MCQ Questions for Class 11 Engineering Physics Laws Of Motion Quiz 4

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.

0%
$0$
0%
$\tan ^{-1}\displaystyle \frac{v^{2}}{Rg}$
0%
$\tan ^{-1}\displaystyle \frac{Rg}{v^{2}}$
0%
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 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 (

$\theta \le \phi$ )

0%
$T = mg \cos \theta$ , for all values of $\theta$
0%
$T = mg \cos \theta$ , only for $\theta = \phi$
0%
$T = mg$ , for $\displaystyle \theta =\cos^{-1} \left[\frac{1}{3}(2cos\phi +1)\right]$
0%
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 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

0%
$\sqrt{u^2 - 2gL}$
0%
$\sqrt{2gL}$
0%
$\sqrt{u^2 - gL}$
0%
$\sqrt{2(u^2 - gL)}$

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 circular track of radius

$100$ m is designed for an average speed

$54$ km/h. Find the angle of banking.

0%
$\tan ^{ -1 }{ \left( \dfrac { 3 }{ 20 } \right) }$
0%
$\tan ^{ -1 }{ \left( \dfrac { 9 }{ 40 } \right) }$
0%
$\tan ^{ -1 }{ \left( \dfrac { 3 }{ 10 } \right) }$
0%
None of these

Explanation

$\tan { \theta } = \dfrac { { v }^{ 2 } }{ rg }$

$= \dfrac { 15\times 15 }{ 100\times 10 } =\dfrac { 9 }{ 40 }$

$\theta =\tan ^{ -1 }{ (\dfrac { 9 }{ 40 } ) }$

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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$mg$
0%
$3 mg$
0%
$\sqrt {10} mg$
0%
$4 mg$

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 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

0%
${ T }_{ A } = { T }_{ B }$
0%
${ T }_{ X } - { T }_{ Y } =6mg$
0%
${ T }_{ Y } - { T }_{ X } =6mg$
0%
${ T }_{ Y } > { T }_{ X } \neq 6mg$

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

0%
$mg$ and $T$ only.
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$mg$ , $T$ and an additional force of $mv^2/r$ directed inwards.
0%
$mg$ , $T$ and an additional force of $mv^2/r$ directed outwards.
0%
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.

Which of the following must be true for the sum of the magnitude of the momenta of the individual particles in the system?

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It must be zero
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It could be non-zero, but it must be a constant
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It could be non zero, and it might not be a constant
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It could be zero, even if the magnitude of the total momentum is not zero

Let

$\theta$ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is m, then tension in the string is mg

$\cos\theta$

0%
always
0%
never
0%
at the extreme position
0%
at the mean position

Explanation

For simple pendulum

$\dfrac { m{ v }^{ 2 } }{ r } =T-mg\cos { \phi }$

But when

$\phi =\theta$ , i.e., the bob is at extreme position. its velocity is zero, hence the equation becomes:

$T-mg\cos { \theta }=0$

$\Rightarrow \quad T = mg\cos { \theta }$

Water in a bucket is whirled in a vertical circle with a string to it. The water does not fallen even when the bucket is inverted at the top of its path . We conclude that in this position:

0%
$\displaystyle mg = \frac{mv^2}{r}$
0%
$mg$ is greater than $\displaystyle \frac{mv^2}{r}$
0%
$mg$ is not greater than $\displaystyle \frac{mv^2}{r}$
0%
$mg$ is not less than $\displaystyle \frac{mv^2}{r}$

Explanation

Its because a centrifugal force

$\dfrac { m{ v }^{ 2 } }{ r }$ acts on the water which is away from the center and opposite to force of gravity

$mg$ . If the

$mg$ is greater then

$\dfrac { m{ v }^{ 2 } }{ r }$ water will spill out. but if

$\dfrac { m{ v }^{ 2 } }{ r }$ is greater then

$mg$ water will not spill out of the bucket.

Note: for keeping the water in bucket while rotating it in a vertical circle you will have to keep the velocity above a certain value to make

$\dfrac { m{ v }^{ 2 } }{ r }$ greater then

$mg$ .

