MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 2

A wagon of mass 10 tons moving at a speed of 12 kmph collides with another wagon of mass 8 tons moving on the same track in the same direction at a speed of 10 kmph. If the speed of the first wagon decreases to 8 kmph. Find the speed of the other after collision

0%
18 kmph
0%
25 kmph
0%
5 kmph
0%
15 kmph

Explanation

Conserving momemtum,

$10 \times 12 + 8 \times10 = 10 \times 8 + 8 \times v$

$200 - 80 = 8 \times v$

$\Rightarrow v = 15 \ kmph$

A particle of mass m has a velocity

$-v_{0}\hat{i}$ . while a second particle of same mass has a velocity

$v_{0}j$ . After the particles collide, first particle is found to have a velocity

$\dfrac{-1}{2}v_{0}\hat{i}$ then the velocity of other particle is

0%
$\dfrac{-1}{2}v_{0}\hat{i}+v_{0}\hat{j}$
0%
$\dfrac{1}{2}v_{0}\hat{i}+v_{0}\hat{j}$
0%
$v_{0}\hat{i}+v_{0}\hat{j}$
0%
$-v_{0}\hat{i}+v_{0}\hat{j}$

Explanation

Using momentum conservation
Final momentum

$=$ Initial momentum

$-\dfrac{m}{2}v_{0}\hat{i} + m(v_{x}\hat{i} + v_{y}\hat{j}) = -mv_{0}\hat{i} + mv_{0}\hat{j}$

Comparing

$v_{x} = -\dfrac{1}{2}v_{0}\hat{i}$

$v_{y} = v_{0}\hat{j}$

A 6 kg mass travelling at

$2.5$ ms

$^{-1}$ collides head on with a stationary 4 kg mass. After the collision the 6 kg mass travels in its original direction with a speed of

$1$ ms

$^{-1}$ . The final velocity of 4 kg mass

0%
$1$ ms $^{-1}$
0%
$2.25$ ms $^{-1}$
0%
$2$ ms $^{-1}$
0%
$0$ ms $^{-1}$

Explanation

Use momentum conservation,

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$
But

$m_1=6$ kg

$, m_2=4$ kg,

$u_1=2.5$ ms

$^{-1}, u_2=0$ ms

$^{-1}, v_1=1$ ms

$^{-1}, v_2=?$

$\Rightarrow 6 \times 2.5 +4\times (0) = 6 \times 1 + 4 \times v_2$

$\Rightarrow 9 = 4v \Rightarrow v = 2.25$ ms

$^{-1}$ .

A pilot of mass

$m$ can bear a maximum apparent weight

$7$ times of

$mg$ . The aeroplane is moving in a vertical circle. If the velocity of aeroplane is

$210\ m/s$ while driving up from the lowest point of vertical circle, the minimum radius of vertical circle should be:

0%
$375\ m$
0%
$420\ m$
0%
$750\ m$
0%
$840\ m$

Explanation

Given maximum apparent weight is

$7mg$
Apparent weight at the lowest point is (

$mg + m \dfrac{v^2}{r}$ )
So,

$7mg = mg + m \dfrac{v^2}{r}$
or,

$6mg = m \dfrac{v^2}{r}$
or,

$r = \dfrac{v^2}{6g}$
or,

$r = \dfrac{210^2}{6g}$
or,

$r = 750\ m$
So , minimum radius of circle is

$750\ m$

A body of mass

$2\ kg$ attached at one end of light string is rotated along a vertical circle of radius

$2\ m$ . If the string can withstand a maximum tension of

$140.6\ N$ , the maximum speed with which the stone can be rotated is:

0%
$22\ m/s$
0%
$44\ m/s$
0%
$33\ m/s$
0%
$11\ m/s$

Explanation

$m= 2 Kg \quad R = 2m$

$T_{max} = 1406\ N$
The tension is max at bottom point.
At bottom:

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

$\dfrac{2(140.6- 2X9.81)}{2} = V^{2}$

${V = 10.99\ m/s}$

A body is revolving in a vertical circle of radius r such that the sum of its

$K.E.$ and

$P.E.$ is constant. If the speed of the body at the highest point is

$\sqrt{2gr}$ then the speed of the body at the lowest point will be:

0%
$\sqrt{7gr}$
0%
$\sqrt{6gr}$
0%
$\sqrt{8gr}$
0%
$\sqrt{9gr}$

Explanation

Applying WET between A and B.

$\dfrac{m(\sqrt{2gR})^{2}}{2} + mg(2R) = \dfrac{m}{2}v_{0}^{2}$

$mgr + 2 mgR = \dfrac{m}{2}v_{0}^{2}$

$v_{0} = \sqrt{6gr}$

A simple pendulum is oscillating with an angular amplitude

$60^o$ . If

$m$ is mass of bob and

$T_1$ ,

$T_2$ are tensions in the string, when the bob is at extreme position, mean position respectively then is:

$T_1 = \dfrac{mg}{2}$
B)

$T_2 = 2\ mg$
C)

$T_1 = 0$
D)

$T_2 = 3\ mg$

0%
A and B are true.
0%
A and D are true.
0%
B and C are true.
0%
C and D are true.

