the varying mass system may be considered

the varying mass system is not considered

the varying mass system must be considered

only varying of mass due to velocity is considered

MCQ Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 2

Two particles each of the same mass move due north and due east respectively with the same velocity

$'V'$ . The magnitude and direction of the velocity of the center of mass is:

0%
$\dfrac{V}{{\sqrt 2 }}\,NE$
0%
$\sqrt 2 V\,NE$
0%
$2V\,SW$
0%
$\dfrac{V}{2}\,SW$

Explanation

Velocity of particle 1

$=Vj$
Velocity of particle 2

$=Vi$

$\begin{array}{l}{V_{com}} = \dfrac{{{m_1}{V_1} + {m_2}{V_2}}}{{{m_1} + {m_1}}}\\ = \dfrac{{mVj + mVi}}{{m + m}} = \dfrac{V}{2}(i + j)\\mag. = \sqrt {\dfrac{{{V^2}}}{4} + \dfrac{{{V^2}}}{4}} = \dfrac{V}{{\sqrt 2 }}\\\left[ {ans.\frac{V}{{\sqrt 2 }},NE} \right]\end{array}$

A hollow sphere rolls down a

$30^o$ incline of length 6m without slipping. The speed of center of mass a the bottom of plane is

0%
$6 m/s^{-1}$
0%
$3 ms^{-1}$
0%
$6\sqrt{2} ms^{-1}$
0%
$3\sqrt{2} ms^{-1}$

Explanation

apply consecration of energy

${E_i} = mgh$

${E_i} = m10(6\sin 30^\circ )$

${E_i} = 30\,m$

${E_f} = {1 \over 2}m{v^2} + {1 \over 2}l{\omega ^2}$

${E_f} = {1 \over 2}m{R^2}{\omega ^2} + {1 \over 2}\left( {{2 \over 3}m{R^2}} \right){\omega ^2}$

${E_f} = {5 \over 6}m{R^2}{\omega ^2}$

${E_i} = {E_f}$

${R^2}{\omega ^2} = 36$

${v_{cm}} = 6$

Moment of inertia about axis 1 is :

0%
$\dfrac { { M2 }^{ 2 } }{ 12 }$
0%
$\dfrac { { Mb }^{ 2 } }{ 12 }$
0%
$\dfrac { M1^{ 2 } }{ 3 }$
0%
$\dfrac { Mb^{ 2 } }{ 3 }$

If the K.E. of a body is increased by

$300\%$ , its momentum will increase by _______.

0%
$100\%$
0%
$150\%$
0%
$\sqrt {300}\%$
0%
$175\%$

The centre of mass of a system of particles is at the origin. It follows that:

0%
the number of particles to the right of the origin is equal to the left of origin.
0%
the total mass of the particles to the right of the origin is same as total mass to the left of the origin.
0%
the number of particles on the X-axis should be equal to the number of particles on the Y-axis .
0%
if there is a particle on the +ve X-axis, there should be atleast one particle on the -ve X-axis.
0%
None of these.

The centre of mass of a system particles does not depend on :

0%
masses of the particles
0%
Internal forces on the particle
0%
Position of the particles
0%
Relative distance between the particles

SI. unit of moment of force is

0%
$N$
0%
$N\quad cm$
0%
$kgfm$
0%
$N \quad m$

Explanation

Moment of force= Force x Distance of force from axis of rotation.

so Unit of moment of force

$N.m$

Write true or false for the following statements :
The center of mass of gravity for a small body lie at the same point.

0%
True
0%
False

In classical system:

0%
the varying mass system is not considered
0%
the varying mass system must be considered
0%
the varying mass system may be considered
0%
only varying of mass due to velocity is considered

Explanation

In classical systems varying mass may be considered. One common example is the burning of fuel in a rocket due to which its mass keeps on decreasing with time. In such a situation newton's second law is modified a little.

$F= \dfrac{dP}{dt}$

$\Rightarrow F=\dfrac{d(mv)}{dt}$

$\Rightarrow F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$

$\Rightarrow F=ma+v\dfrac{dm}{dt}$

Unit of which physical quantity is same as that of unit of impulse of force?

