MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 12

A weightless thread can withstand tension upto

$30\;N$ . A stone of mass

$0.5\ kg$ is tied to it and is revolved in a circular path of radius

$2\ m$ in a vertical plane. If

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

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

Explanation

Given :

$T_{max} =30$ N

$m = 0.5 kg$

$r =2$ m

$g = 10 m/s^2$

Tension in the string is maximum when the stone is at the lowest position of the vertical motion.

Let the maximum angular velocity be

$w_{max}$ .

Using circular motion equation :

$mrw^2_{max} =T_{max} - mg$

$\therefore$

$0.5 \times 2 \times w^2_{max} =30 - 0.5 \times 10$

$w^2_{max} = 30 - 5 =25$

$\implies w_{max} = 5$

$rad/s$

A weightless thread can bear tension upto

$3.7\;kg$ wt. A stone of mass

$500\;gm$ is tied to it and revolved in a circular path of radius

$4m$ in a vertical plane. If

$g=10\;ms^{-2}$ , then the maximum angular velocity of the stone will be:

0%
$\;16\;rad/s$
0%
$\;\sqrt{21}\;rad/s$
0%
$\;2\;rad/s$
0%
$\;4\;rad/s$

Explanation

Given :

$T_{max} =3.7 kg wt = 37$ N

$m =500$ gm

$= 0.5 kg$

$r =4$ m

$g = 10 m/s^2$

Tension in the string is maximum when the stone is at the lowest position of the vertical motion.

Let the maximum angular velocity be

$w_{max}$ .

Using circular motion equation :

$mrw^2_{max} =T_{max} - mg$

$\therefore$

$0.5 \times 4 \times w^2_{max} =37 - 0.5 \times 10$

$w^2_{max} =16$

$\implies w_{max} = 4$

$rad/s$

A weightless thread can bear tension upto

$37 N$ . A stone of mass

$500 g$ is tied to it and revolved in a circular path of radius

$4 m$ in a vertical plane. If

$g=10 {ms}^{-2}$ , then the maximum angular velocity of the stone will be:

0%
$2\ rad\ {s}^{-1}$
0%
$4\ rad\ {s}^{-1}$
0%
$8\ rad\ {s}^{-1}$
0%
$16\ rad\ {s}^{-1}$

Explanation

Maximum tension in the string will be at the lowest point of the circular path, where the tension will support the weight as well as provide the centripetal force for circular motion.

So, we have:

$T=mg+ml{ \omega }^{ 2 }$

$\Rightarrow ml{ \omega }^{ 2 }=T-5$

For maximum angular speed, the tension must be

$37N$

So, the maximum angular speed is

$\sqrt { \dfrac { T-5 }{ ml } } =\sqrt { \dfrac { 37-5 }{ 0.5\times 4 } } = 4 rad/s$

Thus,

${ \omega }_{ max }= 4 rad/s$

Stone tied at one end of light string is whirled round a vertical circle. If the difference between the maximum and minimum tension experienced by the string wire is

$2\ kg\ wt$ , then the mass of the stone must be

0%
$1\ kg$
0%
$6\ kg$
0%
$1/3\ kg$
0%
$2\ kg$

Explanation

$\textbf{Step 1: FBD [Ref. Fig.]}$

Tension will be maximum at bottom and minimum at top

$\textbf{Step 2: Velocities at highest and lowest points}$

Minimum velocity required at bottom point to complete circle

$= v$

$v = \sqrt{5gR}$

$....(1)$

$\textbf{Step 3: Calculations }$

For revolving in complete circle tension at highest point

$T\geq 0$

$\Rightarrow T_{\min} = 0$

Applying Newton's second law on the stone along centripetal direction (Considering direction towards center as positive)

$\Sigma F_c = ma_c$

$=m\left (\cfrac{v^2}{R} \right)$

Using equation

$(1)$ in the following

$\Rightarrow T_{max} - mg = \dfrac{mv^2}{R} = 5 mg$

$\Rightarrow T_{\max} = 6 mg$ [T is maximum at bottom]

So,

$T_{\max} - T_{\min} = 6mg$

$\Rightarrow 6mg = 2g$ Given:

$T_{\max} - T_{\min} = 2g$

$\Rightarrow m = \dfrac{2}{6} = \dfrac{1}{3} kg$

Hence

$C$ is the correct option.

