rotational as well as translational motion

MCQ Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 5

Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is:

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4 : 1
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2 : 1
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1 : 2
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1 : 16

Explanation

Kinetic energy

$K = \dfrac{p^2}{2m}$

Two masses are given as-

$m_1 = 1 \ gm$ and

$m_2 = 4 \ gm$

But

$K_1 = K_2$

$\therefore$

$\dfrac{p^2_1}{2m_1} = \dfrac{P^2_2}{2m_2}$

$\dfrac{P_1^2}{2(1)} = \dfrac{P^2_2}{2(4)}$

$\implies$

$P_1:P_2 = 1:2$

A sphere moving at some instant with horizontal velocity

$v_c$ towards right and angular velocity in the anti-clockwise sense. If

$|v_c| = |\omega R|$ The instantaneous center of rotation is:

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at the bottom of the sphere
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at the top of the sphere
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at the centre of the sphere
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Anywhere inside the sphere

Any point on the circumference of a rigid body which is rolling without slipping undergoes :

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a circular path
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an elleptic path
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a cycloid path
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an parabolic path

Explanation

Any point on the rigid body, undergoing rolling without slipping, will have linear and angular speed, such that

$linear\ speed = distance\ from\ CoM \times angular\ speed$

Such particles follows cycloid path.

Rotation as well as translation motion is an example of

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constrained motion
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unconstrained motion
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perpetual motion
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None of these

$ML^2T^{-2}$ is the dimensional formula for

0%
moment of inertia
0%
pressure
0%
elasticity
0%
couple acting on a body

Explanation

$\left[ MOI \right] =\left[ M{ R }^{ 2 } \right] =\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ 0 } \right] \\ \left[ Pressure \right] =\left[ N/{ M }^{ 2 } \right] =\left[ { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 } \right] \\ \left[ Couple \right] =\left[ N.{ M } \right] =\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 } \right]$

Drum A undergoes

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rotational motion
0%
translational motion
0%
rotational as well as translational motion
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None of these

Explanation

The

$\omega$ of drum B provides enough length of connecting wire for translational motion

Rolling without slipping is an example of

0%
Rotation
0%
Translation
0%
Rotation with translation
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None of these

Two bodies,

$A$ and

$B$ initially, at rest, move towards each other under mutual force of attraction. At the instant when the speed of

$A$ is

$v$ and that of

$B$ is

$2v$ , the speed of the center of mass of the bodies is

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$3 v$
0%
$2 v$
0%
$1.5 v$
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Zero

The motion in which all points of a moving body move uniformly in the same line or direction is known as

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Rolling
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Rotatory
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Both A and B
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Translational

A diatomic molecule is formed by two atoms which may be treated as mass points

$m_1$ , and

$m_2$ joined by a massless rod of length r. Then, the moment of inertia of the molecule about an axis passing through the centre of mass and perpendicular to rod is :

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Zero
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$(m_1+m_2)r^2$
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$\dfrac{(m_1+m_2)}{m_1m_2}r^2$
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$\left ( \dfrac{m_1m_2}{m_1+m_2} \right )r^2$

Explanation

Let point C be the centre of mass situated at distance x from atom of mass mi and at distance (r - x) from mass

$m_2$ . By definition of centre of mass, we have

$m_1x= m_2(r - x)$

$x =\dfrac{m_2}{ (m_1 + m_2)}r$
Moment of inertia about the centre of mass is

$I=m_1x^2+m_2(r-x)^2$
or

$I=m_1\left ( \dfrac{m_2r}{m_1+m_2} \right )^2+m_2\left ( r-\dfrac{m_2r}{m_1+m_2} \right )^2$

$m_1 \left ( \dfrac{m_2r}{m_1+m_2} \right )^2+m_2\left ( \dfrac{m_1r}{m_1+m_2} \right )^2$

$\dfrac{m_1m_2}{m_1+m_2}r^2$

Sagar was playing with the spinning top, after some time he realized that motion of top is same as motion of earth spinning about its own axis. Identify the type of motion?

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Rotational motion
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Oscillatory motion
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Rectilinear motion
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None of these

The radius of gyration of a disc of mass

$100 g$ and radius

$5 cm$ about an axis passing through its centre of gravity and perpendicular to the plane is (in cm)

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$0.5$
0%
$2.5$
0%
$3.54$
0%
$6.54$

Explanation

Let radius of gyration be

$k$

$\therefore m{ k }^{ 2 }=m{ r }^{ 2 }\\ k=\sqrt { \dfrac { { R }^{ 2 } }{ 2 } } =\sqrt { \dfrac { 25 }{ 2 } } =\dfrac { 5 }{ \sqrt { 2 } } \\ =3.54cm$

Which of the following doesn't represent rotatory motion?