Explanation

Angular Magnitude is given

$\theta=60^o$

As we see in figure, At extreme position:

$T_A=mg\cos\theta$

$T_A=mg\cos 60^o$

$T_A=T_1=\dfrac{mg}{2}$

As we see in figure, At middle position:

$T_B=mg+F_c$ where

$F_c$ is centrifugal force

$T_B=mg+\dfrac{mv^2}{R}$ where

$v$ is velocity of pendulum.

Using energy conservation for extreme position and middle position

$v=\sqrt{gR}$

putting values :

$T_B=T_2=mg+\dfrac{mgR}{R}=2mg$

Hence option A is correct.

A body of mass m is rotated at uniform speed along a vertical circle with the help of light string. If

$T_{1} and \ T_{2}$ are tensions in the string when the body is crossing highest and lowest point of the vertical circle respectively, then which of the following expressions is correct?

0%
$T_{2}-T_{1}=6mg$
0%
$T_{2}-T_{1}=4mg$
0%
$T_{2}-T_{1}=2mg$
0%
$T_{2}-T_{1}=mg$

Explanation

Centripetal Force is:

$\dfrac { m{ v }^{ 2 } }{ r }$
At the highest point:

${ T }_{ 1 }=\dfrac { m{ v }^{ 2 } }{ r } -mg$
At the lowest point:

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

$\therefore \quad { T }_{ 2 }-{ T }_{ 1 }=\dfrac { m{ v }^{ 2 } }{ r } +mg-(\dfrac { m{ v }^{ 2 } }{ r } -mg)=2mg$

The bob of a simple pendulum at rest position is given a velocity

$V$ in horizontal direction so that the bob describes vertical circle of radius equal to length of pendulum

$l$ . If the tension in string is

$4$ times weight of bob when the string is horizontal, the velocity of bob when it is crossing highest point of vertical circle is:

0%
$\sqrt{\dfrac{gl}{2}}$
0%
$\sqrt{gl}$
0%
$\sqrt{\dfrac{3gl}{2}}$
0%
$\sqrt{2gl}$

Explanation

At pt B.

$T = \dfrac{mv^{2}} {L} and T = 4mg.$

$4\dfrac{mg}{m}L = v^{2}$

$V= \sqrt{4gL}$

Now WET between B and C.

$mgL = \dfrac{1}{2}mv^{2} - \dfrac{m}{2}v_{0}^{2}$

$mgL = 2mgL - \dfrac{m}{2}v_{0}^{2}$

$v_{0}= \sqrt{2gL}$

A ball of mass 0.6kg attached to a light inextensible string rotates in a vertical circle of radius 0.75m such that it has speed of 5 m/s when the string is horizontal. Tension in string when it is horizontal on other side is:

$(g=10ms^{-2})$

0%
$30N$
0%
$26N$
0%
$20N$
0%
$6N$

Explanation

Given:

$m=0.6 kg; R=0.75 m$

Force equilibrium

$T=\dfrac{mv^{2}}{R}$

$=\dfrac{(0.6)(5)^{2}}{(0.75)}$

$=20 N$

A simple pendulum is oscillating with an angular amplitude

$60^{0}$ . If mass of bob is

$50\ gram$ , the tension in the string at mean position is:

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

0%
$0.5\ N$
0%
$1\ N$
0%
$1.5\ N$
0%
$2\ N$

Explanation

$H = R(1 - cos\ 60^0)$

$H = \dfrac {R}{2}$

Apply Work Energy Theorem,

$mg\dfrac {R}{2} = \dfrac {1}{2}mV^2$

$V = \sqrt {Rg}$

At mean position,

$T - mg = \dfrac {mV^2}{R}$

$T = 2\ mg$

$= 2 \times 0.05 \times 10$

$= 1\ N$

A water bucket of mass '

$m$ ' is revolved in a vertical circle with the help of a rope of length '

$r$ '. If the velocity of the bucket at the lowest point is

$\sqrt{7gr}$ . Then the velocity and tension in the rope at the highest point are:

0%
$\sqrt{3gr},2mg$
0%
$\sqrt{2gr},mg$
0%
$\sqrt{gr},mg$
0%
$Zero$ , $Zero$

Explanation

$V_{b} = \sqrt{7gr}$

Work Energy Theorem applied between top and bottom.