2115133_280642_ans_40548d08555448c6834960c21299b13c.png

If a particle of mass

$M$ is tied to a light inextensible string fixed at point

$P$ and particle is projected at A with velocity

$V_A\, =\, \sqrt{4 gL}$ as shown. Find tension in the string at

$B$ . (Assume particle is projected in the vertical plane.)

0%
$Mg$
0%
$3Mg$
0%
$2Mg$
0%
$7Mg$

Explanation

Taking point A as reference where gravitational potential energy is zero,

Total energy of particle at point A is,

$E_A = \dfrac{1}{2}MV_A^2 = 2MgL$

Let velocity at point B as

$V_B$ .

So, Total energy at point B is,

$E_B = mgL + \dfrac{1}{2}MV_B^2$

By conservation of energy,

$E_A=E_B$

$\implies \dfrac{1}{2}MV_B^2 = mgL\implies \dfrac{MV_B^2}{L}=2Mg$

So centripetal force point B is

$2Mg$

At point B,

$T+Mg=\dfrac{MV_B^2}{L}=2Mg\implies T=Mg$

So Tension in the string is

$Mg$ , at point B.

A force

$F=-6{x}^{3}$ is acting on a block moving along x-axis. Work done by this force is:

0%
Positive in displacing the block from $x=3$ to $x=1$ .
0%
Positive in displacing the block from $x=-3$ to $x=-1$ .
0%
Negative in displacing the block from $x=0$ to $x=4$ .
0%
Zero in displacing the block from $x=-2$ to $x=+2$ .

Explanation

For $(A):$ Displacement $< 0$ and $F<0$ so $W>0$

For $(B):$ Displacement $>0$ and $F>0$ so $W>0$

For $(C):$ Displacement $>0$ and $F<0$ so $W<0$

For $(D):$ $W=\int _{ -2 }^{ 2 }{ -6{ x }^{ 3 } } dx=\left( -\cfrac { 6 }{ 4 } \right) { \left( { x }^{ 4 } \right) }_{ -2 }^{ 2 }=0$

A body crosses the topmost point of a vertical circle with critical speed. What will be its centripetal acceleration when the string is horizontal. :-

0%
$\;g$
0%
$\;2g$
0%
$\;3g$
0%
$\;6g$

Explanation

The critical speed of the body at the topmost point is equal to

$\sqrt{gR}$ i.e

$v = \sqrt{gR}$

Let the velocity of the body when the string is horizontal be

$v_2$

Using work-energy theorem :

$W = \Delta K.E$

$\therefore$

$mgR = \dfrac{1}{2} m v_2^2 - \dfrac{1}{2}mv^2$

$mgR = \dfrac{1}{2} m v_2^2 - \dfrac{1}{2}m(gR)$

$\implies v_2^2 = 3gR$

Centripetal acceleration

$a_r = \dfrac{v_2^2}{R} = 3g$

A small sphere is given vertical velocity of magnitude

$v_{0} = 5m/s$ and it swings in a vertical plane about the end of massless string. The angle

$\theta$ with the vertical at which string will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere, is

$[g = 10 m/s^{2}]$ .

0%
$\cos^{-1}\left (\dfrac {2}{3}\right )$
0%
$\cos^{-1}\left (\dfrac {1}{4}\right )$
0%
$60^{\circ}$
0%
$30^{\circ}$

Work done in time

$t$ on a body of mass

$m$ which is accelerated from rest to a speed

$v$ in time as a function of time

$t$ is given by

0%
$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }\quad$
0%
$m\cfrac { v }{ { t }_{ 1 } } { t }^{ 2 }$
0%
$\cfrac { 1 }{ 2 } m{ \left( \cfrac { v }{ { t }_{ 1 } } \right) }^{ 2 }{ t }^{ 2 }$
0%
$\cfrac { 1 }{ 2 } m\cfrac { v }{ { t }_{ 1 }^{ 2 } } { t }^{ 2 }$

