A block of mass m is connected to the lower end of a massless vertically suspended spring. The ball is displaced slightly downward and released; it oscillates up and down Match the column I and Column II A In the system of the block, the spring and the earth, what form(s) of energy will be there during the motion of the block? p Kinetic energy B In the system of the block and the spring, what form(s) of energy will be there during the motion? q Elastic potential energy C While oscillating, as the block passes its equilibrium position, what form(s) of energy will be there in the system of block, spring and earth? r Gravitational potential energy D The block is disconnected from the spring and released from some height above the earth, so that, it falls freely towards earth, what form(s) of energy will be there during its motion? s Electromagnetic energy

A - p,q,r B - p,q C - p,r D - p,r

A - p,q,r B - p,q C - p,r D - p

A - p,q,r B - p,q C - r D - p,r

A - p,q,r B - q C - p,r D - p,r

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

The work-energy theorem states that the change in:

0%
kinetic energy of a particle is equal to the work done on it by the net force
0%
kinetic energy of a particle is equal to the work done by one of the forces acting on it
0%
potential energy of a particle is equal to the work done on it by the net force
0%
potential energy of a particle is equal to the work done by one of the forces acting on it
0%
total energy if a particle is equal to the work done on it by the net force

Explanation

According to work-energy theorem, change in kinetic energy of a particle is equal to the work done on it by the net force., i.e.,

$W=F\cdot S=\displaystyle\frac{1}{2}mv^2$ .

A solid sphere is placed on a smooth horizontal plane. A horizontal impulse

$I$ is applied at a distance

$h$ above the central line as shown in the figure. Soon after giving an impulse, the sphere starts rolling. The ratio

$h/R$ would be:

0%
$\cfrac { 1 }{ 2 }$
0%
$\cfrac { 2 }{ 5 }$
0%
$\cfrac { 1 }{ 4 }$
0%
$\cfrac { 1 }{ 5 }$

Explanation

After applying horizontal impulse $J$ horizontal velocity of sphere becomes $v$ as initial linear momentum was $0$

$J=Mv$

Angular impulse about C.M

$=hMv=Iw$

$hMv=\cfrac{2}5M{R}^{2}w$

$w=\cfrac { 5vh }{ 2{ R }^{ 2 } }$

While rolling $w=\cfrac { 5vh }{ 2{ R }^{ 2 } }$

$\cfrac { h }{ R } =\cfrac { 2 }{ 5 }$

When a rubber-band is stretched by a distance

$x$ , it exerts a restoring force of magnitude

$F=ax+b{x}^{2}$ where

$a$ and

$b$ are constants.The work done in stretching the unstretched rubber band by

$L$ is:

0%
$\cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { { bL }^{ 3 } }{ 3}$
0%
$\cfrac { 1 }{ 2 } \left( \cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { { bL }^{ 3 } }{ 2 } \right)$
0%
$a{ L }^{ 2 }+{ bL }^{ 3 }$
0%
$\cfrac { 1 }{ 2 } \left( a{ L }^{ 2 }+{ bL }^{ 3 } \right)$

Explanation

Work done by streching dx length is,

$dw=Fdx$

$=(ax+bx^2)dx$

$w=\int _{ 0 }^{ L }{ { \left( ax+b{ x }^{ 2 } \right) }dx } \\ =\overset { L }{ \underset { 0 }{ \left[ \dfrac { { ax }^{ 2 } }{ 2 } +\dfrac { { bx }^{ 3 } }{ 3 } \right] } } \\ =\dfrac { { aL }^{ 2 } }{ 2 } +\dfrac { { bL }^{ 3 } }{ 3 }$

The diagram below shows the path taken by a ball when Sundram kicks it. The potential energy of the ball is highest at ______________

0%
P
0%
Q
0%
R
0%
S

When the speed of a body is doubled, its kinetic energy becomes

0%
Double
0%
Half
0%
Quadruple
0%
One-fourth

Explanation

The kinetic energy of a body is given by:

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

Hence, Kinetic energy depends on the velocity as:

$KE\propto v^2$

So, if we double the velocity, then

$KE$ becomes four times.