$mg(2R) = \dfrac{1}{2}m (V_{b} )^{2} - \dfrac{1}{2}m (V_{T} )^{2}$

$2mgr = \dfrac{7}{2}mgr - \dfrac{1}{2}m (V_{T} )^{2}$

$V_{T} = \sqrt{3gr}$

At top point:

$T = \dfrac{mV_{T}^{2}}{r} - mg$

$=m \dfrac{3gr}{r} -mg$

$= 2mg$

The bob of a simple pendulum of mass

$'m'$ is performing oscillations such that the tension in the string is equal to twice the weight of the bob while it is crossing the mean position. The tension in the string when the bob reaches extreme position is :

0%
$\dfrac{mg}{2}$
0%
$mg$
0%
$\dfrac{3mg}{2}$
0%
zero

Explanation

At bottom

$T = 2mg$

$T - mg = \dfrac{mV^2}{R}$

$2mg - mg = \dfrac{mV^2}{R}$

$\sqrt {Rg} = V$
Applying Work Energy Theorem between pt. (1) and (2)

$mg R(1-cos\theta) = \dfrac{1}{2}mV^2$

$mg R(1-cos\theta) = \dfrac{mRg}{2}$

here,

$\theta$ is the angle made by the particle with the axis at its extreme point,

$1-cos\theta = \dfrac{1}{2}$

$\theta = 60^0$
at,

$\theta = 60^0$

$T = mgcos\theta$

$\therefore T = mg\dfrac{1}{2}$

A simple pendulum consists of a light string from which a spherical bob of mass,

$M$ , is suspended. The distance between the point of suspension and the center of bob is

$L$ . At the lowest position, the bob is given tangential velocity of

$\sqrt{5gL}$ . The K.E of the bob when the string becomes horizontal is:

0%
$0$
0%
$\dfrac{MgL}{2}$
0%
$\dfrac{3MgL}{2}$
0%
$\dfrac{5MgL}{2}$

Explanation

From Work Energy Theorem,

$(WD)_G = K_f - K_i$ , where

$(WD)_G$ is work done by gravity

$-MgL = \dfrac{-1}{2}M\times 5gL + K_i$

$K_i = \dfrac {3}{2}MgL$

The bob of a simple pendulum is given a velocity in horizontal direction when the bob is at lowest position ,so that the bob describes vertical circle of radius equal to length of pendulum and tension in the string is

$10 \ N$ when the bob is at an angle

$60^0$ from lowest position of vertical circle. The tension in the string when the bob reaches highest position is (The mass of bob is

$100$ gram.

$g = 10\ ms^{–2}$ ):

0%
$9\ N$
0%
$7 \ N$
0%
$5.5\ N$
0%
$3.5 \ N$

Explanation

For a body of mass

$'m'$ moving in a vertical circle, Tension in the string at highest point and Tension at any angle

$\theta$ with vertical is related as:-

$T-{ T }_{ H }=3mg(1+cos\theta )$

Substituting the values

$10-{ T }_{ H }=3(0.1)(10)(1+cos60^0)=4.5\\ \Rightarrow { T }_{ H }=5.5\ N$

lf the two bodies moving at right angles collide and their initial momenta are

$\vec{\mathrm{P}}_{1}$ and

$\vec{\mathrm{P}}_{2}$ , their resultant momentum after collison is :

0%
$\vec{\mathrm{P}_{1}}-\vec{\mathrm{P}}_{2}$
0%
$\vec{\mathrm{P}}_{1}\sim\vec{\mathrm{P}}_{2}$
0%
$\sqrt{\vec{\mathrm{P}_{1}}^{2}+\vec{\mathrm{P}}_{2}^{2}}$
0%
$\sqrt{\vec{\mathrm{P}_{1}}^{2}-\vec{\mathrm{P}}_{2}^{2}}$

Explanation

By momentum conservation,
Resultant momentum after collision = Resultant momentum before collision =

$\sqrt{\vec {P_1}^2 + \vec {P_2}^2}$

The length of a simple pendulum is '

$L$ '. Its bob from rest position is projected horizontally with a velocity

$\sqrt{\dfrac{7gL}{2}}$ . The maximum angular displacement of bob, such that the string does not slack, is:

0%
$30^o$
0%
$60^o$
0%
$120^o$
0%
$150^o$

Explanation

At point where tension become just zero, the centripetal force must be balanced by gravitational force

$mg cos\theta = \dfrac{mv_{1}^{2}}{L}$ ------ (1)
BY conservation of energy

$\dfrac{1}{2}m \dfrac{7gL}{2} = \dfrac{mv_{1}^{2}}{2}+mgL(1+cos\theta)$

$\sqrt{2gL} \sqrt{\dfrac{3}{4}-cos\theta} = v_{1}$
Put

$v_{1}$ in (1)

$mg$

$cos\theta =$

$\dfrac{m\times2gL}{L}\left ( \dfrac{3}{4}-cos\theta \right ).$

$cos\theta =\dfrac{3}{2}- 2cos\theta$

$cos\theta=\dfrac{1}{2}$

$\theta=60^{0}$

$\therefore$ Total maximum angular displacement

$= 180^0-60^0$

$=120^{0}$

The length of a ballistic pendulum is

$1 m$ and mass of its block is

$1.9 kg$ . A bullet of mass

$0.1 kg$ strikes the block of ballistic pendulum in horizontal direction with a velocity