Explanation

Ans : We know

$W = F \times s$

where,

$F = ma$

By using Kinematics,

$s = \dfrac{1}{2}at^{2}$ , if u =0

So,

$W = (ma).(\dfrac{1}{2}at^{2})$

$W = \dfrac{1}{2}ma^{2}t^{2}$ where

$a = v/t_{1}$

$\boxed{W = \dfrac{1}{2}m(\dfrac{v}{t_{1}})^{2}t^{2}}$

A mass tied to a string moves in a vertical circle and at the point

$P$ its speed is

$5m/s$ . At the point

$P$ the string breaks. The mass will reaches height above

$P$ of nearly

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

0%
$1m$
0%
$0.5m$
0%
$1.27m$
0%
$1.25m$

A stone of mass

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

$L = \dfrac{10}{3}\ m$ , whirling in a circular path in a vertical plane. The ratio of maximum tension in the string to the minimum tension in the string is

$4$ . If g is taken to be

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

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

Explanation

$The\quad ratio\quad to\quad the\quad mamimum\quad { T }_{ 2 }{ \quad to\quad the\quad minimum\quad tension\quad T }_{ 1 }\quad is\quad -\\ \frac { { T }_{ 2 } }{ { T }_{ 1 } } =4\\ { T }_{ 2 }=4{ T }_{ 1 }\\ Now\quad the\quad difference\quad between\quad the\quad two\quad 6mg;\\ Hence\quad we\quad have\quad \\ { T }_{ 2 }-{ T }_{ 1 }=6mg\\ \therefore 4{ T }_{ 1 }-{ T }_{ 1 }=6mg\\ 3{ T }_{ 1 }=6mg\\ { T }_{ 1 }=2mg\\ Now\quad tension\quad at\quad the\quad top\quad of\quad the\quad circle\quad is-\\ { T }_{ 1+mg }=\frac { { mv }_{ 1 }^{ 2 } }{ r } =\frac { { mv }_{ 1 }^{ 2 } }{ L } \\ 2mg+mg=\frac { { mv }_{ 1 }^{ 2 } }{ \frac { 10 }{ 3 } } \\ { v }_{ 1 }^{ 2 }=3g\frac { 10 }{ 3 } =10g\\ { v }_{ 1 }=\sqrt { 10g } \\ =\sqrt { 10\times 10 } =10m/s$

An athlete in the Olympic games covers a distance of

$100$ m in

$10$ s. His kinetic energy can be estimated to be in the range. (Assume weight = 60kg)

0%
$200J-500J$
0%
$2\times 10^5J-3\times 10^5J$
0%
$20000J-50000J$
0%
$2000J-5000J$

Explanation

Velocity

$V=\dfrac{distance}{time}=\dfrac{100}{10}=10 m/s$

Assuming his mass to be 60kg, his kinetic energy is

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

$= \dfrac{1}{2}\times 60\times 100$

$=60\times50$

$=3000 J$

Hence, Option D

From

$x=0$ to

$x=6$ , the force experienced by an object varies according to the function

$F\left(x\right)=\sqrt{6x-{x}^{2}}$ , as shown above. What is the work done by this force as the object moves from

$x=0$ to

$x=3$ ?
Assume all numbers are given in standard units.

0%
$0 J$
0%
$4.5 J$
0%
$7.1 J$
0%
$9.0 J$
0%
$14.0 J$

Explanation

There is a simpler method to arrive at the solution.

Conceptually, you are integrating force over displacement, so if you are given the graph, the integration is equal to the area under the curve. The plot given is one of a semicircle.

The function

$F\left( x \right) =\sqrt { 6x-{ x }^{ 2 } }$ is one of a semicircle.

It may be easier to see this when it is in the form

$F(x)=\sqrt{9+6x-x^2-9}$ , which corresponds to

$F(x)=3^2-(x-3)^2$ which is the upper half of a circle centered at

$x = 3$ with radius

$3$ .

The area of a circle is given by

$A=\pi r^2$ , and in this case you are looking for the area under half of the top half.