Hence, option C is correct.

A force acts on a

$3 g$ particle in such a way that position of the particle as a function of time is given by

$x=3t-4t^2+t^3$ , where x is in metre and t is in sec. The work done during the first

$4s$ is

0%
570 mJ
0%
450 mJ
0%
490 mJ
0%
528 mJ

Explanation

$\dfrac{dx}{dt}=3-8t+3t^2$

$dx=(3-8t+3t^2)dt$

$\dfrac{d^2}{dx^2}=a=-8+6t$

Force

$F=ma=m(-8+6t)$

Work done

$F.dx$

$F.(3-8t+3t^2)dt$

$\int _{ o }^{ w }{ dw\quad =\int _{ 0 }^{ 4 }{ m\left( -8+6t \right) \left( 3-8t+3{ t }^{ 2 } \right) } } dt\\ w=m\int _{ 0 }^{ 4 }{ \left( -24+82t-72{ t }^{ 2 }+18{ t }^{ 3 }\quad \right) dt } \\ =m\overset { 4 }{ \underset { 0 }{ \left[ -24t+41{ t }^{ 2 }-24{ t }^{ 3 }+\dfrac { 3 }{ 2 } { t }^{ 4 } \right] } } \\ =m(176)\\ =3\times { 10 }^{ -3 }\times 176\\ =528\quad mJ$

When a rubber -band is stretched by a distance x,it exerts a restoring force of magnitude

$F= ax + bx^2$ where a and b are constants. the work done is stretching the unstretched rubber band by L is :

0%
$\dfrac {aL^{2}}{2}+\dfrac{bL^{3}}{3}$
0%
$\dfrac{1}{2} \begin {pmatrix} \dfrac{aL^2}{2} + \dfrac{bL^3}{3}\end {pmatrix}$
0%
$aL^2 + bL^3$
0%
$\dfrac {1}{2} ( aL^2 + bL^3)$

A triangular block ABC of mass m and sides 2a lies on a smooth horizontal plane as shown in the figure. Three point masses of mass m each strike the block at A, B and C with speeds v as shown. After the collision, the particles come to rest. Then the angular velocity acquired by the triangular block is (I is the moment of inertia of the triangular block about G, perpendicular to the plane of the block)

0%
$\displaystyle 2mva \frac{(1 + \sqrt{3})}{\sqrt{3} l}$ clockwise
0%
$\displaystyle \frac{2mva}{ l}$ clockwise
0%
$\displaystyle \frac{2 \sqrt{3} mva}{ l}$ clockwise
0%
None of these.

Explanation

$h=\dfrac { 2 }{ 3 } \sqrt { 3 } a\dfrac { 2 }{ \sqrt { 3 } } a\\ I\omega =3mvh\\ =2\sqrt { 3 } mva\\ \omega =\dfrac { 2\sqrt { 3 } mva }{ l }$

A raindrop of mass

$1\ g$ falling from a height of

$1\ km$ hits the ground with a speed of

$50\ ms^{-1}$ . Which of the following statements is correct?
(Take

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

0%
The loss of potential energy of the drop in $10\ J$
0%
The gain in kinetic energy of the drop is $1.25\ J$
0%
The gain in kinetic energy of the drop is not equal to the loss of potential energy of the drop
0%
All of these

Explanation

$K.E$ at ground =loss of potential energy

$\dfrac{1}{2}mv^2$ =loss of potential energy

$\dfrac{1}{2}\times 1 \times 10^{-3} \times 50^2$ =loss of potential energy

$\dfrac{1}{2}\times 10^{-3} \times 2500$ =loss of potential energy

$1.25 j$ loss of potential energy=Gain of

$K.E$

Which of the following statements is incorrect?

0%
Kinetic energy may be zero, positive or negative
0%
Power, energy and work are all scalars
0%
Potential energy may be zero, positive or negative
0%
Ballistic pendulum is a device for measuring the speed of bullets

Explanation

The kinetic energy of a body of mass

mm which is moving with velocity

vv is

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

Mass is always positive and if the velocity is either positive or negative, the kinetic energy is always positive due to the square of velocity.

