at position 3 is greatest, the tension in the string is greatest at this position.

Therefore option A is correct.

A ball of mass

$m$ is attached with the string of length

$R$ , rotating in circular motion, with instantaneous velocity

$v$ and centripetal acceleration

$a$ .

Calculate the centripetal acceleration of the ball if the length of the string is doubled?

0%
${a}/{4}$
0%
${a}/{2}$
0%
$a$
0%
$2a$
0%
$4a$

Explanation

Centripetal acceleration

$a = \dfrac{v^2}{R}$

Now the length of the string is doubled keeping the instantaneous velocity constant i.e

$R' = 2R$

$\therefore$ New centripetal acceleration

$a' = \dfrac{v^2}{R'} = \dfrac{v^2}{2R} =\dfrac{a}{2}$

A car standing at the top of a hill would be an example of what type of energy?

0%
Kinetic
0%
Topical
0%
Potential
0%
Beneficial
0%
Residual

Explanation

The car standing at the top of hill has attained a height above the ground level. Thus it has attained a gravitational potential energy given by

$mgh$

where

$m$ is the mass of the car,

$g$ is the gravitational constant,

$h$ is the height above ground that it has attained.

This energy can convert into the kinetic energy of the car if it starts coming down the hill, that is lose its potential energy.

$110 kg$ panda is riding on a

$3.0 m$ long swing whose mass can be considered negligible. The highest point of its arc occurs when the swing makes a

${20}^{o}$ angle with the vertical.
What is the magnitude of the total tension in the ropes of the swing at that point?

0%
$0N$
0%
$103 N$
0%
$369 N$
0%
$1013 N$
0%
$1078 N$

Explanation

Given :

$m =110$ kg

Tension in the string

$T = mg$

$cos20^o = 110 \times 9.8 \times 0.94 = 1013$ N

The position function of a particle is given by

$x\left(t\right)=k{t}^{{5}/{2}}$ , where

$k$ is a constant.
If the particle starts at rest and is propelled through some distance

$d$ so that the trajectory matches

$x\left(t\right)$ , the work done on the particle is proportional to which power of

$t$ ?

0%
${t}^{5}$
0%
${t}^{3}$
0%
${t}^{{5}/{2}}$
0%
${t}^{{3}/{2}}$
0%
Not enough information

Explanation

$x=kt^{5/2}$

Speed of the particle is

$v=\dfrac{dx}{dt}=\dfrac{5}{2}kt^{3/2}$

The acceleration of the particle becomes

$a=\dfrac{dv}{dt}=\dfrac{15}{4}kt^{1/2}$

Force acting on particle is given by

$ma=\dfrac{15}{4}mkt^{1/2}$

Hence the work done is given by

$F.x\propto t^{3}$

A stone ties to a rope is rotated in a vertical circle with uniform speed. If the difference between the maximum and minimum tension in the rope is

$20\ N$ , mass of the stone in

$kg$ is:
(

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

0%
$1.0$
0%
$1.5$
0%
$0.5$
0%
$0.75$

Explanation

Let the mass of the stone be

$m$ , and the angular speed with which it rotates be

$\omega$ .

Thus at the bottom most point,

$T_b-mg=m\omega^2 l$

And at the top most point,

$T_t+mg=m\omega ^2 l$

$\implies T_b-T_t=2mg=20$

$\implies mg=10$

$\implies m=1\ kg$

In vertical circular motion, the ratio of kinetic energy to potential energy at the horizontal position is ____________.

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

Explanation

We assume that the object in motion just manages to complete the vertical circle. In that case,

$v_{top} = \sqrt{gR}$
Total energy at the topmost point

$=\dfrac12mgR + 2mgR = \dfrac52mgR$

At the horizontal position

$PE = mgR$

Total energy

$= \dfrac52mgR$

$\therefore KE = (\dfrac52-1) = \dfrac32mgR$

$\therefore \dfrac{KE}{PE} = \dfrac32$

$50 kg$ acrobat is swinging on a rope with a length of

$15 m$ from one horizontal platform to another. Both platforms are at equal height.
If the maximum tension the rope can support is

$1200 N$ , which of the following answer best represents the maximum velocity the acrobat can reach without breaking the rope?

0%
$25 {m}/{s}$
0%
$19 {m}/{s}$
0%
$20 {m}/{s}$
0%
$15 {m}/{s}$
0%
$5 {m}/{s}$

Explanation

Given :

$T =1200$ N

$m = 50$ kg

$L =15$ m

Let the maximum velocity of the acrobat be

$v$ .