Thus,

$W=\dfrac{1}{4}A=\dfrac{1}{4}\pi (3)^2\approx7.0658\ J$ , that is choice "C".

A spherical ball A of mass

$4\ kg$ , moving along a straight line strikes another spherical ball B of mass

$1\ kg$ at rest. After the collision, A and B move with velocities

$v_{1}\ ms^{-1}$ and

$v_{2}\ ms^{-1}$ respectively making angles of

$30^{\circ}$ and

$60^{\circ}$ with respect to the original direction of motion of ball A. The ratio

$\dfrac {v_{1}}{v_{2}}$ will be:

0%
$\dfrac {\sqrt {3}}{4}$
0%
$\dfrac {4}{\sqrt {3}}$
0%
$\dfrac {1}{\sqrt {3}}$
0%
$\sqrt {3}$

Explanation

Since there is no momentum in the line perpendicular to the direction of motion of ball A before collision, after collision also there would be no momentum along that line. So by conservation of momentum along that line we have:

$0=4{ v }_{ 1 }\sin { 30° } -{ v }_{ 2 }\sin { 60° } \Rightarrow \dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\dfrac { \sqrt { 3 } }{ 4 }$

So, option A is the correct answer.

Two blocks are connected by a massless string that passes over a frictionless peg as shown in fig. One end of the string is attached to a mass

$m_1 = 3kg$ , i.e. a distance

$R = 1.20 m$ from the peg. The other end of the string is connected to a block of mass

$m_2 = 6 kg$ resting on a table. When the 3 kg block be released at

$\displaystyle \theta = \frac{\pi}{k}$ , the 6 kg block just lift off the table? Find the value of k.

0%
$2$
0%
$3$
0%
$4$
0%
$6$

Explanation

Tension in the string will be maximum when

$m_1$ is at lowest position. For

$m_2$ to just lift, tension in the string at this position should equal the weight of

$m_2$ at this position.

Using conservation of energy,

$\dfrac{m_1 R^2 \omega^2}{2} = m_1 g R (1-\cos \theta)$

$m_1 \omega^2 R = 2m_1 g (1 - \cos \theta) \quad ........(i)$

From the FBD of

$m_1$ ,

$T_1 - m_1g = m_1 \omega^2 R \quad ...........(ii)$

From

$(i)\ \& \ (ii),$

$T_1 = m_1g + 2m_1g (1-\cos \theta)$

$T_1 = m_1g (3-2\cos\theta) \quad .........(iii)$

From the FBD of

$m_2$ ,

$T_2 + N_2 = m_2 g \quad ........(iv)$

When the block just leaves the surface,

$N_2 = 0 \quad .....(v)$

Also, tension throughout the string is constant,

$T_2=T_1 \quad ........(vi)$

Using

$(iii), \ (iv), \ (v) \ \& \ (vi),$

$m_2g = m_1g(3-2cos\theta)$

$\cos \theta = \dfrac{3 - m_2/m_1}{2}$

Substituting values,

$\cos \theta = 1/2$

$\theta = \pi/6$

$\implies k = 6$

A car of mass

$m$ starts moving so that its velocity varies according to the law

$v=\beta \sqrt { s }$ , where

$\beta$ is a constant, and

$s$ is the distance covered. The total work performed by all the forces which are acting on the car during the first

$t$ seconds after the beginning of motion is:

0%
${ m\beta }^{ 4 }{ t }^{ 2 }/8$
0%
${ m\beta }^{ 2 }{ t }^{ 4 }/8$
0%
${ m\beta }^{ 4 }{ t }^{ 2 }/4$
0%
${ m\beta }^{ 2 }{ t }^{ 4 }/4$