A balloon has

$5g$ of air. A small hole is pierced into it. The air escapes at a uniform rate with a velocity of

$4cm/s$ . If the balloon shrinks completely in

$2.5$ seconds, then the mean force acting on the balloon is:

0%
$2dyne$
0%
$50dyne$
0%
$8dyne$
0%
$8N$

Explanation

$Mean force = \dfrac {change\ in\ momentum}{time}$

Units are given in CGS unit, Hence force will be in

$dyne$

hence

$Mean\ force = \dfrac{5\times 4}{2.5} = 8\ dyne$

The magnitude of force (in

$N$ ) acting on a body varies with time

$t$ (in

$\mu s)$ as shown.

$AB, BC$ and

$CD$ are straight line segments. The magnitude of total impulse of force on the body from

$t=4\mu s$ to

$t=16\mu s$ is :

0%
${ 10 }^{ -3 }Ns$
0%
${ 10 }^{ -2 }Ns$
0%
${ 10 }^{ -4 }Ns$
0%
$5\times { 10 }^{ -4 }Ns$

Explanation

The magnitude of impulse is area under the curve of

$F-t$ diagram

$I=$ area from

$4$ to

$6 \mu s +$ area from

$6$ to

$16 \mu s$

$I= \dfrac{1}{2}(200+800) \times (6-4)+\dfrac{1}{2} \times 800 \times (16-6)$

$I=1000+4000 N\mu s$

$I=5 \times 10^{-3} Ns$

A particle P of mass m attached to a vertical axis by two strings AP and BP of length 1m each. The separation

$AB = l$ . P rotates around the axis with an angular velocity

$\omega$ . The tension in the two strings are

$T_1$ and

$T_2$ .

0%
$T_1 = T_2$
0%
$T_1 + T_2 = m\omega^2l$
0%
$T_1 - T_2 = 2mg$
0%
BP will remain taut only if $\omega \ge \sqrt{2g/l}$

Explanation

As the triangle

$ABP$ is equilateral. all the angles are of

$60^o$ .

Balancing the forces in vertical direction,

$T_1sin30^o-T_2sin30^o=mg\Rightarrow T_1-T_2=2mg\ -(1)$

The radius of rotation of circle,

$r=lcos30^o$

Balancing forces in horizontal direction,

$T_1cos30^o+T_2cos30^o=m\omega^2r\Rightarrow T_1+T_2=m\omega^2l\ -(2)$

For BP to be taut,

$T_2$ should be greater than 0.

From

$(1)-(2)$ ,

$2T_2=m\omega^2l-2mg>0\Rightarrow \omega \geq \sqrt{2g/l}$

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

0%
$\sqrt{u^2-2gl}$
0%
$\sqrt{2gl}$
0%
$\sqrt{u^2-gl}$
0%
$\sqrt{2(u^2-gl)}$

A bead

$A$ can slide freely along a smooth rod bent in the form of a half circle of Radius

$R$ . The system is set in rotation with a constant angular velocity

$\omega$ about a vertical axis

$OO'$ . Find the angle

$\theta$ corresponding to steady position of the bead.

0%
$\cos ^{ -1 }{ \left( \cfrac { R{ \omega }^{ 2 } }{ g } \right) }$
0%
$\cos ^{ -1 }{ \left( \cfrac { g }{ R{ \omega }^{ 2 } } \right) }$
0%
$\sin ^{ -1 }{ \left( \cfrac { g }{ R{ \omega }^{ 2 } } \right) }$
0%
$\sin ^{ -1 }{ \left( \cfrac { R{ \omega }^{ 2 } }{ g } \right) }$

Explanation

The bead is in steady position if the net force along the tangent is zero. i,e,

$mg\sin\theta-mw^2r\cos\theta=0$
or

$g\sin\theta=w^2(R\sin\theta)\cos\theta$ where

$r=R\sin\theta$
or

$\cos\theta=\frac{g}{w^2R}$
or

$\theta=\cos^{-1}(\frac{g}{w^2R})$

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

$T$ .

0%
$T$ will decrease, $\omega$ will decrease
0%
$T$ will decrease, $\omega$ will remain constant
0%
$T$ and $\omega$ will remain unchanged
0%
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 stone of mass

$1000g$ tied to a light string of length

$10/3m$ is whirling in a vertical circle. If the ratio of the maximum tension to minimum tension is

$4$ and

$g=10{ ms }^{ -2 }$ , then the speed of stone at the highest point of circle is :

0%
$20{ ms }^{ -1 }$
0%
$10\sqrt { 3 } { ms }^{ -1 }$
0%
$5\sqrt { 3 } { ms }^{ -1 }$
0%
$10{ ms }^{ -1 }$

Explanation

Minimum tension is at topmost point of the circle
Maximum tension is bottom point of the circle.