A force F is related to the position of a particle by the relation

$F =(10x^2)N$ . The work done by the force when the particle moves from x=2m to x=4 m is

0%
$\dfrac{56}{3}J$
0%
$560 J$
0%
$\dfrac {560}{3}J$
0%
$\dfrac{3}{560} J$

Explanation

$F=10 x^{2}N$

We know that work done

$=\displaystyle\int_{x_{1}}^{x_{2}}F.dx$

$\omega=\displaystyle\int_{2}^{4}10 x^{2}dx$

$\Rightarrow \omega=10\displaystyle\int_{2}^{4}x^{2}dx$

$=10\left[\dfrac{x^{3}}{3}\right]_{2}^{4}$

$=10\left[\dfrac{64}{3}-\dfrac{8}{3}\right]=\dfrac{560}{3}\ J$

Which of the following is not an example of potential energy?

0%
A vibrating pendulum at its maximum displacement from the mean position
0%
A body at rest at some height from the ground
0%
A wound clock-spring
0%
A vibrating pendulum when it is just passing through the mean position

If S denotes sound energy, E denotes electrical energy, M denotes magnetic energy, the correct representation of recording and reproduction in an audio tape recorder is?

0%
E $\rightarrow$ M $\rightarrow$ E $\rightarrow$ S
0%
S $\rightarrow$ E $\rightarrow$ M $\rightarrow$ E $\rightarrow$ S
0%
E $\rightarrow$ S $\rightarrow$ M $\rightarrow$ S
0%
S $\rightarrow$ M $\rightarrow$ E $\rightarrow$ M $\rightarrow$ S

Explanation

Correct representation of recording and reproduction in an audio tape recorder is :-

$\underset { Energy }{ Sound } \rightarrow \underset { Energy }{ Electrical } \rightarrow \underset { Energy }{ Magnetic } \rightarrow \underset { Energy }{ Electrical } \rightarrow \underset { Energy }{ Sound }$

A body of mass

$0.5\ kg$ travels in a straight line with velocity

$v = ax^{1/2}$ where

$a = 4 m^{}s^{-2}$ . The work done by the net force during its displacement from

$x = 0$ to

$x = 2\ m$ is

0%
$1.5\ J$
0%
$50\ J$
0%
$10\ J$
0%
None of these

Which of the diagram shown in figure represents variation of total mechanical energy of a pendulum oscillating in air as function of time?

A force

$F$ acting on an object varies with distance

$x$ as shown in the figure. The work done by the force in moving the object from

$x = 0$ and

$x = 20\ m$ is

0%
$500\ J$
0%
$1000\ J$
0%
$1500\ J$
0%
$2000\ J$

Explanation

Area under force displacement curve is the work done in that interval

Area under the given figure

$=$ Area of surface

$+$ Area of triangle.

Work done

$=10\times 100+\cfrac { 1 }{ 2 } \times 10\times 100$

$=1000+500\\ =1500J$

A body of mass starts moving from rest along x-axis so that its velocity varies as

$v = a\sqrt {s}$ 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 seconds after the start of the motion is:

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

Explanation

Velocity of body

$v = a \sqrt s$

acceleration

$a' = \dfrac{dv}{dt} = \dfrac{a}{2 \sqrt s} \dfrac{ds}{dt} = \dfrac{a^2}{2}$

displacement

$s' = \dfrac{1}{2} a' t^2 = \dfrac{1}{2} \dfrac{a^2}{2} t^2$

Work done

$W = force \times displacement$

$W = ma' \times s' = \dfrac{1}{8} ma^4 t^2$

A particle is dropped from a height of

$20\ m$ onto a fixed triangular wedge of inclination

$30^{\circ}$ . The time gap between first two successive collisions is (Assume that collision is perfectly elastic).