Explanation

$v=\beta \sqrt s$

$\dfrac{ds}{dt}=\beta \sqrt s$

On integration

$2\sqrt s=\beta t$

$s=\dfrac{\beta^2 t^2}{4}$

$v=\dfrac{\beta^2 t}{2}$

On differentiation with respect to t

$a=\dfrac{\beta^2}{2}$

Work =F.s=mas=

$\dfrac{m\beta^4 t^2}{8}$

By applying a force

$\vec{F} = (3xy - 5z)\hat{j} + 4z\hat{k}$ a particle is moved along the path

$y=x^{2}$ from point

$(0,0,0)$ to point

$(2,4,0)$ . The work done by the

$F$ on the particle is

0%
$\dfrac{280}{5}$
0%
$\dfrac{140}{5}$
0%
$\dfrac{232}{5}$
0%
$\dfrac{192}{5}$

A small block of mass m is pushed on a smooth track from position A with a velocity

$2\sqrt5$ times the minimum velocity required to reach point D. The block will leave the contact with track at the point where normal force between them becomes zero.
At what angle

$\theta$ with horizontal does the block gets separated from the track?

0%
$sin^{-1}(\frac{1}{3})$
0%
$sin^{-1}(\frac{3}{4})$
0%
$sin^{-1}(\frac{2}{3})$
0%
never leaves contact with the track

The bob

$A$ of a pendulum of mass

$m$ released from horizontal to the vertical hits another bob

$B$ of the same mass at rest on a table as shown in figure. If the length of the pendulum is

$1\ m$ , what is the speed with which bob

$B$ starts moving. (Neglect the size of the bobs and assume the collision to be elastic) (Take

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

0%
$4.47\ ms^{-1}$
0%
$5.47\ ms^{-1}$
0%
$6.47\ ms^{-1}$
0%
$3.47\ ms^{-1}$

Explanation

As the collision is elastic and two balls have the same mass, therefore ball

$A$ transfers its entire momentum to the ball

$B$ and ball

$A$ does not rise at all. After a collision, ball

$A$ comes to rest and ball

$B$ moves with speed of ball

$A$ .

$\therefore$ The speed with which bob

$B$ starts moving is

$v = \sqrt {2gh}$

$= \sqrt {2\times 10\ ms^{-2} \times 1m} = \sqrt {20}\ ms^{-1}$

$= 4.47\ ms^{-1}$

$1\ kg$ stone at the end of

$1\ m$ long string is whirled in a vertical circle at a constant speed of

$4\ ms^{-1}$ . The tension in the string is 6 N when the stone is

0%
At the top of the circle
0%
At the bottom of the circle
0%
Half way down
0%
None of above

Explanation

$\textbf{Step 1 - Draw FBD }$

$\textbf{Step 2 - Apply Newton second law -}$

$\sum F=ma$

along the radially -

$F_{net} = Ma_{R}$

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

$T = mg \cos \theta + \dfrac {mv^{2}}{R}$

$\textbf{Step3:Put values to obtain answer}$

As given that -

$T = 6N, m = 1\ kg, g = 10\ m/s^{2}, v = 4\ m/s$ and

$R = 1\ m$

So, put these all values in the above equation -

$6 = 1\times 10\times \cos \theta + \dfrac {1\times (4)^{2}}{1}$

$6 = 10\cos \theta + 16$

$10\cos \theta = -10$

$\cos \theta = -1$

$\theta = 180^{\circ}$

Means, stone is to be on top of the circle.

So, the tension in the string is

$6N$ when -

the stone is at the top of circle.

Hence, option (A) is correct.

Two bodies

$A$ and

$B$ have masses

$20\ kg$ and

$5\ kg$ respectively. If they acquire the same kinetic energy. Find the ratio of thier velocities.

0%
$\dfrac {1}{2}$
0%
$2$
0%
$\dfrac {2}{5}$
0%
$\dfrac {5}{6}$

Explanation

Given:

$m_{A} = 20 kg$

$m_{B} = 5 kg$

Kinetic Energy of both bodies A and B are the same.

Thus,

$\dfrac{1}{2} m_{A} V_{A}^{2} = \dfrac{1}{2} m_{B} V_{B}^{2} \\$

$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{m_{B}}{m_{A}} \\$

$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{5}{20}\\$

$\dfrac{V_{A}}{V_{B}} = \dfrac{1}{2}$

So, option A is correct.