$100ms^{–1}$ and got embedded in the block. After collision the combined mass (block & bullet) swings away from lowest point. The tension in the string when it makes an angle

$60°$ with vertical is

$(g=10ms^{-2})$ :

0%
$20\ N$
0%
$30\ N$
0%
$40\ N$
0%
$50\ N$

Explanation

Momentum Conservation

$0.1\times 100= \left ( 1.9+0.1 \right )V_{o}$

$V_o = 5m s^{-1}.$

$H= L(1-cos60^{\circ}). =\dfrac{1}{2}= \dfrac{1}{2}m.$

Energy Conservation

$\dfrac{1}{2}(2)(5)^{2} = \dfrac{1}{2} \times 2 \times V^{2}+2\times10 \times \dfrac{1}{2}$

$25-10= V^{2}.$

$V=\sqrt{15}mvs$

By Force Balance

$T-mg\ cos\ 60^{\circ}= m\dfrac{V^{2}}{L}.$

$T= 2\times10\times\dfrac{1}{2}+2\times\dfrac{15}{1}$

$= 40N.$

A body is moving in a vertical circle of radius '

$r$ ' by a string. If the ratio of maximum to minimum speeds is

$\sqrt{3}:1$ , the ratio of maximum to minimum tensions in the string is:

0%
3 : 1
0%
5 : 1
0%
7 : 1
0%
9 : 1

Explanation

Let

$v_1$ and

$v_2$ be the velocities at the lowest and the highest points.

$\therefore \dfrac {v_1}{v_2}=\dfrac {\sqrt 3}{1}$
Conserving energy between the lowest and highest points gives

${v_1}^2={v_2}^2+2gh$

$\Rightarrow 3{v_2}^2={v_2}^2+2gh$

$\Rightarrow {v_2}^2=gh$

$\Rightarrow {v_1}^2=3gh$

$h$ is distance between the lowest and the highest points which is

$2r$ .

$h=2r$

Tensions at the lowest and highest point

$T_1=\dfrac {m{v_1}^2}{r}+mg$

$T_2=\dfrac {m{v_2}^2}{r}-mg$

$\Rightarrow \dfrac {T_1}{T_2}=\dfrac {{v_1}^2/r+g}{{v_2}^2/r-g}$

$\Rightarrow \dfrac {T_1}{T_2}=\dfrac {6gr/r+g}{2gr/r-g}$

$\Rightarrow \dfrac {T_1}{T_2}=7/1$

A simple pendulum of length

$50 cm$ is suspended from a fixed point

$O$ and a nail is fixed at a point

$P$ which is vertically below

$O$ at some distance. The bob is released when string is horizontal. The bob reaches lowest position then it describes vertical circle whose centre coincides with point

$P$ . The minimum distance between

$O$ and

$P$ is:

0%
20 cm
0%
25 cm
0%
30 cm
0%
40 cm

Explanation

When the ball just reaches the bottom, from conservation of energy,

$v^2=2gL$

But when it reaches the bottom, it describes a vertical circle. hence

$v^2=5gr$

where

$r$ is the radius of the circle.

Hence

$\dfrac{r}{L}=\dfrac{2}{5}$

$\implies r=20cm$

Hence minimum distance between O and P=

$50cm-20cm=30cm$

Mass of the bob of a simple pendulum of length

$L$ is

$m$ . If the bob is projected horizontally from its mean position with velocity

$\sqrt{4gL}$ , then the tension in the string becomes zero after a vertical displacement of :

0%
$L/3$
0%
$3L/4$
0%
$4L/3$
0%
$5L/3$

Explanation

Let the angle

$\theta$ be when tension becomes zero.

Under this condition

$mg sin\theta = \dfrac{mv^{2}}{L}$ ....... eq(1)

Also, by conservation of energy:

$\dfrac { m (\sqrt {4gL})^{2}}{2} = mg(L+L sin \theta) + \dfrac{mv^{2}}{2}$

$\dfrac{mv^{2}}{L} = 2gmL - 2gmLsin \theta$ ------- eq(2)

From eq (1) and eq(2):

$mg sin\theta = 2mg - 2mg sin\theta$

$sin \theta = 2/3$

So, the tension in the string becomes zero after a vertical displacement of:

$L+Lsin\theta = 5L/3$

A mass

$0.1\ kg$ is rotated in a vertical circle using the cord of length

$1\ m$ , when the cord makes an angle

$60^0$ with the vertical, the speed of the mass is

$3\ m/s$ the resultant radial acceleration of mass in that position is:

0%
$9m/s^{2}$
0%
$4.9m/s^{2}$
0%
$4.1m/s^{2}$
0%
zero

Explanation

Radial acceleration at that position ,

$a_r = \dfrac{v^2}{r} - g cos \theta$

$a_r = 9 - 4.9 = 4.1\ m/s^2$

Two bodies A, B of masses

$m_1, m_2$ are knotted to a mass-less string at different points rotated along concentric circles in horizontal plane. The distances of A, B from common centre are 50cm, 1m. If the tensions in the string between centre to A and A to B are in the ratio 5:4, then the ratio of

$m_{1}$ to

$m_{2}$ is:

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

Explanation

The tension will be more between center to A as it has to bear the centripetal force of 2 bodies.

applying force balance on the masses,

$T_{1}-T_2=m_1 \omega^{2}r_1$ .........(1)

$T_2=m_2 \omega^{2}r_2$ .........(1)

$\Rightarrow \dfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\dfrac { 4 }{ 5 } \quad$

Now:

${ r }_{ 1 } = 0.5$ ,

${ r }_{ 2 } =1$

$\dfrac { 5 }{ 4 } = \dfrac { { m }_{ 1 }(0.5) }{ { m }_{ 2 }+1 }$

$\dfrac{m_1}{m_2} = \dfrac{2\times 1}{4} = \dfrac{1}{2}$

Water stored in a dam possesses:

0%
no energy
0%
electrical energy
0%
kinetic energy
0%
potential energy

When velocity of a moving object is doubled its:

0%
acceleration is doubled
0%
momentum becomes four times more
0%
kinetic energy is increased to four times
0%
potential energy is increased

Explanation

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

$K.E \propto v^2$

$New\ K.E = \dfrac{1}{2}m(2v)^2 = 4 \times K.E$

$\therefore$ The K.E. of a body will increase to 4 times if its velocity doubles.

Instantaneous power of constant force acting on a particle moving in a straight line under the action of this force

0%
is constant
0%
increases linearly with time
0%
decreases linearly with time
0%
either increases or decreases linearly with time

Explanation

Power is defined as rate of work done

$\implies$ Power,

$P = F.v$

Even though the force is constant, due to the force the velocity increases because of which its power increases.

Also from equations of motion we know that

$v = u + at$

$\implies$ instantaneous power increases linearly with time

In a hydro power plant:

0%
Potential energy possessed by stored water is converted into electricity
0%
Kinetic energy possessed by stored water is converted into potential energy
0%
Water is heated to produce electricity
0%
Water is converted into steam to produce electricity

Which of the following physical quantities is different from others?

0%
Work
0%
Kinetic energy
0%
Force
0%
Potential energy

Explanation

Kinetic energy is the energy possessed by the body by virtue of its motion and potential energy is the energy possessed by the virtue of its position and shape. Energy is required to do work.

Hence work, kinetic energy, potential energy all can be measured using same unit, i.e joule.

So, force is the physical quantity here which is different from others as given by product of mass and acceleration. S.I unit of force is

$N$ (newton).

A stone of mass

$1 kg$ tied to a light inextensible string of length

$L=\dfrac{10}{3}m$ is whirling in a circular path of radius

$L$ in a vertical plane. If the ratio of the maximum tension in the string to the minimum is

$4$ and if

$g$ is taken to be

$10 \ m/s^2$ , then speed of stone at the highest point of the circle is

0%
$20 m/sec$
0%
$10\sqrt{3} \ m/sec$
0%
$5\sqrt{2} \ m/sec$
0%
$10 m/sec$

Explanation

Let '

$v$ ' be the velocity at the top point
Velocity at the bottom point is

$= v^2 + 2gd$

$= v^2 + 4gr$
Maximum tension in string appears when the stone is at the bottom of the circle.

$T_{max} = \dfrac{m(v^2 + 4gr)}{r} + mg$

Minimum tension appears in a string when the stone is at the top of the circle.

$T_{min} = \dfrac{mv^2}{r} - mg$

$\dfrac{T_{max}}{T_{min}} = \dfrac{\dfrac{v^2 + 4gr}{r} + g}{\dfrac{v^2}{r} - g}$

$= 4$

On solving the equation we get,

$v$

$= 10\ m/s$

A car is accelerated on a leveled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car

0%
does not change
0%
becomes twice to that of initial
0%
becomes 4 times that of initial
0%
becomes 16 times that of initial

An 800 g ball is pulled up a slope as shown in the diagram. Calculate the potential energy it gains.

0%
1.96 J
0%
1.568 J
0%
7.84 J
0%
1.225 J

A coconut fruit hanging high in a palm tree has ......... owing to its location.

0%
Free energy
0%
Kinetic energy
0%
Activation energy
0%
Potential energy

A particle of mass

$m_0$ , travelling at speed

$v_0$ , strikes a stationary particle of mass

$2m_0$ . As a result, the particle of mass

$m_0$ is deflected through

$45^o$ and has a final speed of

$\dfrac {v_0}{\sqrt 2}$ . Then the speed of the particle of mass

$2m_0$ after this collision is

0%
$\dfrac {v_0}{2}$
0%
$\dfrac {v_0}{2\sqrt 2}$
0%
$\sqrt 2v_0$
0%
$\dfrac {v_0}{\sqrt 2}$

Explanation

From conservation of momentum,

$\vec{p_i}=\vec{p_f}$ ;
Since the particle of mass

$m_0$ moves at angle

$45^0$ , we have the unit vector

$(\dfrac{1}{\sqrt2}\hat{i}+\dfrac{1}{\sqrt2}\hat{j})$ to show the direction.