0%
Force
0%
Acceleration
0%
Momentum
0%
Velocity

The moment of inertia of a flywheel is

$0.2 kgm^{2}$ which is initially stationary. A constant external torque

$5 Nm$ acts on the wheel. The work done by this torque during

$10 sec$ is:

0%
$1250 J$
0%
$2500 J$
0%
$5000 J$
0%
$6250 J$

Explanation

Change in momentum

$L=\int\tau.dt$ (basic property of torque)

$=5\times10=50$

$Work\: done=change\: in \:kinetic\: energy$

$=\dfrac{L^2}{2I}=\dfrac{(50)^2}{2\times 0.2}=\dfrac{25000}{4}$

$=6250 J$

A constant torque acting on a uniform circular wheel changes its angular momentum from

$L$ to

$4L$ in

$4$ seconds. The torque acting on it is:

0%
$4L$
0%
$12L$
0%
$\dfrac{3}{4} L$
0%
$\dfrac{4}{3} L$

Explanation

As we know that,

$\tau=\dfrac{\Delta{L}}{{\Delta}t}$

So,

$\tau=\dfrac{4L-L}{4}=\dfrac{3L}{4}$

A particle of mass

$m$ moves uniformly with a speed

$v$ along a circle of radius

$r$ . A and B are two points on the circle, such that the arc

$AB$ subtends an angle

$\theta$ at the center of the circle. The magnitude of change in momentum as the particle moves from

$A$ to

$B$ is given by:

0%
$2 mv sin (\dfrac{\theta}{2})$
0%
$2 mv cos (\dfrac{\theta}{2})$
0%
$zero$
0%
$2 mv tan (\dfrac{\theta}{2})$

Explanation

Assume particle is at lowest point of the circle

$P_i= mv\hat{i}$

After it rotates

$\theta$ angle

$P_f= mv\cos \theta \hat{i}+mv\sin \theta \hat{j}$

Change in momentum

$=P_f-P_1 =$

$mv\cos \theta \hat{i}+mv\sin \theta \hat{j} -mv\hat{i}$

$= mv(-1+\cos \theta) \hat{i}+mv\sin \theta \hat{j}$

$|\Delta P|= mv\sqrt{(-1+\cos \theta)^2+\sin^2 \theta \ }$

$|\Delta P|= mv\sqrt{1+\cos^2 \theta-2\cos \theta+\sin^2 \theta }$

$|\Delta P|= 2mv\sin (\dfrac{ \theta }{2})$

The kinetic energy of a moving body is given by

$K = 2v^{2}$ ,

$K$ being in joules and

$v$ in m/s. Its momentum when traveling with a velocity of

$2\ ms^{-1}$ will be :

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

Explanation

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

$\Rightarrow m=4kg$

$\vec{p}=m.\vec{v}$

$=4\times 2$

$=8\ kg m/s$

A torque of

$0.5 Nm$ is required to drive a screw into a wooden frame with the help of a screw driver. If one of the two forces of couple produced by screw driver is

$50 N$ , the width of the screw driver is:

0%
$0.5 cm$
0%
$0.75 cm$
0%
$1 cm$
0%
$1.5 cm$

Explanation

Let width be

$x$ . So the torque due to couple is

$2\times50\times x/2=50x$

$50x = 0.5\Rightarrow x=0.01m=1\ cm$

A ceiling fan is rotating about its own axis with uniform angular velocity

$\omega$ . The electric current is switched off then due to constant opposing torque its angular velocity is reduced to

$\dfrac{2\omega }{3}$ as it completes

$30$ rotations. The number of rotations further it makes before coming to rest is :

0%
$18$
0%
$12$
0%
$9$
0%
$24$

Explanation

$\omega_1^2-\omega_2^2=2\alpha \theta$ (governing equation as per question)

$\therefore \omega ^2-\left ( \dfrac{2\omega }{3} \right )^2=2\times \alpha \times 30$

$\dfrac{5\omega ^2}{9}=60\alpha$
similarly let the no of rotations further made be