A spring lies along the x-axis attached to a wall at one end and a block at the other end. The block rests on a friction less surface at x =A force of constant magnitude F is applied to the block that begins to compress the spring, until the block comes to a maximum displacement

$x_{max}$ .
During the displacement, which of the curves shown in the graph best represents the work done on the spring block system by the applied force?

0%
1
0%
2
0%
3
0%
4

Explanation

$'F'$ is a constant force

Work done by

$F=F(displacement)\\ \quad=Fx\\ \Rightarrow a$ straight line graph

$(1)$

1174391_985655_ans_1c00760a0625444f8d2c50b1ef59ccc7.png

An elastic spring of unstretched length

$L$ and force constant

$K$ , is stretched by a small length

$x$ . It is further stretched by another small length

$y$ . calculate the work done during the second stretching is:

0%
$\cfrac { ky }{ 2 } \left( x+2y \right)$
0%
$\cfrac { k }{ 2 } \left( 2x+y \right)$
0%
$ky(x+2y)$
0%
$\cfrac { ky }{ 2 } \left( 2x+y \right)$

Explanation

Initial stretching of the spring is

$x$ .
Initial potential energy

$U_i = \dfrac{1}{2}kx^2$
Final stretching of the spring is

$x+y$ .
Final potential energy of the spring

$U_f = \dfrac{1}{2}k(x+y)^2 =$

$\dfrac{1}{2}kx^2 +\dfrac{1}{2}ky^2+\dfrac{1}{2}k(2xy)$
Work done

$W = U_f-U_i =$

$\dfrac{1}{2}kx^2 +\dfrac{1}{2}ky^2+\dfrac{1}{2}k(2xy) - \dfrac{1}{2}kx^2$

$\implies \ W = \dfrac{ky}{2}(2x+y)$

A string wraps around a fat pipe as a bob attached to the string is made to move in a circular path in the horizontal. Assuming the speed is somehow held constant as the radius diminishes due to the wrapping, how will the centripetal force change?

0%
It will stay the same.
0%
It will diminish.
0%
It will increase
0%
None of the above (this is a jokeits got to be one of the three above)

Under the action of a force,

$2kg$ body moves such that its position

$x$ as a function of time is given by

$x={t}^{3}/3$ where

$x$ is in meters and

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

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

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

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

Explanation

Given that-

$F\propto s$

$\implies F=ks$ where

$k=$ proportionality constant

Now, we know that, Work done

$W=F.s$

$\implies W=ks^2$

$\implies F\propto s^2$

Answer-(B)

A particle of mass

$m$ is moving horizontally with a constant velocity

$v$ towards a rigid wall that is moving in opposite direction with a constant speed

$u$ . Assuming elastic impact between the particle and wall the work done by the wall in reflecting the particle is equal to:

0%
$\left( \dfrac{1}{2} \right) m{ \left( u+v \right) }^{ 2 }\quad \quad$
0%
$\left( \dfrac{1}{2} \right)4 mv\left( { u }+{ v } \right)$
0%
$\dfrac{1}{2}muv$
0%
None of these

Explanation

The correct answer is option (B).

Hint: Make a pictorial representation and then calculate the speed.

Step 1: Make a representation of the given data.

Let speed of the wall be

$v_w=u$

Let the speed after collision be

$v'$

According to the above diagram,

$\dfrac{v'-v_w}{-v-v_w}=-e$

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

$v'=v+2u$

Step 2: Calculate the work done

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

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

$W=\dfrac{1}{2}(2u)(2v+2u)$

$W=\dfrac{1}{2}4mu(u+v)$

Hence, the work is $W=\dfrac{1}{2}4mu(u+v)$

The correct answer is option (B).

A man is standing on a plank which is placed on a smooth horizontal surface. There is sufficient friction between the feet of man and plank. Now man starts running over plank, correct statement is/are

0%
Work done by friction on the man with respect to the ground is negative.
0%
Work done by friction on the man with respect to the ground is positive.
0%
Work done by friction on the plank with respect to the ground is positive.
0%
Work done by friction on the man with respect to the plank is zero.