$\therefore m_0v_0\hat{i}=m_0\dfrac{v_0}{\sqrt 2}(\dfrac{1}{\sqrt2}\hat{i}+\dfrac{1}{\sqrt2}\hat{j}) +2m_0\vec{v}$

$\therefore \vec{v}=\dfrac{v_0}{4}\hat{i}-\dfrac{v_0}{4}\hat{j}$

$\therefore\left |\vec{v} \right |=\dfrac{v_0}{2\sqrt2}$

what is the work done by a force

$4N$ in moving the body from d

$=$ 1m to 4m?

0%
6 J
0%
16 J
0%
12 J
0%
8 J

Explanation

$W=F.S \rightarrow 4(4-1)J = 12J$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

Work done by a force is given by

$W = F.S$

$W = FS\cos\theta$

where force is acting along positive x-direction but displacement is in negative x-direction.

$\therefore \cos\theta = -1$

Which means work done is negative.

A spring of natural length

$l$ is compressed vertically downward against the floor so that its compressed length becomes

$\displaystyle\frac{l}{2}$ . On releasing, the spring attains it's natural length. If

$k$ is the stiffness constant of the spring, then the work done by the spring on the floor is

0%
$zero$
0%
$\displaystyle\frac{1}{2}kl^{2}$
0%
$\displaystyle\frac{1}{2}k(\displaystyle\frac{l}{2})^{2}$
0%
$kl^{2}$

Explanation

Only the center of mass of the spring changes, but the floor has no displacement.

$W=F.x$

$W=F\times0$

$W=0$
Hence the work done by the spring on the floor is zero.
Option A.

A heavy ball moving with speed

$v$ collides with a tiny ball. The collision is elastic, then immediately after the impact, the second ball will move with a speed approximately equal to-

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$v$
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$2v$
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$\dfrac {v}{2}$
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$\dfrac {v}{3}$

Explanation

Using the conservation of momentum and KE, the velocities of the balls after collision is derived.

$m_1=$ mass of the 1st ball

$m_2=$ mass of the 2nd ball

$v_1=$ velocity of the 1st ball before collision

$=v$ ;

$v_2=$ velocity of the 2nd ball before collision

$=0$

${v_2}^{'}= \left(\dfrac{2m_1}{m_1+m_2}\right) v_1 + \left(\dfrac{m_1-m_2}{m_1+m_2}\right)v_2$ ;

$m_1 > > m_2$ ;
Substituting

$v_1=v$ and

$v_2=0$ ,we get

${v_2}'=2v$

A particle moves along the x-axis from

$x=0$ to

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

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

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360 J
0%
85 J
0%
185 J
0%
135 J

Explanation

Work done is given by,

$\displaystyle W=\int_0^5 Fdx$

$\displaystyle =\int_0^5(7-2x+3x^2)dx$

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

$=135 \ J$

A ball is projected upwards. As it rises, there is increase in its:

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Momentum
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Retardation
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Kinetic energy
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Potential energy

Explanation

When a ball is projected upwards it's height increases.

As height increases,

$v$ velocity decreases (Kinetic Energy Decreases) so Potential energy increases.

$Potential$

$Energy$ =

$mgh$

A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector

$\overrightarrow{a}$ is correctly shown in

Explanation

The bob has both radial as well as tangential acceleration when it is at a displacement less than its maximum displacement.

From figure

$a_t =g sin\theta$ and

$a_r = \dfrac{T}{m} - g cos\theta$

Thus resultant acceleration

$\vec{a} = \vec{a_r} + \vec{a_t}$ points in the direction as shown in the figure.

A stone tied to a string of length

$L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at its lowest position and has a speed

$u$ .The magnitude of change in its velocity as it reaches a position, where the string is horizontal is

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$\sqrt{u^2 - 2gL}$
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$\sqrt{2gL}$
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$\sqrt{u^2 - gL}$
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$\sqrt{2(u^2 - gL)}$

Explanation

Here

$v^2 - u^2 = -2gl$ ....(i)

$v^2 = u^2-2gl$

Since the velocities are mutually perpendicular, change in velocity

$\triangle{v} = \sqrt{u^2 + v^2}$ ....(ii)

$=\sqrt{u^2 + u^2 - 2gl}$

(substituting the value of

$v^2$ from (i))

$\triangle{v} = \sqrt{2(u^2 - gl)}$

A simple pendulum has a bob of mass m and swings with an angular amplitude

$\phi$ . The tension in the thread is T. At a certain time, the string makes an angle