$n$

$\left( \dfrac{2\omega }{3} \right )^2=2\alpha n$

$\dfrac{4\omega^2 }{9}=2n\alpha$

$\Rightarrow \dfrac{5}{4}=\dfrac{30}{n}$

$n=24$

A wheel of moment of inertia

$5\times 10^{-3}kgm^{2}$ is making

$20 \ rev/s$ . The torque required to stop it in

$10 \ s$ is :

0%
$2\pi \times 10^{-2}Nm$
0%
$2\pi \times 10^{2}Nm$
0%
$4\pi \times 10^{-2}Nm$
0%
$4\pi \times 10^{2}Nm$

Explanation

$\omega =20\times 2\pi$

$\tau = I\omega$

So,

$\tau \Delta t=5\times 10^{-3}\omega$

$\tau =\dfrac{5\times 10^{-3}\times 20\times 2\pi}{10}=2\pi\times 10^{-2} Nm$

A wheel of radius

$R$ is free to rotate about its natural axis and

$I$ is its moment of inertia. If a tangential force

$F$ is applied on the wheel along its rim, then angular acceleration of wheel is:

0%
$IFR$
0%
$\dfrac{FR}{I}$
0%
$\dfrac{IF}{R}$
0%
$\dfrac{IR}{F}$

Explanation

The torque is given using the relation

$\vec{\tau} = \vec{F} \times \vec{R}$
or

$\tau=I\alpha$
or

$\alpha=\dfrac{FR}{I}$

A flywheel making

$120$ r.p.m is acted upon by a retarding torque producing angular retardation of

$\pi \ rad/s^{2}$ . Time taken by it to come to rest is:

0%
$1$ s
0%
$2$ s
0%
$3$ s
0%
$4$ s

Explanation

$T=I \alpha$

$\omega =\dfrac{120\times 2\pi }{60}$

$\omega =4\pi$

$\omega =\omega _{0}+\alpha t$

$0=4\pi -\pi t$

$t=4 s$

A body of mass 'M' collides against a wall with a velocity 'v' and retraces its path with the same velocity, the change in momentum is

0%
$zero$
0%
$2 Mv$
0%
$Mv$
0%
$- Mv$

Explanation

$\Delta P = Final\:\:mom - Initial\:\:mom$

$= Mv - ( - Mv)$

$= 2 Mv$

$K$ is radius of gyration of a thin square plate about an axis perpendicular to the plane of plate and passing through its centre. The radius of gyration of plate about a parallel axis to the first and passing through a corner of plate is :

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

Explanation

MI about an axis perpendicular to the plane of plate and passing through its center is

$I_{cm}=\dfrac{Ml^2}{6}$ ,

Also

$MK^2=Ml^2/6\Rightarrow K^2=l^2/6$ ...(1)

Distance of corner from the center of plate

$d=\dfrac{l}{\sqrt{2}}$

Using parallel axis theorem

$I_{1}=I_{cm}+Md^2=\dfrac{2Ml^2}{3}$

Also

$MK_1^2=\dfrac{2Ml^2}{3}\Rightarrow K_1^2=2l^2/3$ ...(2)

From equation 1 and 2 we get

$K_1=2K$

The radius of a flywheel is

$10 cm$ . It is rotated at the rate of

$25 rads^{-2}$ . When a constant force

$40 N$ is applied on the rim of wheel along its tangent. The moment of inertia of wheel is:

0%
$0.01 kg m^{2}$
0%
$0.1 kg m^{2}$
0%
$0.16 kg m^{2}$
0%
$1.6 kgm^{2}$

Explanation

Torque on wheel

$={F}\times{R}$ as it is applied at tangent

$=40\times0.1$

$=4Nm$
Torque

$\tau =I\alpha$

$4=I\times25$

$\Rightarrow I=\dfrac{4}{25}=0.16kgm^2$

A circular disc of radius

$10 \ cm$ is free to rotate about an axis passing through its centre without friction and its moment of inertia is

$\dfrac{1}{2}\pi \ kg\ m^{2}$ . A tangential force of

$20 N$ is acting on the disc along its rim. Starting from rest, the number of rotations made by the disc in

$10 \ s$ is:

0%
$50$
0%
$100$
0%
$150$
0%
$200$

Explanation

Torque on the disc

$\tau ={F}\times{R}=20\times0.1=2Nm$
Given moment of inertia

$I=\dfrac{1}{2\pi }kgm^2$

$\tau =I\alpha$

$\Rightarrow \alpha =4\pi rads^{-2}$
No. of rotations made

$=\dfrac{\left (\dfrac{1}{2}\:\alpha\: t^2 \right )}{2\pi }$ (by the 1st law of uniform angular rotation)

$=\dfrac{\dfrac{1}{2}\times 4\pi \times 100 }{2\pi }$

$=100$

$I$ is moment of inertia of a thin rod about an axis perpendicular to the length of rod and passing through one end of rod. If half length of this rod from free end is cut and removed, then moment of inertia of remaining rod about same axis is :

0%
$\dfrac{I}{2}$
0%
$\dfrac{I}{4}$
0%
$\dfrac{I}{8}$
0%
$\dfrac{I}{16}$

Explanation

Moment of inertia about the end point

$\Rightarrow \dfrac{Ml^{2}}{3}=I$
But now length

$L$ is halved so mass will be reduced to

$\dfrac{M}{2}$
So, new

$I$

$\Rightarrow \left ( \dfrac{M}{2} \right )\dfrac{\left ( \dfrac{L}{2} \right )^{2}}{3}=\dfrac{1}{8}\left ( \dfrac{ML^{2}}{3} \right )$

$=\dfrac{I}{8}$

A thin rod of mass

$M$ and length

$L$ is rotating about an axis perpendicular to the length of rod and passing through a point of rod at a distance

$\dfrac{L}{6}$ from centre of rod. Its moment of inertia is:

0%
$\dfrac{ML^{2}}{24}$
0%
$\dfrac{ML^{2}}{18}$
0%
$\dfrac{ML^{2}}{12}$
0%
$\dfrac{ML^{2}}{9}$

Explanation

Moment of inertia about the center of rod is

$I_{cm}=\dfrac{ML^2}{12}$

Now using parallel axis theorem

$I=I_{cm}+Md^2$ here

$d=\dfrac{L}{6}$

$I=\dfrac{ML^2}{12}+\dfrac{ML^2}{36}=\dfrac{ML^2}{9}$

A thin rod of mass

$96 g$ and of length

$1 m$ is made to rotate about an axis perpendicular to the length of rod passing through a point at a distance

$25 cm$ from one end of rod. The moment of inertia of that rod is:

0%
$1.4\times 10^{-2}kgm^{2}$
0%
$0.7\times 10^{-2}kgm^{2}$
0%
$1.4\times 10^{-3}kgm^{2}$
0%
$0.7\times 10^{-3}kgm^{2}$

Explanation

$M=96gm$

$M=0.096kg.$

Ml of rod about the axis passing from the point which is 25 cm away from one end of the rod.

$\Rightarrow \dfrac{Ml^2}{12}+Mr^2$

$\Rightarrow \dfrac{0.096(1)^2}{12}+0.096(0.25)^2$

$\Rightarrow 0.096\left ( \dfrac{1}{12}+\dfrac{1}{16} \right )\Rightarrow 0.096\times \dfrac{7}{48}$

$\Rightarrow 14\times 10^{-3}kgm^2$ or

$1.4\times 10^{-2}kgm^2$

A wheel of radius

$10 cm$ can rotate freely about an axis passing through its centre. A force

$20 N$ acts tangentially at the rim of wheel. If moment of inertia of wheel is

$0.5 kgm^{2}$ , its angular acceleration is:

0%
$2 \ rad/s^{2}$
0%
$2.5 \ rad/s^{2}$
0%
$4 \ rad/s^{2}$
0%
$5 \ rad/s^{2}$

Explanation

Torque on the wheel

$\tau =\vec{F}\times \vec{R}$
as F is applied at rim.