$\dfrac{Max. Tension}{Min. Tension} = \dfrac { \dfrac { m{ { v }_{ b }^{ 2 } } }{ r } +mg }{ \dfrac { m{ v }_{ t }^{ 2 } }{ r } -mg } =4$
and
Conservation of energy is :

$\dfrac { { v }_{ b }^{ 2 }-{ v }_{ t }^{ 2 } }{ 2 } =2gr$
Using both equations,

${ v }_{ t }^{ 2 }=3gr$
So

${v}_{t} = \sqrt { 3gr } = 10\ m/s$

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 :

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

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 toy car is tied to the end of an unstretched string of a length,

$a$ . When revolved, the toy car moves in a horizontal circle of radius

$2a$ with time period,

$T$ . If it is now revolved in a horizontal circle of radius

$3a$ with a period

$T'$ with the same force, then

0%
$\displaystyle T' = \dfrac{\sqrt{3}}{2} T$
0%
$\displaystyle T' = \sqrt {\dfrac{3}{2}} T$
0%
$T' = T$
0%
$\displaystyle T' = \dfrac{3}{2} T$

Explanation

$\displaystyle m(2a)\left(\dfrac{2\pi}{T}\right)^2 = m(3a) \left(\dfrac{2\pi}{T'}\right)^2$ .

So,

$\dfrac{T'}{T}=\sqrt{\dfrac{3}{2}}$

A body of mass

$1\ kg$ is rotating in a vertical circle of radius

$1\ m$ . What will be the difference in its kinetic energy at the top and bottom of the circle?

Take

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

0%
$50\ J$
0%
$30\ J$
0%
$20\ J$
0%
$10\ J$

Explanation

According to work energy theorem,

$\Delta K.E.=W$ and here work is done by the gravitational force.

$\Rightarrow \Delta K.E.=W = mg\ (2r) = 1 \times 10 \times 2(1)=20\ J$

The velocity of a particle at highest point of the vertical circle is

$\sqrt{3rg}$ . The tension at the lowest point, if mass of the particle is

$m$ , is

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

Explanation

Let the velocity at lowest point be

$v$ . then according to law of conservation of energy.
energy at lowest point= energy at highest point

$\Rightarrow \dfrac{1}{2}mv^2=mg(2r)+\dfrac{1}{2}m(\sqrt{3rg})^2 \\ \Rightarrow v=\sqrt{7rg}$

At lowest point,

$\dfrac { m{ v }^{ 2 } }{ r } =T-mg \\ \Rightarrow T=mg+ \dfrac { m{ v }^{ 2 } }{ r }=mg+ \dfrac { m (\sqrt{7rg} )^{ 2 } }{ r }$

$\Rightarrow T=mg+7mg=8mg$

A truck has a width of

$2.8\ m$ . It is moving on a circular road of radius

$48.6\ m$ . The height of centre of mass is

$1.2\ m$ . The maximum speed so that it does not over turn will be nearly

0%
$24.3\ m/s$
0%
$27.2\ m/s$
0%
$13.4\ m/s$
0%
$15.7\ m/s$

Explanation

$\displaystyle \frac{mgl}{2} =\dfrac{mv^2}{R} h$

(overturning torque

$=$ restoring torque)

$\displaystyle v=\sqrt{\dfrac{gRl}{2h}}=\sqrt{\dfrac{10\times 50\times 1.4}{1.2}}$

$=24.3\ m/s$

At a curved path of the road, the road bed is raised a little on the side away from the centre of the curved path. The slope of the road bed is given by

0%
$\tan{\theta} = rg/v^2$
0%
$\tan{\theta} = r/gv^2$
0%
$\tan{\theta} = v^2/rg$
0%
$\tan{\theta} = v^2g/r$

Explanation

As shown in the given diagram, the road bed has been raised by an angle

$\theta$ .
Applying Newton's laws in horizontal and vertical directions. we get

$N\sin { \theta } =\dfrac { m{ v }^{ 2 } }{ r }$ ...

$(I)$ and

$N\cos { \theta } =mg$ ...

