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

A body of mass

$10\ kg$ moves according to the relation

$x=t^{2}+2t^{3}$ . The work done by the force in the first

$2s$ is.

0%
$7840\ J$
0%
$1960\ J$
0%
$3920\ J$
0%
$4840\ J$

A force of

$5N$ acts on a

$15kg$ body initially at rest. The work done by the force during the first second of motion of the body is:

0%
$5J$
0%
$\cfrac{5}{6}J$
0%
$6J$
0%
$75J$

The maximum energy stored in the spring in the subsequent motion will be:

0%
$15 v_0^2$
0%
$5 v_0^2$
0%
Zero
0%
$10 v_0^2$

Explanation

It is given that 2 kg and 3 kg blocks are placed on a smooth horizontal surface and connected by spring which is unstretched initially. Let v is the final velocity then from conservation of momentum:

$5v=2(-4{{v}_{0}})+3({{v}_{0}})$

$5v=-8{{v}_{0}}+3{{v}_{0}}$

$v=-{{v}_{0}}...........(1)$

Let x is the maximum elongation in the spring. From conservation of energy:

$\dfrac{1}{2}(2){{(-4{{v}_{0}})}^{2}}+\dfrac{1}{2}3{{({{v}_{0}})}^{2}}=\dfrac{1}{2}k{{x}^{2}}+\dfrac{1}{2}5{{({{v}_{0}})}^{2}}\,(from\,\,equation(1))$

$16{{v}_{0}}^{2}+\dfrac{3}{2}v_{0}^{2}=\dfrac{1}{2}k{{x}^{2}}+\dfrac{5}{2}v_{0}^{2}$

$15v_{0}^{2}=\dfrac{1}{2}k{{x}^{2}}$

The energy stored in spring is $15v_{0}^{2}$

A block of mass

$10kg$ is moving in x-direction with a constant speed of

$10m/sec$ . It is subjected to a force

$F=-0.1x$ Joules/meter during its travel from

$x=20$ meters to

$x=30$ meters. Its final kinetic energy will be-

0%
$475$ joules
0%
$450$ joules
0%
$275$ joules
0%
$250$ joules

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 velocity of a car increase from

$54 km/hr$ to

$72km/hr$ . How much is the work done if the mass of the car is

$1500kg$ ?

0%
$231250 J$
0%
$131250 J$
0%
$431250 J$
0%
$13125 J$

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 smooth ball

$A$ collides elastically with an another identical ball

$B$ with velocity

$10 m/s$ at an angle of

$30^o$ from the line joining their centres

$C_1$ and

$C_2$ , then mark INCORRECT statement :

0%
Velocity of ball $A$ after collision is $15 m/s$
0%
Velocity of ball $B$ after collision is $5.4 m/s$
0%
Both the balls move at right angles an collision
0%
KE will not be conserved here, because collision is not head on

A massive disc of radius

$R$ is moving with a constant velocity

$u$ on an frictionless table. Another small disc collides with it elastically with a speed of

$v_{0} = 0.3\ m/s$ , the velocities of the discs are parallel. The distance

$d$ shown in the figure is equal to

$R/2$ , friction between the discs is negligible.
For which

$u$ (in m/s) will the small disc move perpendicular to its original motion after the collision?

0%
$1$
0%
$0.1$
0%
$2$
0%
$0.2$

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 $3.92\;k{\rm{J}}$

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 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%
$280J$

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 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 $F=-\dfrac{dU}{dx}$

An object is dropped from height

$h=2R$ on the surface of earth. Find the speed with which it will collide with ground, by neglecting effect of air. [

$R$ is radius of earth. Take mass of earth as

$M$ ]

0%
$2 \sqrt {\dfrac {GM} {3R}}$
0%
$\sqrt {\dfrac {2GM} {R}}$
0%
$\sqrt {\dfrac {GM} {R}}$
0%
$\sqrt {\dfrac {GM} {2}}$

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%
$100kg$ $m/s$

In the given circuit, if point C is connected to earth and a potential of +2000V is given to a point A . Then the potential at point B is

0%
1500 V
0%
1000 V
0%
5000 V
0%
500 V

$540$ 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 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 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

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%
$6.1\times10^{-23}J$

The diagram shows a barrel of weight

$1.0 \times 10^3$ N on a frictionless slope inclined at

$30^0$ to the horizontal. A force is applied to the barrel to move it up the slope at constant speed. The force is parallel to the slope. What s the work done in moving the barrel a distance of 5.0 m up the slope?

0%
$1.0 \times 10^4$ J
0%
$2.5 \times 10^3$ J
0%
$4.3 \times 10^3$ J
0%
$5.0 \times 10^3$ 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 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 man raises 1 kg wt. to a height of 100 cm and holds it there for 30 minutes. How much work has he performed?