$\Rightarrow \tau =20\times 0.1=2Nm$
Moment of Inertia

$I=0.5kgm^2$

$\tau =I\alpha$

$\Rightarrow \alpha=\dfrac{\tau }{I}=\dfrac{2}{0.5}=4rad/s^2$

A circular disc is rotating about its own natural axis. A constant opposing torque

$2.75 N m$ is applied on the disc due to which it comes to rest in

$28$ rotations. If moment of inertia of disc is

$0.5 kgm^{2}$ , the initial angular velocity of disc is:

0%
$210 rpm$
0%
$280 rpm$
0%
$360 rpm$
0%
$420 rpm$

Explanation

we know that

$\tau =I\alpha$

$\Rightarrow \alpha =\dfrac{ \tau }{I}=\dfrac{2.75Nm}{0.5}=5.5\:rads^{-2}$
given no of rotations

$=28$

$\therefore Total\:angle\:\theta =28(2\pi )=56\pi$

$\omega^2=2\alpha \theta$

$\Rightarrow \omega^2=2\times 5.5 \times 56\pi$

$\omega=43.99\:rad/s$

$=43.99\times \dfrac{60}{2\pi }rpm=420\;rpm$

A ceiling fan is rotating at the rate of

$3.5 rps$ and its moment of inertia is

$1.25 kgm^{2}$ . If the current is switched off, the fan comes to rest in

$5.5 s$ . The torque acting on the fan due to friction is:

0%
$2.5 \ Nm$
0%
$5 \ N m$
0%
$7.5 \ N m$
0%
$10 \ N m$

Explanation

$\tau =I\alpha$

$\alpha =\dfrac{d\omega }{dt}=\dfrac{3.5\times 2\pi }{5.5}=4$

$\therefore \tau =I\alpha =1.25\times4=5 \ Nm$

Identify the correct order in which the ratio of radius of gyration to radius increases for the following bodies.
I) rolling solid sphere
II) rolling solid cylinder
III) rolling hollow cylinder
IV) rolling hollow sphere

0%
I, II, IV, III
0%
I, III, II, IV
0%
II, I, IV, III
0%
II, I, III, IV

Explanation

For solid sphere, the radius of gyration is

$k=\sqrt{2R^2/5}=\sqrt{2/5}R$

For solid cylinder, the radius of gyration is

$k=\sqrt{R^2/2}=\sqrt{1/2}R$

For hollow cylinder, the radius of gyration is

$k=\sqrt{R^2}=R$

For hollow sphere, the radius of gyration is

$k=\sqrt{2R^2/3}=\sqrt{2/3}R$

Arranging in increasing order,

$I<II<IV<III$

The radius of gyration of rod of length

$L$ and mass

$M$ about an axis perpendicular to its length and passing through a point at a distance

$\dfrac{L}{3}$ from one of its ends is:

0%
$\dfrac{\sqrt{7}}{6}L$
0%
$\dfrac{L}{9}$
0%
$\dfrac{L}{3}$
0%
$\dfrac{\sqrt{5}}{2}L$

Explanation

Moment of inertia of rod about cm ,

$Icm=\dfrac{ML^2}{12}$

Distance of point from cm d=

$l/2-l/3=l/6$ ,

So, Moment of inertia at that point

$I=Icm+Md^2=\dfrac{ML^2}{12}+\dfrac{ML^2}{36}=\dfrac{ML^2}{9}$

$ML^2/9=MK^2\Rightarrow K=L/3$

A thin rod of length

$L$ and mass

$M$ is bent at the middle point O at an angle of

$60^o$ . The moment of inertia of the rod about an axis passing through O and perpendicular to the plane of the rod will be:

0%
$\dfrac{ML^{2} }{6}$
0%
$\dfrac{ML^{2} }{12}$
0%
$\dfrac{ML^{2} }{24}$
0%
$\dfrac{ML^{2} }{3}$

Explanation

When a rod is bend, then the mass of each half is

$m=M/2$ and length

$l=L/2$

MI of the road about the point O is

$I=2\dfrac{ml^2}{3}$

So,

$I=2\dfrac{(M/2)(L/2)^2}{3}=\dfrac{ML^2}{12}$

Identify the increasing order of the radius of gyration of following bodies of same radius:
I) About natural axis of circular ring.
II) About diameter of circular ring.
III) About diameter of circular plate.
IV) About diameter of solid sphere.