$\theta$ with the vertical (

$\theta \le \phi$ )

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$T = mg \cos \theta$ , for all values of $\theta$
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$T = mg \cos \theta$ , only for $\theta = \phi$
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$T = mg$ , for $\displaystyle \theta =\cos^{-1} \left[\frac{1}{3}(2cos\phi +1)\right]$
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T will be larger for smaller values of $\theta$

Explanation

At any point of time the forces on the bob are Tension , gravitational

and the centripetal force

We have

$mg(lcos\theta -lcos\phi )=\ 1/2\ m{ v }^{ 2 }\\ \dfrac { m{ v }^{ 2 } }{ l } =\ 2mg(cos\theta -cos\phi )\\ Also\ T -mgcos\theta =\dfrac { m{ v }^{ 2 } }{ l } = 2mg(cos\theta \ -\quad cos\phi )\\ or\ T=\ mg(3cos\theta -\ 2cos\phi )$

For

$\theta$ =

$\cos ^{ -1 }{ \dfrac { 1 }{ 3 } (2cos\phi +1) }$

$T=mg$

A pendulum hangs from the ceiling of a jeep moving with a speed,

$v$ , along a circle of radius,

$R$ . Find the angle with the vertical made by the pendulum.

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$0$
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$\tan ^{-1}\displaystyle \frac{v^{2}}{Rg}$
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$\tan ^{-1}\displaystyle \frac{Rg}{v^{2}}$
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None of the above

Explanation

Considering accelerations,

$a_{r}=\displaystyle \dfrac{v^{2}}{R}$

$\therefore tan\theta =\displaystyle \dfrac{v^2/R}{g}$

$=\displaystyle \dfrac{v^{2}}{Rg}$

A simple pendulum has a string of length

$l$ and bob of mass

$m$ . When the bob is at its lower position, it is given the maximum horizontal speed necessary for it to move in a circular path about the point of suspension. The tension in the string at the lowest position of the bob is

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$mg$
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$3 mg$
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$\sqrt {10} mg$
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$4 mg$

Explanation

Lets A is the topmost point of the circle and B be the lowest point of the circle.

Let

$v_2$ and

$v_1$ be the velocities at B and A respectively.
Applying the principle of conservation of energy between A and B.

$\displaystyle \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2 = mg2l$

$\displaystyle \frac{mv_2^2}{l} = \frac{mv_1^2}{l} + 4mg$ .....(1)

At B,

$\displaystyle\frac{mv_2^2}{l} = T -mg$ .....(2)

At A.

$\displaystyle \frac{mv_1^2}{l} = mg$ ....(3)

Solving (1), (2) and (3)

$mg+4mg=T-mg$ or

$T=6mg$

A stone tied to string of length

$l$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at its lowest position and has a speed

$u$ . The magnitude of the change in velocity as it reaches a position, where the string is horizontal is

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$\sqrt{u^2-2gl}$
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$\sqrt{2gl}$
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$\sqrt{u^2-gl}$
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$\sqrt{2(u^2-gl)}$

A simple pendulum rotates in a horizontal plane with an angular velocity

$\omega$ about a fixed point P in a gravity free space. There is a negative charge at P. The bob gradually emits photo electrons(Leave the energy and momentum of incident photons and emitted electrons). The total force acting on the bob is

$T$ .

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$T$ will decrease, $\omega$ will decrease
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$T$ will decrease, $\omega$ will remain constant
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$T$ and $\omega$ will remain unchanged
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The elastic strain in the string will decrease

Explanation

Answer is C and D

As no forces act on the system,

ω remains constant. Hence,

$T=m\omega^2l$ remains constant. However, T is the sum of electrostatic force of attraction on the bob and the force exerted by the string due to elastic strain in it. As the positive charge on the bob increases due to photo emission, the electrostatic force on it increases, and the elastic strain on the string must decrease.

A particle moves along a vertical circle of radius r with a velocity

$\sqrt { 8rg }$ at Y.

${ T }_{ A }$ ,

${ T }_{ B }$ ,

${ T }_{ x }$ ,

${ T }_{ y }$ respresent tension at A, B, X and Y, respectively, then

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${ T }_{ A } = { T }_{ B }$
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${ T }_{ X } - { T }_{ Y } =6mg$
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${ T }_{ Y } - { T }_{ X } =6mg$
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${ T }_{ Y } > { T }_{ X } \neq 6mg$

Explanation

When the string is horizontal only tension provides the radial acceleration, thus,