$(II)$
by dividing

$(I)$ by

$(II)$ , we get

$\tan { \theta } =\dfrac { { v }^{ 2 } }{ rg }$

A stone of mass

$1\ kg$ is tied of the end of a string

$1\ m$ long. It is whirled in a vertical circle. If the velocity of stone at the top is

$4\ m/s$ . What is the tension in the string at the lowest point?

Take

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

0%
$6\ N$
0%
$66\ N$
0%
$5.2\ N$
0%
$76\ N$

Explanation

Let the velocity at lowest point is

$v$ and at highest point be

$u$ , then according to law of conservation of energy.
energy at lowest point= energy at highest point.

$\Rightarrow\dfrac{1}{2}mv^2=mg(2r)+\dfrac{1}{2}mu^2 \\ \Rightarrow v=\sqrt{u^2+4rg}=\sqrt{4^2+(4\times 1 \times 10)}=\sqrt{56}$
At lowest point,

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

$\Rightarrow T=mg+ \dfrac { m{ v }^{ 2 } }{ r }=mg+ \dfrac { m (\sqrt{56} )^{ 2 } }{ r }$

$\quad$

$(r=1)$

$\Rightarrow T=mg+56\ m=66 \times m =66 \times 1 = 66\ N$ .

For traffic moving at

$60 km/hr$ along a smooth circular track of radius

$0.1 km$ , the correct angle of banking should be:

0%
$\displaystyle tan^{-1} \left [ \frac{(100 \times 9.8)}{\left ( \frac{50}{3} \right )^2} \right ]$
0%
$\displaystyle tan^{-1} \left [ \frac{\left ( \frac{50}{3} \right )^2}{(100 \times 9.8)} \right ]$
0%
$tan^{-1} \displaystyle \left ( \frac{60^2}{0.1} \right )$
0%
$tan^{-1} \sqrt{(60 \times 0.1 \times 9.8)}$

Explanation

Relation between banking angle, speed and radius of circular track is given by

$\theta =\tan ^{ -1 }{ \left[ \dfrac { v^{ 2 } }{ rg } \right] }$
here

$v=60 \mathrm { km/hr}=60 \times \dfrac{5}{18} \mathrm{ m/s}=\dfrac{50}{3} \mathrm{ m/s}$
and

$r=0.1 \mathrm{km}=100\mathrm{m}$

$\Rightarrow \theta =\tan ^{ -1 }{ \left[ \dfrac { \left( \dfrac { 50 }{ 3 } \right) ^{ 2 } }{ (100\times 9.8) } \right] }$

If a particle of mass m moving with velocity

$v_1$ is subject to an impulse

$I$ which produces a final velocity

$v_2$ , then

$I$ is given by :

0%
$m(v_1 - v_2)$
0%
$m(v_1 + v_2)$
0%
$m(v_2 - v_1)$
0%
$\dfrac{m\left(v_2^2 - v_1^2\right)}{2}$

Explanation

Impulse is equal to the change in momentum caused by the force.

$I=m\times v_{2} - m\times v_{1}$

A small sphere is attached to a cord and rotates in a vertical circle about a point

$O$ . If the average speed of the sphere is increased, the cord is most likely to break at the orientation when the mass is at

0%
Bottom point $B$
0%
Top point $A$
0%
Point $D$
0%
Point $C$

Explanation

Tension in the cord is maximum (for a given average speed of rotation) when the mass,

$m$ , is at the bottom points B, as the gravitational force is in the downward direction and tension of the cord is directly opposing it.

A cricket ball of mass 500 gm strikes a bat normally with a velocity 30 m/s and rebounds with a velocity 20 m/s in the opposite direction. The impulse of the force exerted by the ball on the bat is:

0%
0.5 N.s
0%
1.0 N.s
0%
25 N.s
0%
50 N.s

Explanation

Impulse is given by:

$I=m\Delta v=0.5\times(20-(-30))=0.5\times50$

$I=25 Ns$

A body

$X$ of mass

$5 kg$ is moving with velocity

$20 m s^{-1}$ while another body Y of mass

$20 kg$ is moving with velocity

$5 m s^{-1}$ . Compare the momentum of the two bodies.