0%
$2\ \sec$
0%
$2\sqrt {3}\ \sec$
0%
$4\ \sec$
0%
$4\sqrt {3}\ \sec$

A body of mass

$0.5\ kg$ travels in a straight line with velocity

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

$a = 5\ m^{-1/2}s^{-1}$ . The work done by the net force during its displacement from

$x = 0$ to

$x = 2m$ is

0%
$1.5\ J$
0%
$50\ J$
0%
$10\ J$
0%
$100\ J$

Explanation

Given:

$m = 0.5\ kg, v = kx^{3/2}$ where,

$k = 5\ m^{-1/2}s^{-1}$
Acceleration,

$a = \dfrac {dv}{dt} = \dfrac {dv}{dx}\dfrac {dx}{dt} = v\dfrac {dv}{dx} \left (\because v = \dfrac {dx}{dt}\right )$
As

$v^{2} = k^{2}x^{3}$
Diferentiating both sides with respect to

$x$ , we get

$2v \dfrac {dv}{dx} = 3k^{2}x^{2}$

$\therefore Acceleration, a = \dfrac {3}{2} k^{2}x^{2}$
Force,

$F = Mass\times Acceleration = \dfrac {3}{2} mk^{2}x^{2}$
Work done,

$W = \int Fdx = \int_{0}^{2}\dfrac {3}{2} mk^{2}x^{2}dx$

$W = \dfrac {3}{2} mk^{2}\left [\dfrac {x^{3}}{3}\right ]_{0}^{2} = \dfrac {3}{6} \times 0.5\times 5^{2} \times [2^{3} - 0] = 50\ J$ .

A sphere

$P$ of mass

$m$ and velocity

$v_{i}$ undergoes an oblique and perfectly elastic collision with an identical sphere

$Q$ initially at rest. The angle

$\theta$ between the velocities of the spheres after the collision shall be

0%
$0$
0%
$45^{\circ}$
0%
$90^{\circ}$
0%
$180^{\circ}$

Explanation

According to the law of conservation of linear momentum, we get

$m\vec {v_{i}} + m\times 0 = m\vec {v_{Pf}} + m\vec {v_{Qf}}$

where

$\vec {v_{Pf}}$ and

$\vec {v_{Qf}}$ are the final velocities of spheres

$P$ and

$Q$ after collision respectively.

$\vec {v_{i}} = \vec {v_{Pf}} + \vec {v_{Qf}}$

$(\vec {v_{i}}\cdot \vec {v_{i}}) = (\vec {v_{Pf}} + \vec {v_{Qf}})\cdot (\vec {v_{Pf}} + \vec {v_{Qf}})$

$= \vec {v_{Pf}} \cdot \vec {v_{Qf}} \cdot \vec {v_{Qf}} + 2\vec {v_{Pf}} \cdot \vec {v_{Qf}}$

$v_{i}^{2} = v_{Pf}^{2} + v_{Qf}^{2} + 2v_{Pf} v_{Qf} \cos \theta .... (i)$

According to the conservation of kinetic energy, we get

$\dfrac {1}{2}mv_{i}{2} = \dfrac {1}{2} mv_{Pf}^{2} + \dfrac {1}{2}mv_{Qf}^{2} \Rightarrow v_{i}^{2} = v_{Pf}^{2} + v_{Qf}^{2} ....(ii)$

Comparing (i) and (ii), we get

$2v_{Pf} v_{qf}\cos \theta = 0$

$\Rightarrow \cos \theta = 0$ or

$\theta = 90^{\circ}$ .