From figure,

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

$\therefore$

$\dfrac{50 \times v^2}{15} =1200 - 50 \times 9.8$

$\implies v =\sqrt{213} \approx 15$

$m/s$

Under the action of a force

$F=Cx$ , the position of a body changes from

$0$ to

$x$ . The work done is :

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

Explanation

Work done

$W = \int^x_0 Fdx$

$W = \int^x_0 Cxdx$

$W = C\times \dfrac{x^2}{2}\bigg|^x_0$

$\therefore$

$W = \dfrac{C}{2}\times (x^2-0 )$

$\implies W = \dfrac{1}{2}Cx^2$

Calculate the work done in moving a cart from

$x=a$ to

$x=b$ whose mass is

$3kg$ and the given surface is frictionless.

$(Take\ g=9.8m/s^{2})$

0%
$0$
0%
$3g(a-b)$
0%
$3g$
0%
$3g(a+b)$

Fifteen joules of work is done on object A, which is attached to an uncompressed spring, so that all the work done goes into compressing the spring.
Sixty joules of work is done on object B, which is attached to an uncompressed spring that is identical to the spring to which object A is attached; all the work done goes into compressing the spring.
How does the compression of the spring for object B compare to the compression of the spring for object A after this work is done?

0%
The compression of the spring for object B is four times as much as it is for object A
0%
The compression of the spring for object B is sixteen times as much as it is for object A
0%
The compression of the spring for object B is two times as much as it is for object A
0%
The compression of the spring for object B is less than two times as much as it is for object A (but not the same amount)
0%
The compression of the spring for object B is the same as it is for object A

Explanation

The work done in compressing the spring is stored as the potential energy of spring , given by

$U=1/2kx^{2}$

where

$x=$ compression of spring ,

$k=$ spring constant ,

for object A , potential energy =work done

$1/2kx_{A}^{2}=15$ ...................................eq1

for object B , potential energy =work done

$1/2kx_{B}^{2}=60$ ....................................eq2

dividing eq2 by eq1 ,

$x_{B}^{2}/x_{A}^{2}=60/15=4$

$x_{B}/x_{A}=2$

$x_{B}=2x_{A}$

A simple pendulum of length

$L$ carries a bob of mass

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

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

The correct formula to find the velocity of a body with kinetic energy '

$k$ ' is:

0%
$V=\dfrac{2k}{m}$
0%
$V=\sqrt{\dfrac{2k}{m}}$
0%
$V=\dfrac{4{k}^{2}}{m}$
0%
$V=\dfrac{1}{2}km$

Explanation

Kinetic energy of a body is given by

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

$v^2$ =

$\dfrac{2k}{m}$

$v=\sqrt {\dfrac{2k}{m}}$ .

Hence, answer is option B.

Kinetic energy of a body depends upon its:

0%
mass
0%
velocity
0%
density
0%
both A and B

Explanation

Kinetic energy of the body

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

$\implies$

$K \propto m$ and

$K\propto v^2$

Thus kinetic energy of a body depends on the mass of the body as well as its velocity.

A body is acted upon by a force which is proportional to the distance covered. If distance covered by denoted by x, then work done by the force will be proportional to:

0%
$x$
0%
$x^2$
0%
$x^{\dfrac{3}{2}}$
0%
$x^4$

Explanation

Here, Force

$F\propto x$ or

$F=cx$ where c is propotionality constant

Work done

$W=Fx=cx\times x=cx^2$

So,

$W\propto x^2$

A small stone tied to an inextensible string of negligible mass is rotated in a circle of radius

$2 m$ in a vertical plane. Find the speed at a horizontal point on the circle. (in

$m/s$ )

0%
$7.67$
0%
$2.36$
0%
$6.32$
0%
$8.32$

Explanation

$v= \sqrt {3gr}$

$v= \sqrt {3 * 9.8* 2} = 7.67 m/s$

A stone of mass

$10 kg$ tied with a string of length

$0.5 m$ is rotated in a vertical circle. Find the total energy of the stone at the highest position. (in

$J$ )

0%
$123.36$
0%
$122.5$
0%
$154.3$
0%
$185.6$

Explanation

assume that potential energy at lowest point is zero and stone is rotating with critical velocity

so the magnitude of velocity at highest point is

$v=\sqrt { gl }$ where

$l=$ length of string

total energy

$E=K.E.+P.E.$

$\Rightarrow K.E.=\dfrac { 1 }{ 2 } m{ v }^{ 2 }=\dfrac { 1 }{ 2 } 10\times { (\sqrt { 9.8\times 0.5 } ) }^{ 2 }=24.5J$

$P.E.=mgh=mg(2l)=10\times 9.8\times \left( 2\times 0.5 \right) =98J$

so the total energy

$E=98+24.5=122.5J$

$2kg$ object is moving at

$3m/s$ . A

$4N$ force is applied in the direction of motion and then removed after the object has travelled an additional

$5m$ . The work done by this force is:

0%
$10 J$
0%
$15 J$
0%
$20 J$
0%
$25 J$

Explanation

intial velocity

$u=3m/s$

final velocity

$v^2=u^2+2as=3^2+(2 \times 2 \times 5)=29$

$\implies v=5.3m/s$

$KE=\dfrac{1}{2}m(v^2-u^2)=\dfrac{1}{2} \times 2 \times {((5.3)^2-3^2)}=20J$

$200 g$ mass is whirled in a vertical circle making

$60$ revolutions per minute. What is the tension in the string at the top of the circle if the radius of the circle is

$0.8 m$ ? (in

$N$ )

0%
$6.3$
0%
$2.36$
0%
$4.35$
0%
$4.23$

Explanation

$w = 60 \space \text {revolutions per minute }= 60 * 2\pi / 60 \space\space\text {rad/ sec} = 2\pi \space\space rad/ sec$

$200 gm = 0.2 kg$

$T = mrw^2 - mg$

$T = 0.2* 0.8 * (2\pi)^2 - 0.2* 9.8$

$T = 4.35 N$

A force of

$10 N$ is applied on a object of mass

$1 kg$ for

$2 s$ , which was initially at rest. What is the work done on the object by the force?

0%
$200 J$
0%
$20 J$
0%
$16 J$
0%
$180 J$

Explanation

Force acting on the object

$F = 10 N$

Mass of object

$m = 1kg$

Acceleration of the object

$a = \dfrac{F}{m}$

$\therefore$

$a = \dfrac{10}{1} = 10 m/s^2$

Initial speed of the object

$u = 0 m/s$

Distance covered by it in

$2s$ ,

$S = ut+\dfrac{1}{2}at^2$ where

$t = 2s$

$\therefore$

$S = 0+\dfrac{1}{2}\times 10 \times 2^2 = 20 m$

Work done

$W = FS$

$\implies$

$W = 10\times 20 = 200J$

$500 g$ particle tied to one end of a string is whirled in a vertical circle of circumference

$14 m$ .If the tension at the highest point of its path is

$2 N$ , what is its speed? (in m/s)

0%
$4.365$
0%
$5.369$
0%
$5.546$
0%
$4.331$

Explanation

as the diagram shows

balancing the force at point

$P$

given that

$T=2N$ ,

$m=0.5kg$ and

$R=\dfrac { 14 }{ 2\pi }=2.23m$

$\Rightarrow T+mg=m\dfrac { { v }^{ 2 } }{ R }$

$\Rightarrow v=\sqrt { R\dfrac { (T+mg) }{ m } }$

$\Rightarrow v=\sqrt { 2.23\times \dfrac { (2+0.5\times 10) }{ 0.5 } } =5.546m/s$

Two wires are stretched through same distance. The force constant of second wire is half as that of the first wire. The ratio of work done to stretch first wire and second wire will be

0%
$2 : 1$
0%
$1 : 2$
0%
$3 : 1$
0%
$1 : 3$

Explanation

$W=\dfrac { 1 }{ 2 } k{ x }^{ 2 }$
If both wires are stretched through same distance, then

$W\propto k$
It is given that

${ k }_{ 2 }=\dfrac { { k }_{ 1 } }{ 2 }$

$\therefore \dfrac { { W }_{ 1 } }{ { W }_{ 2 } } =\dfrac { { k }_{ 1 } }{ { k }_{ 2 } }$

$=\dfrac { { k }_{ 1 } }{ \dfrac { { k }_{ 1 } }{ 2 } } =\dfrac { 2 }{ 1 } =2 : 1$

Two springs have their force constant as

$k_1$ and

$k_2(k_1 > k_2)$ . When, they are stretched by the same force, then,

0%
More work is done in case of first spring
0%
Less work is done in case of first spring
0%
More work is done in case of second spring
0%
Loss work is done in case of second spring

Explanation

Work done

$W=\frac{F}{2k}$
Given,

$F_1=F_2=F$

$\therefore W\propto \frac{1}{k}$
Given,

$k_1 > k_2$

$\Rightarrow W_2 > W_1$
Hence, more work is done is case of second spring.

The work done in joules in increasing the extension of a spring of stiffness

$10\ N/cm$ from

$4\ cm$ to

$6\ cm$ .

0%
$1$
0%
$10$
0%
$50$
0%
$100$

Explanation

Given Spring stiffness

$K= 10 N/cm= 1000 N/m$ , Initial Position

$x_1= 4 cm= 0.04 m$ , Final position

$x_2= 6 cm= 0.06 m$

Work done by spring

$= \dfrac{1}{2}K(x_2^2-x_1^2)$

$\Rightarrow \text{Work done}= \dfrac{1}{2}\times 1000\times (0.06^2- 0.04^2)= 1.0 J$

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.