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

Explanation

For body X,

mass:

$m_X=5\ kg$

velocity:

$v_X= 20\ m/s$

$\therefore$ momentum:

$p_X=m_X \times v_X=5 \times 20 = 100\ kg\ m/s$

For body Y,

mass:

$m_Y=20\ kg$

velocity:

$v_Y= 5\ m/s$

$\therefore$ momentum:

$p_Y=m_Y \times v_Y = 20 \times 5 = 100\ kg\ m/s$

So,

$p_X:p_Y=1:1$

Find the maximum speed at which a truck can safely travel without toppling over, on a curve of radius

$250m$ . The height of the centre of gravity of the truck above the ground is

$1.5 m$ and the distance between the wheels is

$1.5m$ , the truck being horizontal.

0%
$\displaystyle 15\:ms^{-1}$
0%
$\displaystyle 25\:ms^{-1}$
0%
$\displaystyle 30\:ms^{-1}$
0%
$\displaystyle 35\:ms^{-1}$

Explanation

The Truck will topple about the outer wheels. let the height of the center of gravity be h, distance between the tyres be d, and radius of curvature be R, for toppling the maximum speed is given by,(equating the torque by gravity and torque by centrigal force about the outer wheel)

$\dfrac{m{ { v }_{ max } }^{ 2 } }{ R } h=mg\dfrac { d }{ 2 }$

$\Rightarrow { v }_{ max }=\sqrt { \dfrac { Rgd }{ 2h } } =35.35m/s$

A car of mass

$600 kg$ is moving with a speed of

$10 m s^{-1}$ while a scooter of mass

$80 kg$ is moving with a speed of

$50 m s^{-1}$ . Compare their momentum.

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

Explanation

As we know that momentum is nothing but product of mass and velocity. Let us substitute the given values.

Given that mass of the car as 600 kg and velocity as 10 m/sec.

Hence momentum of the car is

$P=600\times 10=6000 kg m/sec$ .

We have mass of the scooter as 80 kg where as velocity is 50 m/sec.

Hence momentum is

$P=80\times 50=4000 kg m/sec$ .

Therefore, momentum are in the ratio

$3:2$ .

A simple pendulum is vibrating with angular amplitude of

$\theta=90^{o}$ as shown in figure.
For what value of

$\theta$ is the acceleration directed

$(i)$ Vertically upwards

$(ii)$ Horizontally

$(iii)$ Vertically downwards

0%
$0^{o},90^{o} \ cos^{-1}\displaystyle\frac{1}{\sqrt {3}},\ 90^{o}$
0%
$\cos^{-1}\displaystyle\frac{1}{\sqrt {3}},\ ,\ 0^{o},90^{o}$
0%
$0^{o},90^{o}, \cos^{-1}\displaystyle\frac{1}{\sqrt{2}},$
0%
$\cos^{-1}\displaystyle\frac{1}{\sqrt {3}},\ 90^{o},\ 0^{o}$

Explanation

Acceleration of a simple pendulum is

$a = -ω^{2}\times A\times cos (ωt + θ)$ , where ω is angular velocity,

$A$ is amplitude,

$t$ is time and θ is angle.

If θ is

$0$ , then

$a = -ω^{2}\times A\times cos (ωt)$ so direction is vertically upwards, when θ is

$cos^{-1} (1/3^{1/2})$ then direction is horizontally, when θ is

$90^o$ , then

$a = ω^{2}\times A\times sin (ωt)$ so direction is vertically downwards.

A pendulum bob is raised to a height,

$h$ , and released from rest. At what height will it attain half of its maximum speed?

0%
$\displaystyle\frac{3h}{4}$
0%
$\displaystyle\frac{h}{2}$
0%
$\displaystyle\frac{h}{4}$
0%
$0.707h$

Explanation

By energy conservation,

$\dfrac{1}{2}mv^2_{max} = mgh$
Let at height

$h'$ , it attain velocity

$v'=\dfrac{v_{max}}{2} = \sqrt{gh/2}$
Now,

$mgh = \dfrac{1}{2}mv'^2 + mgh'$

$mgh = \dfrac{mgh}{4} + mgh'$

$mgh'=\dfrac{3}{4}mgh$

$\therefore h' = \dfrac{3h}{4}$

A simple pendulum of length

$l$ has maximum angular displacement

$\displaystyle \theta .$ Then maximum kinetic energy of a bob of mass

$m$ is

0%
$\displaystyle \frac{1}{2}mgl$
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$\displaystyle \frac{1}{2}mgl\cos \theta$
0%
$\displaystyle mgl\left ( 1-\cos \theta \right )$
0%
$\displaystyle \frac{1}{2}mgl\sin \theta$

Explanation

At point B,

$v=0$ and

$P.E$ is maximum.