0%
II, III, IV, I
0%
III, II, IV, I
0%
III, IV, II, I
0%
II, IV, III, I

Explanation

(1) about Neutral axis of circular ring

$\to$

$MR^{2}MRg^{2}$

$Rg=R$

(2) about diameter of circular ring

$\to$

$\dfrac{MR^{2}}{2}=MRg^{2}$

$Rg=\dfrac{R}{\sqrt{2}}=0.707 R$

(3) about diameter of circular plate

$\to$

$\dfrac{MR^{2}}{4}=MRg^{2}$

$Rg=\dfrac{R}{2}=0.5 R$

(4) about diameter of solid sphere

$\to$

$\dfrac{MR^{2}}{5}=MRg^{2}$

$Rg=\sqrt{\dfrac{2}{5}}R=0.63 R$

Ans.

$(III)<(IV)<(II)<(I)$

A body of mass

$m$ slides down an incline and reaches the bottom with a velocity

$V$ . If the same mass were in the form of a ring which rolls down the same incline, the velocity of its centre of mass at bottom of the plane is:

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

Explanation

Let

$h$ the height of the point on the inclined plane at which the body starts to slide to reach the bottom.

For the body of mass

$m$ :

The initial kinetic energy of the body is zero.

Using work-energy theorem,

$W = \dfrac{1}{2}mV^2$

$\therefore$

$mgh = \dfrac{1}{2}mV^2$

We get

$gh = \dfrac{V^2}{2}$ ....(1)

For a ring of mass

$m$ rolling down the plane:

The initial kinetic energy of the ring is zero.

Using work-energy theorem,

$W = K.E_f$

$\therefore$

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

$mgh = \dfrac{1}{2}mv^2+\dfrac{1}{2}(mr^2)w^2$

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

$v = rw$ )

$mgh = mv^2$

$v^2 = gh$

$v^2 = \dfrac{V^2}{2}$ (using 1)

$\implies$

$v = \dfrac{V}{\sqrt{2}}$

Identify the decreasing order of moments of inertia of the following bodies of same mass and radius.
I) about diameter of circular ring
II) about diameter of circular plate
III) about tangent of circular ring perpendicular to its plane
IV) about tangent of circular plate in its plane

0%
III, IV, II, I
0%
IV, III, I, II
0%
IV, III, II, I
0%
III, IV, I, II

Explanation

Moment of inertia about diameter of circular ring

$I_1=\dfrac{mr^2}{2}$

Moment of inertia about diameter of circular plate

$I_2=\dfrac{mr^2}{4}$

Moment of inertia about tangent of circular ring perpendicular to its plane

$I_3=mr^2+mr^2=2mr^2$

Moment of inertia about tangent of circular plate in its plane

$I_4=mr^2/4+mr^2=\dfrac{5mr^2}{4}$

$I_3>I_4>I_1>I_2$

A particle of mass

$M$ is moving in a horizontal circle of radius

$R$ with a uniform speed

$v$ . When it moves from one point to a diametrically opposite point, its

0%
Momentum does not change
0%
Momentum changes by $2M v$
0%
Kinetic energy does not change
0%
Kinetic energy changes by $Mv^{2}$

Explanation

Initial momentum

$= Mv$

Final momentum

$= -Mv$

Change in momentum

$= 2 Mv$

Initial Kinetic energy = final kinetic energy =

$\dfrac{Mv^{2}}{2}$

Three identical thin rods each of length

$20 cm$ and of mass

$30 g$ are joined in the form of an equilateral triangular frame. The moment of inertia of triangular frame about an axis perpendicular to the plane of frame and passing through its centre is:

0%
$1.5\times 10^{-4}\ kgm^{2}$
0%
$3\times 10^{-4}\ kgm^{2}$
0%
$6\times 10^{-4}\ kgm^{2}$
0%
$1.2\times 10^{-4}\ kgm^{2}$

Explanation

For rods of mass

$m$ and length

$l$ ,

The distance between center of the rod and axis passing through the centroid is

$l/2\sqrt 3$

Using parallel axis theorem, its MI is

$ml^2/12+ml^2/12=ml^2/6$

So total MI of system is

$3\times ml^2/6=ml^2/2$

Here,

$m=30\times10^{-3}kg;l=0.2m$

so MI is

$30\times10^{-3}\times0.2^2/2=6\times10^{-4}kgm^2$

A circular ring starts rolling down on an inclined plane from its top. Let

$V$ be velocity of its centre of mass on reaching the bottom of inclined plane. If a block starts sliding down on an identical inclined plane but smooth, from its top, then the velocity of block on reaching the bottom of inclined plane is:

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

Explanation

For ring applying directly work-energy theorem,

loss in gravitational potential energy = gain in rotational kinetic energy +gain in translational kinetic energy

$mgh = \frac {1}{2}mV^2 + \frac {1}{2}mR^2\left ( \frac {V}{R}\right )^2$

$mgh = mV^2$

$V = \sqrt {gh}$

For the block, Work Energy Theorem,

$mgh = \frac {1}{2}mV_1^2$

$V_1 = \sqrt {2gh}$

$= \sqrt {2} V$

$I$ is moment of inertia of a thin rod of uniform thickness about an axis perpendicular to its length and passing through its centre. The rod at one of two sides of axis is cut and removed. The moment of inertia of remaining rod about same axis is :

0%
$\dfrac{I}{2}$
0%
$\dfrac{I}{3}$
0%
$\dfrac{I}{4}$
0%
$\dfrac{I}{6}$

Explanation

Let initial mass and length be

$m$ and

$l$ .

So MI about an axis perpendicular to its length and passing through its center is

$I=ml^2/12$

After cutting into half, mass is

$m/2$ and length is

$l/2$

So MI about the same axis is

$(m/2)\times(l/2)^2/3=ml^2/24=I/2$

$R$ is radius and

$K$ is radius of gyration then in which of the following cases of rolling bodies match the ratio

$K^2:R^2$ .

List-1	List-2
a) Solid sphere	e) $1:1$
b) Solid cylinder	f) $2:3$
c) Hollow sphere	g) $1:2$
d) Hollow cylinder	h) $2:5$

0%
a - f; b - e; c - h; d - g
0%
a - f; b - e; c - g; d - h
0%
a - h; b - g; c - e; d - f
0%
a - h; b - g; c - f; d - e

Explanation

For solid sphere

$I=\dfrac{2mR^2}{5}=MK^2$

$\dfrac{K^2}{R^2}=\dfrac{2}{5}$

For solid cylinder

$I=\dfrac{mR^2}{2}=MK^2$

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

For hollow sphere

$I=\dfrac{2mR^2}{3}=MK^2$

$\dfrac{K^2}{R^2}=\dfrac{2}{3}$

For hollow cylinder

$I=mR^2=MK^2$

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

A solid cylinder of mass

$M$ and radius

$R$ rolls without slipping down an inclined plane of length

$L$ and height

$h$ . What is the speed of its centre of mass when the cylinder reaches its bottom -

0%
$\sqrt{2gh}$
0%
$\sqrt{\dfrac{3}{4}gh}$
0%
$\sqrt{\dfrac{4}{3}gh}$
0%
$\sqrt{4gh}$

Explanation

Let speed of centre of mass is

$v$

A solid cylinder rolls without slipping down so

$v=R\omega$

Moment of inertia of cylinder,

$I= \dfrac{1}{2}MR^2$

Using conservation of energy,

$\Rightarrow mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{1}{2}I\omega^2$

$\Rightarrow mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{1}{2}MR^2\left (\dfrac{v}{R}\right )^2$

$\Rightarrow v=\sqrt{\dfrac{4}{3}gh}$

A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is

$90^{0}$ . The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is :

0%
$\frac{ML^{2}}{6}$
0%
$\frac{\sqrt{2}ML^{2}}{24}$
0%
$\frac{ML^{2}}{24}$
0%
$\frac{ML^{2}}{12}$

A rod rests on a friction less surface. Two forces each of magnitude F are applied in the opposite direction on the edges of the rod as shown in the figure below. Which of the following quantities are nonzero and constant:
(i) angular momentum (ii) angular acceleration (iii) Total force (iv) total torque (v) total linear momentum (vi) total kinetic energy (vii) moment of inertia (viii) translation kinetic energy

0%
i, ii, iii,v, vi
0%
ii, iv and vii
0%
ii, iii, iv, vii,viii
0%
i, iii, v,vi, viii