A force

$F = -K(x\hat {i} + y\hat {j})$ (where

$K$ is a 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 to the point

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

$\vec {F}$ on the particle is

0%
$-2Ka^{2}$
0%
$2Ka^{2}$
0%
$-Ka^{2}$
0%
$Ka^{2}$

Explanation

For motion of the particle from

$(0,0)$ to

$(a,0)$

$\vec{F}=-K(0\hat{i}+a\hat{j})\Rightarrow \vec{F}=-Ka\hat{j}$
Displacement

$\vec{r}=(a\hat{i}+0\hat{j})-(0\hat{i}+0\hat{j})=a\hat{i}$
So work done from

$(0,0)$ to

$(a,0)$ is given by

$W=\vec{f}.\vec{r}=-Ka\hat{j}.a\hat{i}=0$
For motion

$(a,0)$ to

$(a,a)$

$\vec{F}=-K(a\hat{i}+a\hat{j})$ and displacement

$\vec{r}=(a\hat{i}+a\hat{j})-(a\hat{i}+0\hat{j})=a\hat{j}$
So work done from

$(a,0)$ to

$(a,a)$

$W=\vec{F}.\vec{r}$

$=-K(a\hat{i}+a\hat{j}).a\hat{j}=-Ka^{2}$
So total work done

$=-Ka^{2}$

Mark the correct statement(s).

0%
The work-energy theorem is valid only for particles
0%
The work-energy theorem is an invariant law of physics.
0%
The work-energy theorem is valid only in inertial frames of reference.
0%
The work-energy theorem can be applied in non-inertial frames of reference too.

A particle is projected vertically upwards with a speed of 16

$ms^{-1}$ . After some time, when it again passes through the point of projection, its speed is found to be 8

$ms^{-1}$ . It is known that the work done by air resistance is same during upward and down ward motion. Then the maximum height attained by the particle is (take g = 10

$ms^{-2}$ ).

0%
$8 m$
0%
$4.8 m$
0%
$17.6 m$
0%
$12.8 m$

Explanation

From work- energy theorem, for upward motion against resistance

$\frac{1}{2}m{{v}^{2}}\,=mgh\,+\,W.................(1)$

For downward motion,

$\frac{1}{2}m{{v}^{2}}\,=\,mgh-W............(2)$

From equation (1) and (2)

$\frac{1}{2}m({{16}^{2}}-{{8}^{2}})\,=2mgh$

$\frac{1}{2}(256-64)\,=\,2\times 10\times h$

$h=\,4.8m$

two blocks of masses

$m_1 = 2 kg$ and

$m_2 = 4 kg$ are moving in the same direction with speeds

$\nu_1 = 6 m/s$ and

$\nu_2 = 3 m/s$ , respectively on a frictioneless surface as shown in the figure. An ideal spring with spring constant

$k = 30000 N/m$ is attached to the back side of

$m_2$ . Then the maximum compression of the spring after collision will be:

0%
$0.06 m$
0%
$0.04 m$
0%
$0.02 m$
0%
none of these

Explanation

$m_{1}=2\ kg ,m_{2}=4\ kg$

$v_{1}=6\ m/s, v_{2}=3\ m/s$

$k=30000\ N/m$

From momentum conesrvation

$2\times 6+5\times 3=2v_{1}+5v_{2}$

$2v_{1}+5v_{2}=27$

$v_{1}=v_{2}=v=\dfrac{27}{7}m/s$ for max compression

From Energy conservation

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

$\dfrac{1}{2}m_{1}v_{1}^{2}+\dfrac{1}{2}m_{2}v_{2}^{2}=\dfrac{1}{2}m_{2}v^{2}+\dfrac{1}{2}m_{2}v^{2}+\dfrac{1}{2}kx^{2}$

$\dfrac{1}{2}\times 2\times 6\times 6+\dfrac{1}{2}\times 4\times 3\times 3=\dfrac{1}{2}\times 2\times \dfrac{27\times 27}{7\times 7}+\dfrac{1}{2}\times \dfrac{4\times 27\times 27}{7\times 7}+\dfrac{1}{2}\times 30,000\times x^{2}$

$36+18=14.88+29.75+15000 x^{2}$

$54-44.63=15000 x^{2}\Rightarrow 937=15000 x^{2}$

$x=0.02\ m$

Water is drawn from a well in a 5 kg drum of capacity 55 L by two ropes connected to the top of the drum. The linear mass density of each rope is 0.5

$kgm^{-1}$ . The work done in lifting water to the ground from the surface of water in the well 20 m below is [g = 10

$ms^{-2}$ ].

0%
1.3 X $10^4 J$
0%
1.5 X $10^5 J$
0%
1.4 X 10 X 6J
0%
18J

Explanation

Total mass of bucket with water

$=60$ kg

mass of rope

$=20\times0.5=10$ kg

centre of mass of the rope will be

$10$ m below. So, the next work.