At point A,

$P.E= 0$ and

$K.E$ is maximum.

Height of bob at B,

$AP= l-lcos\theta= l(1-cos\theta)$

Thus

$P.E= mgl(1-cos\theta)$

Applying conservation of energy at A and B,

$K.E_m +0 = 0 + P.E$

$\implies K.E_m=mgl (1-cos\theta)$

The kinetic energy of particle at the lower most position is

0%
$\displaystyle \frac{4mgL}{3}$
0%
$2mgL$
0%
$\displaystyle \frac{8mgL}{3}$
0%
$\displaystyle \frac{2mgL}{3}$

Explanation

$v= 2\sqrt{\dfrac{gL}{3}}$

Also,

$u= 2v \implies u= 4\sqrt{\dfrac{gL}{3}}$

Thus

$K.E_ A= \dfrac{1}{2}mu^2= \dfrac{8mgL}{3}$

With what minimum speed

$v$ must a small ball should be pushed inside a smooth vertical tube from a height

$h$ so that it may reach the top of the tube? Radius of the tube is

$R$ .

0%
$\displaystyle \sqrt{g(R-h)}$
0%
$\displaystyle \sqrt{g(2R-h)}$
0%
$\displaystyle \sqrt{2g(2R-h)}$
0%
$\displaystyle \sqrt{2g}$

Explanation

Hint: If there is no external force acting on a body then Mechanical Energy of the system remains conserved.

That is,

$ME_i = ME_f$

or, $KE_i + PE_i = KE_f + PE_f$

Step 1: Calculate Initial Mechanical Energy of the ball when it is at height $h$ .

Let mass of ball is

$m$ When the ball is at height

$h$ , considering Potential Energy at bottom most point of tube to be zero, Potential Energy of ball is given as,

$PE_i = mgh$

Also when the ball is just pushed let initial velocity of ball is

$v$ , then Initial Kinetic Energy of ball is given by,

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

Step 2: Calculate Final Mechanical Energy of the ball when it is at height $h$ .

When the ball reaches the uppermost point that is at height

$(2R + 2d)$ from surface, the Potential Energy of ball is given by,

$PE_f = mg(2R+2d)$

We want the minimum value of velocity with which ball should be pushed such that it reaches the uppermost point. So let us consider that ball stops at the uppermost point. That is,

$KE_f = 0$

Step 3: Apply Conservation of Mechanical Energy.

Since the surfaces are smooth and no external forces are involved, we can apply conservation of energy. Thus, we get

$KE_i + PE_i = KE_f + PE_f$

$\Rightarrow \dfrac{1}{2}mv^{2} + mgh = 0 + mg(2R+2d)$

$\Rightarrow \dfrac{1}{2}v^{2} + gh = g(2R+2d)$

$\Rightarrow \dfrac{1}{2}v^{2} = g(2R+2d - h)$

It is given that

$d<<R$ , thus we can neglect

$d$ from Right Hand Side.

Therefore,

$\dfrac{1}{2}v^{2} = g(2R- h)$

$\Rightarrow v^{2} = 2g(2R- h)$

$\Rightarrow v = \sqrt{2g(2R- h)}$

This is the required expression.

Thus, minimum velocity with which ball should be pushed is, $v = \sqrt{2g(2R- h)}$ .

Option C is correct.

A bob is suspended from a crane by a cable of length

$5m$ . The crane and load are moving at a constant speed

$v_{0}$ . The crane is stopped by a bumper and the bob on the cable swings out an angle of

$60^{\circ}$ . Find the initial speed

$v_{0}.(g=9.8 m/s^{2})$

0%
$2 ms^{-1}$
0%
$3 ms^{-1}$
0%
$5 ms^{-1}$
0%
$7 ms^{-1}$

Explanation

By using conservation of energy, from the initial point to the highest point

$\frac { 1 }{ 2 } m{ { v }_{ 0 } }^{ 2 }=mgl(1-cos{ 60 }^{ 0 })$

$\Rightarrow { v }_{ 0 }=\sqrt { 2gl(1-cos{ 60 }^{ 0 }) } =\sqrt { gl } =\sqrt { 50 } =7m/s$

A car is travelling along a circular curve that has a radius of

$50 m$ . If its speed is

$16 m/s$ and is increasing uniformly at

$8 m/s$

$^{2}$ . Determine the magnitude of its acceleration at this instant.