A body of mass

$\mathrm{m}=3.513$ kg is moving along the

$\mathrm{x}$ -axis with a speed of 5.00

$\mathrm{m}\mathrm{s}^{-1}$ . The magnitude of its momentum is recorded as :

0%
17.56 kg $\mathrm{m}\mathrm{s}^{-1}$
0%
17.57 kg $\mathrm{m}\mathrm{s}^{-1}$
0%
17. 6 kg $\mathrm{m}\mathrm{s}^{-1}$
0%
17.565 kg $\mathrm{m}\mathrm{s}^{-1}$

Explanation

Momentum of the body is

$mv=3.513kg\times 5.00m/s=17.565kg m/s$

However, the accuracy of the result would be determined by the most inaccurate observation, which is speed with three significant digits. Thus the answer would be expressed in three significant digits, that is,

$17.6\ kg m/s$

Two identical particles each of mass m are projected from points A and B on the ground with same initial speed u making an angle

$\theta$ as shown in the figure, such that their trajectories are in the same vertical plane. The initial velocity of the centre of mass is

0%
ucos $\theta$
0%
$2\mathrm{u}\cos\theta$
0%
usin $\theta$
0%
$2\mathrm{u}\sin\theta$

Explanation

$V_{X(CM)}=\dfrac{m(u \cos \theta)+m(-u \cos \theta)}{2 m}$

$V_{Y(CM)}=\dfrac{m(u \sin \theta)+m(-u \sin \theta)}{2 m}$

$=u \sin \theta$

The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is

0%
$2 : 1$
0%
$\sqrt {5} : \sqrt {6}$
0%
$2 : 3$
0%
$1 : \sqrt {2}$

Explanation

$I_{disk}$ =

$\frac{5}{4}MR^{2}$ =

$MK_{disk}^{2}$

${I_{ring}} = \frac{3}{2}M{R^2} = \ MK_{ring}^2$

$= \dfrac{{{K_{disk}}}}{{{K_{ring}}}} = \sqrt {\dfrac{5}{6}}$

Choose the correct statement about the centre
of mass (CM) of system of two particles.
a. The C.M. lies on the line joining the two particles midway between them
b. The C.M. lies on the line joining them at a point whose distance from each particle is inversely proportional to the mass of that particle
c. The C.M. lies on the line joining them at a point whose distance from each particle is proportional to the square of the mass of the particle
d. The C.M. is on the line joining them at a point whose distance from each particle is proportional to the mass of that particle.

0%
a is correct
0%
b is correct
0%
c is correct
0%
d is correct

Two objects of masses 200gm and 500gm
have velocities of 10i m/s and 3i + 5j m/s
respectively. The velocity of their centre of
mass is

0%
5i 25j
0%
$\displaystyle \frac{5}{7}i-25\mathrm{j}$
0%
$5i+\displaystyle \frac{25}{7}j$
0%
$25i-\displaystyle \frac{5}{7}j$

A boat of mass 60kg is floating in still water. A boy of mass 20kg walks from one end to the other end. If the length of the boat is 3m, the distance through which the boat moves is:

0%
1m
0%
0.75m
0%
0.5m
0%
0.9m

Four identical particles each of mass

$"m"$ are arranged at the corners of a square of side length

$"L"$ . If one of the masses is doubled the shift in the centre of mass of the system w.r.t. diagonally opposite mass.

0%
$\dfrac{L}{\sqrt{2}}$
0%
$\dfrac {3\sqrt {2}L}{5}$
0%
$\dfrac {L}{4\sqrt{2}}$
0%
$\dfrac {L}{5\sqrt{2}}$

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 2 - MCQExams.com

Hint:- Centre of mass depends upon the masses, position and relative distance between the particles.

Explanation:

The resultant of all the internal forces, on any system of particles, is zero. Therefore their centre of mass does not depend upon the forces acting on the particles.

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 2 - MCQExams.com

Hint:- Centre of mass depends upon the masses, position and relative distance between the particles.

Explanation:

The resultant of all the internal forces, on any system of particles, is zero. Therefore their centre of mass does not depend upon the forces acting on the particles.

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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