$W=60\times 10\times 20+10\times 10\times 10$

$=1.3\times { 10 }^{ 4 }$ joule.

A ladder of length l carrying a man of mass m at its end is attached to the basket of a balloon of mass M. The entire system is in equilibrium in the air. As the man climbs up the ladder into the balloon, the balloon descends by a height h.
The work done by the man is

0%
mgl
0%
mgh
0%
mg
0%
mg(l-h)

Explanation

The work done by man

$=mgl.$

Hence, the answer is

$mgl.$

A time-varying force

$F=6t-2{ t }^{ 2 } N$ , at

$t=0$ starts acting on a body of mass

$2 kg$ initially at rest, where t is in second. The force is withdrawn just at the instant when the body comes to rest again. We can see that at

$t=0$ ,the force

$F=0$ . Now answer the following:
Mark the correct statement:

0%
Velocity of the body is maximum when force acting on the body is maximum for the first time.
0%
The velocity of the body becomes maximum when force acting on the body becomes zero again
0%
When force becomes zero again, velocity of the body also becomes zero at that instant.
0%
All of the avove

If the kinetic Energy possessed by a man of

$50$ kg is

$625J$ , the speed of man is

0%
$5m{s^{ - 1}}$
0%
$50m{s^{ - 1}}$
0%
$0.5m{s^{ - 1}}$
0%
$500m{s^{ - 1}}$

The loss of mechanical energy up to that instant is

0%
$32.4 J$
0%
$40 J$
0%
$16.5 J$
0%
$12.5 J$

Explanation

Loss of energy during collision of B and C is

$E_1= \cfrac {1}{2}\times 1 \times a^2-\cfrac {1}{2} \times (1+2)Vo^2=27$ Joule

Loss of energy when impulse is developed in string is

$E_2= \cfrac {1}{2} \times 2 \times 3^2-\cfrac {1}{2}(2+3)(V^1)^2=5.4$ Joule

$\therefore$ Loss of mechanical energy till B reaches highest point is

$\longrightarrow$

$E= E_1+E_2=27+5.4=32.4$ Joule

A plastic ball falls from a height of

$4.9$ metre and rebounds several times from the floor. What is the coefficient of restitution during the impact with the floor if

$1.3$ seconds pass from the first impact to the second one?

0%
$0.9$
0%
$0.1$
0%
$0.7$
0%
$0.8$

Explanation

As,

$t = \sqrt{\dfrac{2h}{g}}$

Velocity of ball just before collision

$= v = \sqrt{2gh}$

After first collision,

Velocity

$= v_{1} = ev = e\sqrt{2gh}$

So,

$t_{1} = \dfrac{v_{1}}{g} = e \sqrt{\dfrac{2h}{g}}$

So, time for second collision will be

$T = t+2t_{1}$

$= (1+2e)\sqrt{\dfrac{2h}{g}}$

Putting,

$T = 1.3\,sec.,h = 4.9\,m$

$1.3 = (1+2e)\sqrt{\dfrac{2\times 4.9}{9.8}}$

$1.3 = 1+2e$

$0.3 = 2e$

$\boxed{e = 0.1}$

A mass of

$M\ kg$ is suspended by a weightless string. The horizontal force that is required to displace it unitl the string makes an angle of

${45}^{o}$ with the initial vertical direction is

0%
$Mg(\sqrt{2}-1)$
0%
$Mg(\sqrt{2}+1)$
0%
$Mg\sqrt {2}$
0%
$\cfrac { Mg }{ \sqrt { 2 } }$

Explanation

Work done by

$F$

$F=\cfrac { l }{ \sqrt { 2 } } =mg\left( l-\cfrac { l }{ \sqrt { 2 } } \right)$

$F=Mg\left( \sqrt { 2 } -1 \right)$

A thin uniform rod of mass

$m$ and length

$l$ is hinged at the lower end of a level floor and stands vertically. It is now allowed to fall, then its upper and will strike the floor with a velocity given by(A)

$\sqrt { mgl }$ (B)

$\sqrt { 3gl }$ (c)

$\sqrt { 5gl }$ (D)

$\sqrt { 2gl }$ Sol.