0%
5.5 m/s $^{2}$
0%
7.5 m/s $^{2}$
0%
9.5 m/s $^{2}$
0%
12.5 m/s $^{2}$

Explanation

Given:

$r =50$ m

$v = 16$ m/s

$a_t =8 m/s^2$

Radial acceleration of the car at that instant,

$a_r = \dfrac{v^2}{r} = \dfrac{16^2}{50} = 5.12$

$m/s^2$

$\therefore$ Net acceleration of the car

$a = \sqrt{a_r^2 + a_t^2} = \sqrt{(5.12)^2 + 8^2} = 9.5 m/s^2$

A stone tied to a string of length

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

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

0%
$\displaystyle \sqrt{(u^{2}-gL)}$
0%
$\displaystyle \sqrt{2(u^{2}-gL)}$
0%
$\displaystyle \sqrt{(u^{2}-2gL)}$
0%
$\displaystyle \sqrt{2(u^{2}-2gL)}$

Explanation

By using conservation of energy,

$\frac { 1 }{ 2 } m{ u }^{ 2 }=mgL+\frac { 1 }{ 2 } m{ v }^{ 2 }$

$\Rightarrow v=\sqrt { { u }^{ 2 }-2gL }$

Change in velocity =

$\sqrt { { u }^{ 2 }-2gL } \hat { i } -u\hat { j }$

Magnitude of change in velocity =

$\sqrt { { u }^{ 2 }-2gL+{ u }^{ 2 } } =\sqrt { 2\left( { u }^{ 2 }-gL \right) }$

Velocity of particle when it is moving vertically downward is

0%
$\displaystyle \sqrt{\frac{10gL}{3}}$
0%
$\displaystyle 2\sqrt{\frac{gL}{3}}$
0%
$\displaystyle \sqrt{\frac{8gL}{3}}$
0%
$\displaystyle \sqrt{\frac{13gL}{3}}$

Explanation

Work-energy theorem for A to C,

$-mgL= \dfrac{1}{2}m(v')^2- \dfrac{1}{2}mu^2$ where

$u= 4\sqrt{\dfrac{gL}{3}}$

$-mgL= \dfrac{1}{2}m(v')^2- \dfrac{8mgL}{3}$

$\implies v'= \sqrt{\dfrac{10 gL}{3}}$

Minimum velocity of the particle is

0%
$\displaystyle 4\sqrt{\frac{gL}{3}}$
0%
$\displaystyle 2\sqrt{\frac{gL}{3}}$
0%
$\displaystyle \sqrt{\frac{gL}{3}}$
0%
$\displaystyle 3\sqrt{\frac{gL}{3}}$

Explanation

The maximum and minimum velocities will be at A and B respectively.

Work-energy theorem,

$-mg(2L)= \dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

Given:

$u:v= 2:1$

$-mg(2L)= \dfrac{1}{2}mv^2-\dfrac{1}{2}m(2v)^2$

$\implies v= 2\sqrt{\dfrac{gL}{3}}$

if AB is a massless string.

0%
$\displaystyle \frac{10L}{27}$
0%
$\displaystyle \frac{20L}{27}$
0%
$\displaystyle \frac{30L}{27}$
0%
$\displaystyle \frac{40L}{27}$

Explanation

When the string has traversed an angle $\theta$ , then

$T-mgcos\theta =\dfrac { m{ v }^{ 2 } }{ l }$ , and conservation of energy gives

$\dfrac { m{ u }^{ 2 } }{ 2 } =\dfrac { m{ v }^{ 2 } }{ 2 } +mgl(1-cos\theta ).$ then the string becomes slacked tension in the string becomes zero. Solving the above two equations to get the value of $\theta$

$\Rightarrow cos\theta =\dfrac { -1 }{ 3 } and { v }^{ 2 }={ u }^{ 2 }-2gl(1-cos\theta )=\dfrac { gl }{ 3 }$ from that point the maximum height reached is $\dfrac { { v }^{ 2 }{ sin }^{ 2 }\left( \pi -\theta \right) }{ 2g }$ maximum height $= l(1-cos\theta )+\dfrac { { v }^{ 2 }{ sin }^{ 2 }\left( \pi -\theta \right) }{ 2g } =\dfrac { 4l }{ 3 } +\dfrac { 4l }{ 27 } =\dfrac { 40l }{ 27 }$

A mass is performing vertical circular motion (see figure). If the average velocity of the particle is increased, then at which point is the maximum breaking possibility of the string:

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

Explanation

Tension at any point in vertical motion is given by :

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

where

$\theta=$ angular displacement from lowest point ,

$l=$ length of string

$m=$ mass of string

It is clear that tension at the lowest point (

$B$ ) is greatest than at other points (

$A,C,D$ ). If we increase average velocity, tension will increase at lowest point, therefore at point B, string has maximum possibility of break.