A small block of mass m is pushed on a smooth track from position A with a velocity

$2\sqrt5$ times the minimum velocity required to reach point D. The block will leave the contact with track at the point where normal force between them becomes zero.
When the block reaches point B, what is the direction (in terms of angle with horizontal) of acceleration of the block?

0%
$tan^{-1}(\frac{1}{2})$
0%
$tan^{-1}(2)$
0%
$sin^{-1}(\frac{2}{3})$
0%
The block never reaches point B.

The position

$x$ of a particle moving along

$x-$ axis at time

$(t)$ is given by the equation

$t=\sqrt x+2$ , where

$x$ is in metres and

$t$ in seconds. Find the work done by the force in first four seconds.

0%
$Zero$
0%
$2\ J$
0%
$4\ J$
0%
$8\ J$

Explanation

$x=(t-2)^2=t^2-4t+4$

$v=2t-4$

$a=2$

F=m.a.x

m=1kg

$2\times (4^2-4\times 4+4)=8J$

The graph below represents the relation between displacement

$x$ and force

$F$ . The work done in displacing an object from

$x=8\ m$ to

$x=16\ m$ is approximately.

0%
$25\ J$
0%
$40\ J$
0%
$8\ J$
0%
$16\ J$

What is the amount of work done in raising a glass of water weighing

$0.5 kg$ through a height of

$50 cm ? (g = 10 m/s ^2$ )

0%
$1 J$
0%
$0.4 J$
0%
$0.20 J$
0%
$2.5 J$

Explanation

The amount of the work done in raising the glass of the water through a height is calculated by the amount of potential energy

$W = mgh$

By substituting the value in the above equation we get

$W = 0.5 \times 10 \times 0.5$

Hence the work done in raising the glass is ${\rm{2}}{\rm{.5J}}$

Two steel spheres approach each other head on with the same speed and collide elastically. After the collision one of the sphere's of radius r comes to rest, the radius of the other sphere is

0%
$\frac{r}{{{{\left( 3 \right)}^{\frac{1}{3}}}}}$
0%
$\frac{r}{3}$
0%
$\frac{r}{9}$
0%
${\left( 3 \right)^{\frac{1}{2}}}r$

An object of mass

$M_1$ moving horizontally with speed u collides elastically with another object of mass

$M_2$ at rest. Select correct statement.

0%
The momentum of system is conserved only in direction PQ.
0%
Momentum of $M_1$ is conserved in direction perpendicular to SR.
0%
Momentum of $M_2$ is conserved in direction perpendicular to CR.
0%
All of these

A uniform elastic string has length

$a_{1}$ when tension is

$T_{1}$ and length

$a_{2}$ when tension is

$T_{2}$ . The amount of work done in stretching it from its natural length to a length

$(a_{1}+ a_{2})$ is ?

0%
$\dfrac {1}{2}\dfrac {(a_{1}T_{1}+a_{2}T_{2})^{2}}{2(T_{1}-T_{2})(a_{1}- a_{2})}$
0%
$\dfrac {1}{2}\dfrac {(a_{1}T_{1}-a_{2}T_{2})^{2}}{2(T_{1}-T_{2})(a_{1}- a_{2})}$
0%
$\dfrac {1}{2}\dfrac {(T_{1}+T_{2})^{2}}{2(T_{1}-T_{2})(a_{1}- a_{2})}$
0%
$\dfrac {1}{2}\dfrac {(T_{1}+ T_{2})}{2(T_{1}-T_{2})(a_{1}- a_{2})}$

If increase in linear momentum of a body is 50%, then change in its kinetic energy is

0%
25%
0%
125%
0%
150%
0%
50%

A stone of mass

$1kg$ is tied to one end of a string of length

$0.5\ m$ . It is whirled in a veretical circular. If the maximum tension in the string is

$58.8N$ , the velocity at the top is

0%
$1.82\ ms^{-1}$
0%
$2.2\ ms^{-1}$
0%
$3.26\ ms^{-1}$
0%
$2.87\ ms^{-1}$

A ball of radius r moving with a speed v collides elastically with another identical stationary ball. The impact parameter for the collision is b as shown in figure.