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

Explanation

Initially rod stand vertically

Its potential energy

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

When it striker the floor, its potential energy will convert into rotational kinetic energy,

$mg\left(\dfrac{1}{2}\right)=\dfrac{1}{2}I\omega^2$

Where,

$I=\dfrac{ml^2}{3}=M.I$ of rod about point A

$\therefore mg\left(\dfrac{l}{2}\right)=\dfrac{l}{2}\left(\dfrac{ml^2}{3}\right)\left(\dfrac{V_B}{l}\right)^2$

$V_V=\sqrt{3gl}$

2077414_1015387_ans_dee8f33bdc1c41f7ace52001f0b31380.png

In an elastic collision between two particles

0%
net kinetic force is zero
0%
the kinetic energy of the system before collision is equal to the kinetic of the system after collision
0%
linear momentum of system before collision = linear momentum after collision
0%
the total energy of the system is never conserved

Explanation

We know,

Net Kinetic force involved in the collision is zero in case of elastic collision.

And as there is no net Kinetic force, net work done during collision=0,

And hence

$\Delta KE=0$ during collision.

Hence there is no change in KE before and after collision.

And as there is no external force involved in collision,

Momentum of the colliding system is conserved.

And hence total momentum before collision=total momentum after collision.

Note; momentum is conserved irrespective of elastic/inelastic collision.

Option

$\textbf{A,B,C}$ is correct answer.

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

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

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

When a rubber-bank is stretched by a distance

$x$ , it exerts a restoring force of magnitude

$F=ax+b{x}^{2}$ where

$a$ and

$b$ are constants. The work done in stretching the unstretched rubber band by

$L$ is:

0%
$\cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { b{ L }^{ 3 } }{ 3 }$
0%
$\cfrac { 1 }{ 2 } \left( \cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { b{ L }^{ 3 } }{ 3 } \right)$
0%
$a{ L }^{ 2 }+b{ L }^{ 3 }$
0%
$\cfrac { 1 }{ 2 } \left( a{ L }^{ 2 }+b{ L }^{ 3 } \right)$

Explanation

$F=ax+b{ x }^{ 2 }$

$dw=Fdx$

$W=\int _{ 0 }^{ L }{ \left( ax+b{ x }^{ 2 } \right) } dx\quad$

$W=\cfrac { a{ L }^{ 2 } }{ 2 } +\cfrac { b{ L }^{ 3 } }{ 3 }$

The work done by a force is equal to :

0%
the area under $f\left( x \right)$ vs $x$ curve and $x$ axis
0%
half the area under $f\left( x \right)$ vs $x$ curve and $x$ axis
0%
the area under $f\left( x \right)$ vs $x$ curve and $F$ axis
0%
half the area under $f\left( x \right)$ vs $x$ curve and $F$ axis

Explanation

he correct option is A.

Therefore,

$W=\int Fdx$

1171484_1017169_ans_5ed1a947868c47ea9c2119a6bf5bcb37.PNG

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 particle moves along

$X-$ axis from

$x=0$ to

$x=1\ m$ under the influence of a force given by

$F=3x^{2}+2x-10$ . Work done in the process is:

0%
$+4\ J$
0%
$-4\ J$
0%
$+8\ J$
0%
$-8\ J$

Explanation

Given,

$F=3x^2+2x-10$

Work done

$=\int { F\cdot ds }$

$=\int _{ 0 }^{ 1 }{ \left( { 3x }^{ 2 }+2x-10 \right) dx }$

$=\left[{ \frac { { 3x }^{ 3 } }{ 3 } +\frac { { 2x }^{ 2 } }{ 2 } -10x }\right]_{ 0 }^{ 1 }$