An instrument box placed on a table is given a push. It is observed that it stops its motion after some time. What is the force that stopped its motion?

0%
Gravitational force
0%
Tension force
0%
Friction force
0%
Mechanical force

A car moves on a circular road, describing equal angles about the centre in equal intervals of times which of the statements about the velocity of car

0%
velocity is constant
0%
magnitude of velocity is constant but the direction of velocity change
0%
both magnitude and velocity change
0%
velocity is directed towards the centre of circle

Explanation

Hint: Apply the formula of angular velocity.

Explanation:

$\bullet$ The car covers equal angles in equal intervals of time, so its angular velocity is constant.

By formula,

$v=\omega r$

$\bullet$ The circular road has a constant

$r$ and according to the question,

$\omega$ is also constant.

Therefore,

$v=\text{constant}$

$\bullet$ Magnitude of the linear velocity is thus constant but since it is moving in a circular road, its direction is constantly changing.

Hence, the magnitude of velocity is constant but the direction of velocity changes.

The correct answer is option (B).

A boy is whirling a stone tied at one end such that the stone is in uniform circular motion.Which of the following statement is correct?

0%
The velocity of stone at A is equal to the velocity of stone B
0%
The speed of stone at A is equal to the speed of stone at B.
0%
The centripetal force is radially outward in the string.
0%
All of the above statements are correct.

A force of

$50 \ dynes$ is acted on a body of mass

$5 g$ which is at rest for an interval of

$3 s$ , then impulse is:

0%
$0.15\times 10^{-13}Ns$
0%
$0.98\times 10^{-3}Ns$
0%
$1.5\times 10^{-3}Ns$
0%
$2.5\times 10^{-3}Ns$

Explanation

$Impulse\: J = F\Delta t = 50 \times 10^{-5}\times 3 = 1.5 \times 10^{-3} Ns$

A turn has a radius of

$10 m$ . If a vehicle goes round it at an average speed of

$18$ km/h, what should be the proper angle of banking?

0%
$\displaystyle \tan ^{-1}\left ( 1/4 \right )$
0%
$\displaystyle \tan ^{-1}\left ( 1/2 \right )$
0%
$\displaystyle \tan ^{-1}\left ( 3/4 \right )$
0%
$\displaystyle \tan ^{-1}\left ( 5/7 \right )$

Explanation

Given :

$r =10$ m

Average speed of the vehicle

$v = 18 km/hr = 18 \times \dfrac{5}{18} = 5$

$m/s$

Let the proper banking angle be

$\theta$

Using

$tan\theta = \dfrac{v^2}{gr}$

$\therefore$

$tan\theta =\dfrac{5^2}{10 \times 10}$

$\implies \theta = tan^{-1} (1/4)$

A boy holds a pendulum in his hand while standing at the edge of a circular platform of radius

$r$ rotating at an angular speed

$\omega$ . The pendulum will hang at an angle

$\theta$ with the verticle such that

0%
$\;tan\,\theta=0$
0%
$\;tan\,\theta=\displaystyle\frac{\omega^2r^2}{g}$
0%
$\;tan\,\theta=\displaystyle\frac{r\omega^2}{g}$
0%
$\;tan\,\theta=\displaystyle\frac{g}{\omega^2r}$

Explanation

A pseudo force

$(ma_r)$ acts on the bob due to the rotation of circular platform.

In bob's frame, the bob is at rest (or equilibrium).

$\therefore$

$T sin \theta = ma_r$ where

$a_r = rw^2$

$\implies$

$T sin\theta = mrw^2$ ........(1)

Also

$T cos\theta =mg$ ..........(2)

Divide (2) by (1) we get,

$tan\theta =\dfrac{rw^2}{g}$

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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