0%
$1\times 9.8J$
0%
$1\times 9.8\times 30\times 60J$
0%
$1\times 9.8\times 30J$
0%
$1\times 9.8\times 30erg$

Explanation

Given

$m=1\ kg\\h=100\ cm=1\ m \\t=30\ mit$

Work dont

$W=$ Force

$\cdot$ Displacement

Force apply to lift the mass = Weight of the body=

$1\times 9.8$ N

From above equation,

$W=1\times 9.8\times 1\ J$

$W=1\times 9.8 \ J$

Option A

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

$M_2$

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

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

what horizontal force f be given to the wedge of mass M so that a block of mass m placed on it remains stationary with respect to the wedge. the magnitude of the force is

0%
Mg cot $\theta$
0%
$(M+m)g{\,\,}sec\theta$
0%
$(M+m)g{\,\,}tan\theta$
0%
$(M+m)g{\,\,}cosec\theta$

Explanation

$a=\dfrac { F }{ m+M } \quad \longrightarrow \left( 1 \right)$

The block will experience a force of

$ma$ and an equal opposite force will be exerted on the wedge.

Now, Since the block does not slide down

$Ma\cos\theta =Mgsina$

$a=g\tan\theta \quad \longrightarrow \left( 2 \right)$

Substituting in

$(1)$

$F=\left( m+M\right) g\tan\theta$

1220267_1220728_ans_f7916f82b6fa40d881a933cd65d06639.jpg

A body of mass

$5$ kg falls from a height of

$20$ meters on the ground and it rebounds to a height of

$0.2$ m. If the loss in potential energy is used up by the body, then what will be the temperature rise? (specific heat of material=

$0.09\ cal\ gm^{-1}\ {^{o}C^{-1}}$ )

0%
$5^0C$
0%
$4^0C$
0%
$8^0C$
0%
None of these

Two identical balls A & B of mass m each are placed on a fixed wedge as shown in figure. Ball B is kept at rest and it is released just before two balls collides. Ball A rolls down without slipping on inclined plane & collide elastically with ball B. The kinetic energy of ball A just after the collision with ball B is :

0%
$\dfrac{mgh}{7}$
0%
$\dfrac{mgh}{2}$
0%
$\dfrac{2mgh}{5}$
0%
$\dfrac{7mgh}{5}$

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%
$L/12$

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 block of mass m =1 kg, moving on a horizontal surface with speed

$v_1 = 2 ms^{-1}$ enters a rough patch ranging from

$x=0.10 m \,to \, x=2.01 m$ The retarding force

$F_r$ on the block in this range is

$F_r= -\frac{k}{x}$ for 0.1 m < x < 2.01 m

$=0$ for x<0.1 m and x > 2.01m
where k=0.5 J
The final kinetic energy of the block is :(given

$ln \,20.1\approx 3$ )

0%
0.5 J
0%
1 J
0%
1.5 J
0%
2 J

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 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

Two identical balls undergo elastic collision Speed of both the balls before colision Is a, maximum possible speed of any ball after collision well be:

0%
$u$
0%
$2u$
0%
$\sqrt{3} u$
0%
$\sqrt{2} u$

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.

A sphere of mass

$m$ moving with velocity

$v$ strikes elastically with a wall moving towards the sphere with a velocity

$u$ . If the mass of the wall is infinitely large, the work done by the wall during collision will be

0%
$mu(u+v)$
0%
$2mu(u+v)$
0%
$2mv(u+v)$
0%
$2m(u+v)$

Explanation

It is given that

$v_{w}=-u$ and

$u_1=v$

let speed of ball after collision be

$v'$

$e=\dfrac{\text{velocity of seperation}}{\text{velocity of approach}}$

$e=\dfrac{v_{wall}+v'}{u_1-v_{wall}}$

$1=\dfrac{-u+v'}{v+u}$

$v'=v+2u$

work done by wall

$=$ change in kinetic energy of ball

$W=\Delta KE$

$W=\dfrac{1}{2}mv'^2-\dfrac{1}{2}mu_1^2$

$W=\dfrac{1}{2}m(v+2u)^2-\dfrac{1}{2}mv^2$

$W=mu(u+v)$

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 neutron moving with velocity u collides with a stationary

$\alpha -particle$ The velocity of the neutron after collision is

0%
$-\dfrac { 3\cup }{ 5 }$
0%
$\dfrac { 3\cup }{ 5 }$
0%
$\dfrac { 2\cup }{ 5 }$
0%
$-\dfrac { 2\cup }{ 5 }$

Explanation

Mass of alpha particle is

$4$ times to neutron

according to the condition alpha paticle is at rest so its velocity is zero

now the final velocity of neutron is calculated from this formula

$U' = (m-1-m_2÷ m_1+m_2) U^1$

by putting

$,$

$U' = (1-4÷1+4)$

$= -3U/5$

Negative sign shows reverse direction of neutro

$.$

Hence,

option

$(A)$ is correct answer.

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 tiny ball of mass

$2\ kg$ is made to move along the positive

$x-$ axis under a force described by the potential energy

$U$ shown below. It is projected towards positive

$x-$ axis from the origin with a speed

$v_0$ . What is the minimum value of

$v_0$ for which the ball will escape infinity far away from the origin.

0%
$8\ m/sec$
0%
$4\ m/sec$
0%
$2\ m/sec$
0%
$It\ is\ not\ possible$

A uniform rod AB of length

$L$ and 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 the collision, is:

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

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

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

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Practice Class 11 Medical Physics Quiz Questions and Answers

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

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Practice Class 11 Medical Physics Quiz Questions and Answers

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