$=(1)^3+(1)^2-10-0$

$=2-10$

$=-8J$

A body of mass

$1kg$ thrown upwards with a velocity of

$10m/s$ comes to rest (momentarily) after moving up by

$4m$ . The work done by air drag in this process is (Take

$g=10m/{ s }^{ 2 }$ )

0%
$-20J$
0%
$-10J$
0%
$-30J$
0%
$0J$

A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly- elastic collision with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value of m is/are :

0%
1 kg only
0%
1 kg , 9kg
0%
1 kg, 6kg
0%
6kg only

Water is dragged from a well of hepth 30 m using bucket of weight 50kg with water If the weight of the rope is 0.4 kg per metre, the amount of work done is

0%
18228J
0%
16464J
0%
14700J
0%
11760J

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 particle moves along

$y=\sqrt { 1-{ x }^{ 2 } }$ betweem the points

$(0,-1)m$ and

$(0,1)m$ under the influence of a force

$\overrightarrow { F } =\left( { y }^{ 2 }\hat { i } +{ x }^{ 2 }\hat { j } \right) N$ . Then

0%
the particle is moving along a semi-ellipse
0%
the particle is moving along a semicircle
0%
work done on the particle by $\overrightarrow { F }$ is $(3/4)J$
0%
work done on the particle by $\overrightarrow { F }$ is $(4/3)J$

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 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

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

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 small particle slides along a track with elevated ends and a flat central part, as shown in figure. The flat part has a length

$3m$ , the curved portions of the track are frictionless, but for the flat part, the coefficient of kinetic friction is

$\mu =0.2$ . The particle is released at a point

$A$ , which is at a height

$h=1.5m$ above the flat part of the track. The position where the particle finally come to rest is:

0%
left to mid-point of the flat part
0%
right to the mid point of the flat part
0%
mid point of the flat part
0%
none of these

Explanation

Initial mechanical energy

$E_{mech}=U_{grav}=mgh=\frac{1}{2}ngL$

Energy loss due to friction in every time going through the flat part

$W_{fric}=-\mu _x mgL$

Consider that the particle moves

$n$ tyrips

$\Delta E=W=n. W_{fric}$

$-\frac{1}{2}mgL=m_x.\mu _xmgL$

$n=\frac{1}{2\mu k}=2.5$

So the particle will stop at the middle part of the flat surface.

1106694_1027393_ans_73d53dc949a44f7a918ccd40c0551c1c.png

A block of mass m is connected to the lower end of a massless vertically suspended spring. The ball is displaced slightly downward and released; it oscillates up and down

Match the column I and Column II

A	In the system of the block, the spring and the earth, what form(s) of energy will be there during the motion of the block?	p	Kinetic energy
B	In the system of the block and the spring, what form(s) of energy will be there during the motion?	q	Elastic potential energy
C	While oscillating, as the block passes its equilibrium position, what form(s) of energy will be there in the system of block, spring and earth?	r	Gravitational potential energy
D	The block is disconnected from the spring and released from some height above the earth, so that, it falls freely towards earth, what form(s) of energy will be there during its motion?	s	Electromagnetic energy

0%
A - p,q,r
B - p,q
C - p,r
D - p
0%
A - p,q,r
B - p,q
C - p,r
D - p,r
0%
A - p,q,r
B - p,q
C - r
D - p,r
0%
A - p,q,r
B - q
C - p,r
D - p,r

A block of mass 5.0 kg slides down an incline of inclination

$30^0$ and length 10 m. Find the work done by the force of gravity in joules?

0%
245
0%
300
0%
350
0%
400

Explanation

We have, the mass of the block,

$M=5kg$

Angle of inclination is

$\theta= 30^0$

The gravitational force acting the block,

$F=mg$

Work done by the force of gravity depends on the height of the object, not on the path length covered by the object.

Height of the object,

$H=10\times sin 30^0$

$=10\times \dfrac{1}{2}=5m$

Since,

Work was done by the force of gravity,

$W=mgh= 5\times 9.8\times5=245J$

956540_1027548_ans_e6a809247ce841b18b3f0f8939881a48.PNG

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

0:0:2

Practice Class 11 Medical Physics Quiz Questions and Answers

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