$\\ { T }_{ A }={ T }_{ B }=\dfrac { m{ v }^{ 2 } }{ r } \\ { T }_{ X }+mg=\dfrac { m{ { v }_{ X } }^{ 2 } }{ r } \ and\ { T }_{ Y }-mg=\dfrac { m{ { v }_{ Y } }^{ 2 } }{ r } \\$

by conservation of

$ME: \dfrac { m{ { v }_{ X } }^{ 2 } }{ 2 } -\dfrac { m{ { v }_{ Y } }^{ 2 } }{ 2 } =-2mgr\Rightarrow { { v }_{ X } }^{ 2 }-{ { v }_{ Y } }^{ 2 }=-4gr\\ Thus,\ { T }_{ X }-{ T }_{ Y }+2mg=\dfrac { m }{ r } \left( { { v }_{ X } }^{ 2 }-{ { v }_{ Y } }^{ 2 } \right) =-4mg\Rightarrow { T }_{ X }-{ T }_{ Y }=-6mg$

A particle of mass,

$m$ , is tied to a light string and rotated with a speed,

$v$ , along a circular path of radius,

$r$ . If

$T=$ tension in the string and

$mg =$ gravitational force on the particle, then the actual forces acting on the particle are

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$mg$ and $T$ only.
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$mg$ , $T$ and an additional force of $mv^2/r$ directed inwards.
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$mg$ , $T$ and an additional force of $mv^2/r$ directed outwards.
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Only a force $mv^2/r$ directed outwards.

Explanation

The force

$mv^2/r$ directed outwards, called centrifugal force, is not a real force.

At A,

$mv_1^2/l = T_1 +mg$

The resultant of

$mg$ and tension (

$T$ ) will provide the centripetal force.

A stone of mass

$1\ kg$ tied to a light inextensible string of length.

$L = \dfrac{10}{3}$ metre is whirling in a circular path of radius,

$L$ , in a vertical plane. If the ratio of maximum tension to the minimum tension is

$4$ and if

$g$ is taken to be

$10\ m/s^{2}$ , the speed of the stone at the highest point of circle is :

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$20\ m/s$
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$10 \sqrt{3}\ m/s$
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$10\ m/s$
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$5\sqrt{2}\ m/s$

Explanation

When the stone is revolved in a vertical circle, tension is maximum at bottom and minimum at top. So we have, Tension at bottom as

$T_B$ and Tension at top as

$T_T$ .

We know,

$\dfrac { { T }_{ B } }{ { T }_{ T } } =4$

$\Rightarrow \dfrac { \dfrac { mv_{ B }^{ 2 } }{ L } +mg }{ \dfrac { mv_{ T }^{ 2 } }{ L } -mg } =4$

$\Rightarrow \dfrac { mv_{ B }^{ 2 } }{ L } +mg=4\left( \dfrac { mv_{ T }^{ 2 } }{ L } -mg \right)$

$\Rightarrow \dfrac { mv_{ B }^{ 2 } }{ L } =\dfrac { 4mv_{ T }^{ 2 } }{ L } -5mg$

$\Rightarrow mv_{ B }^{ 2 } = 4mv_{ T }^{ 2 }-5mgL$ ......

$(I)$

Now using the law of conservation of energy, at top and bottom of the vertical circle, we have:

$\dfrac { 1 }{ 2 } mv_{ T }^{ 2 }\quad +2mgL=\frac { 1 }{ 2 } mv_{ B }^{ 2 }$

$\Rightarrow m{ { v }_{ B } }^{ 2 }=m{ { v }_{ T } }^{ 2 }+4mgL$ ......

$(II)$

from

$(I)$ and

$(II)$ , we have

$4mv_{ T }^{ 2 }-5mgL=m{ { v }_{ T } }^{ 2 }+4mgL\\ \Rightarrow 3mv_{ T }^{ 2 }=9mgL\\ \Rightarrow { v }_{ T }=\sqrt { 3gL } \\ \Rightarrow { v }_{ T }=\sqrt { 3\times 10\times \left( \dfrac { 10 }{ 3 } \right) } =10\ m/s$

A body of mass

$m$ is accelerated to velocity

$v$ in time

$t'$ . The work done by the force as a function of time

$t$ will be

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$\displaystyle \frac{m}{2} \frac{v^2 t^2}{t'^2}$
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$\displaystyle \frac{1}{2} \left ( \frac{mv}{t'} \right )^2 t^2$
0%
$\displaystyle \frac{m}{2} \frac{v}{t'} t^2$
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$\displaystyle \frac{mvt^2}{2t'}$

Explanation

The force as a function of time

$t$ on body of mass

$m$ is

$F = m.a$ ,

$a$ -acceleration
And the work done by the force as a function of time

$t$ is

$W = F.d$ ,

$d$ -distance covered

$W=ma.d$
Here, the body is accelerated to velocity

$v$ in time

$t'$ , hence

$a = \dfrac{v}{t'}$
distance traveled

$\displaystyle d=\int _{ 0 }^{ t }{ \frac { v }{ t' } tdt=\frac { v }{ t' } \frac { { t }^{ 2 } }{ 2 } }$

$\displaystyle W=m a. d = m \frac { {v}^{2} }{ {t'}^{2} } \frac { { t }^{ 2 } }{ 2 }$

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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