0%
The balls must scatter at right angles for $0 < b \le 2r.$
0%
For a head on collision b must be zero, and for an oblique collision $$0
0%
After collisions, ball-1 will comes to rest and ball-2 move at an angle $\sin^{-1} (b/r)$ below the x-axis
0%
After collision ball-1 and 2 will move at angles $cos^{-1} (b/r)$ and $$\sin^{-1} (b/2r) above and below the x-axis respectively.

A mass

$m$ is revolving in a vertical circle at the end of a string of length

$20\ cm$ . By how much times does the tension of the string at the lowest point exceed the tension at the topmost point-

0%
$2\ mg$
0%
$4\ mg$
0%
$6\ mg$
0%
$8\ mg$

Explanation

Let tension

$T_1$ at top most point is given by

$T_1 + mg = \dfrac{mV^2_1}{l}$ ...(1)

similarly, the tension

$T_2$ at the lowest point is given by

$T_2 -mg = \dfrac{mV_2^2}{l}$ ...(1)

here, tension acting in words while balancing centrifugal force and weight of the body acting outwards

$T_2 - T_1 = \dfrac{mv^2_2}{l} +mg - \dfrac{mv_1^2}{l} +mg$

$\dfrac{m}{l} (v^2_2 -V^2_1) + 2mg$

we know from kilometres

$v^2_1 = v^2_2 - 2g(2l)$

$T_2 T_1 = \dfrac{m}{l} [2g(2l)] +2mg$

$= 6mg$

Hence (C) is correct answer

In the figure shown initially spring is in uninstructed state & blocks are at rest. Now

$100N$ force is appiled on block

$A$ and

$B$ as shown in figure. After some time velocity of

$'A'$ becomes

$2m/s$ and that of

$'B'\ 4m/s$ and block

$A$ displaced by amount

$10\ cm$ and spring is streched by amount

$30\ cm$ . then find the work done by sprig force on

$A$ .

0%
$9/3\ J$
0%
$-6\ J$
0%
$6\ J$
0%
$-2 J$

A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 s, what is the magnitude & direction of acceleration of the stone:

0%
$\pi^2 \ m/s^2$ and direction along the tangent to the circle.
0%
$\pi^2 \ m/s^2$ and direction along the radius towards the centre.
0%
$\frac{\pi^2}{4} \ m/s^2$ and direction along the radius towards the centre.
0%
$\pi^2 \ m/s^2$ and direction along the radius away from the centre.

As shown in the given figure the ball is given sufficient velocity at the lowest point to complete the circle. Length of string is

$1m$ . Find the tension in the string, when it is at

$60^{\circ}$ with vertical position.
(Mass of ball

$= 5\ kg$ ).

0%
$160\ N$
0%
$180\ N$
0%
$200\ N$
0%
$225\ N$

One end of a light spring of force constant

$k$ is fixed to the ceiling. The other end is fixed to a block of mass

$m$ . Initially the spring is relaxed. The work done by an external agent to lower the hanging body of mass

$m$ slowly till it comes to equilibrium is:

0%
$3\ \dfrac{m^{2}g^{2}}{2\ k}$
0%
$\dfrac{m^{2}g^{2}}{2\ k}$
0%
$-3 \dfrac{m^{2}g^{2}}{2\ k}$
0%
$-\dfrac{m^{2}g^{2}}{2\ k}$

A stone of mass 1 kg is tied to the end of a string of

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

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

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

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

Mass

$2m$ is kept on a smooth circular track of mass

$m$ , which is kept on a smooth horizontal surface. The circular track is given a horizontal velocity

$\sqrt{2gR}$ towards left and released. The maximum height reached by

$2m$ will be ?

0%
$\dfrac {R}{4}$
0%
$\dfrac {R}{2}$
0%
$\dfrac {R}{3}$
0%
$\dfrac {2R}{3}$

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 stone is tied to a rope is rotated in a vertical circle with unifrom speed. if the difference between maximum and minimum tension in the rope is

$20N$ , mass of the stone in

$Kg$ is

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

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

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%
$75J$

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 12 - 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 12 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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