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0%
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Both A and C

Let 'M' be the mass and 'L' be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case is

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$1$
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$\dfrac{1}{2}$
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$\dfrac{1}{4}$
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$\dfrac{1}{8}$

Explanation

Moment of inertia of the rod about an axis passing through its centre and perpendicular to the length

$I_1 = \dfrac{1}{12}ML^2$

Using

$Mk_1^2 = I_1 = \dfrac{1}{12}ML^2$

We get radius of gyration

$k_1 = \dfrac{L}{\sqrt{12}}$

Moment of inertia of the rod about an axis passing through its one end and perpendicular to the length

$I_2 = \dfrac{1}{3}ML^2$

Using

$Mk_2^2 = I_2 = \dfrac{1}{3}ML^2$

We get radius of gyration

$k_2 = \dfrac{L}{\sqrt{3}}$

Thus ratio of radius of gyration

$\dfrac{k_1}{k_2} = \dfrac{L/\sqrt{12}}{L/\sqrt{3}}$

$\implies$

$\dfrac{k_1}{k_2} = \dfrac{1}{2}$

A body of mass

$m_{1} = 4\ kg$ moves at

$5\hat {i}m/s$ and another body of mass

$m_{2} = 2\ kg$ moves at

$10\hat {i} m/s$ . The kinetic energy of centre of mass is :

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$\dfrac {200}{3}J$
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$\dfrac {500}{3}J$
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$\dfrac {400}{3}J$
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$\dfrac {800}{3}J$

Explanation

$v_{CM} = \dfrac {m_{1}\dfrac {dr_{1}}{dt} + m_{2} \dfrac {dr_{2}}{dt}}{m_{1} + m_{2}}$

$= \dfrac {4\times 5\hat {i} + 2\times 10\hat {i}}{4 + 2}$

$v_{CM} = \dfrac {40\hat {i}}{6} = \dfrac {20}{3}\hat {i}$
The kinetic energy

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

$= \dfrac {1}{2}\times (4 + 2) \times \dfrac {20\times 20}{3\times 3}$

$= \dfrac {1}{2}\times 6\times \dfrac {20\times 20}{3\times 3}$

$K = \dfrac {400}{3}J$ .

Two particles A and B, initially at rest, moves towards each other under a mutual force of attraction. At the instant when the speed of A is $u$ and the speed of B is $2 u$ , the speed of centre of mass is

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Zero
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$u$
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$1.5 u$
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$3 u$

Explanation

As both the particles are at rest initially, thus centre of mass of the system is zero initially i.e.

${V_{cm}}_i = 0$

Since the particles are moving towards each other under a mutual force of attraction i.e. no external force is acting on the system. Thus motion of centre of mass do not get affected and so velocity of centre of mass remains the same at every instant.

Hence velocity of centre of mass at the given instant is also zero i.e.

${V_{cm}}_f = {V_{cm}}_i = 0$

Two particles

$A$ and

$B$ initially at rest, move towards each other under a mutual force of attraction. At the instant when the speed of

$A$ is

$v$ and the speed of

$B$ is

$2v,$ the speed of center of mass of the system is

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$zero$
0%
$v$
0%
$1.5v$
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$3v$

Explanation

Zero, according to newton law of motion if body is at rest and no external force is acting then body will be at rest until external force acts.

As we know both of these are moving due to their mutual attraction so velocity of center of mass will be constant ,i.e.

$zero$ .

Three identical thin rods, each of mass m and length l are joined to form an equilateral triangle. Find the moment of inertia of the triangle about one of its sides.

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$\displaystyle\frac{Ml^2}{2}$
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$\displaystyle\frac{Ml^2}{3}$
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$\displaystyle\frac{Ml^2}{9}$
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$\displaystyle\frac{Ml^2}{12}$

Explanation

$MOI\quad of\quad system={ I }_{ AB }+{ I }_{ BC }+{ I }_{ AC }\\ { I }_{ AB }=0\quad (it\quad is\quad at\quad the\quad axis)\\ { I }_{ BC }=\dfrac { 1 }{ 3 } m({ l }_{ eff. })^{ 2 }=\dfrac { 1 }{ 3 } m({ l }\sin60)^{ 2 }=\dfrac { 1 }{ 3 } m{ l }^{ 2 }\times \dfrac { 3 }{ 4 } \\ \therefore MOI=0+\dfrac { m{ l }^{ 2 } }{ 4 } +\dfrac { m{ l }^{ 2 } }{ 4 } =\dfrac { m{ l }^{ 2 } }{ 2 }$

Two objects

$P$ and

$Q$ initially at rest move towards each other under mutual force of attraction. At the instant when the velocity of

$P$ is

$v$ and that of

$Q$ is

$2v$ , the velocity of centre of mass of the system is

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$v$
0%
$3v$
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$2v$
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$1.5v$
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Zero

If a body of moment of inertia

$2$ kg

$m^2$ revolves about its axis making

$2$ rotations per school, then its angular momentum(in Js) is:

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$2\pi$
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$4\pi$
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$6\pi$
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$8\pi$
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$10\pi$

Explanation

Given,
Moment of inertia of body

$I=2\,kg \,m^2$
Rotation of body in one second

$f=2$ rps
We know that,
Angular momentum

$L=l\omega$

$L=2\times 2\pi f$

$L=2\times 2\pi(2)$

$L=8\pi \ J s$ .

A bicycle wheel rolls without slipping on a horizontal floor. Which one of the following is true about the motion of points on the rim of the wheel, relative to the axis at the wheel's centre?

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Points near the top move faster than points near the bottom
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Points near the bottom move faster than points near the top
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All points on the rim move with the same speed
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All points have the velocity vectors that are pointing in the radial direction towards the centre of the wheel

Explanation

We have,

$V_A=2v\sin\theta /2$
Hence, velocity of point on rim increases with

$\theta$ for

$0^o < \theta < 180^o$ and decreases with

$\theta$ for

$180^o < \theta < 30^o$ .

The radius of gyration of a solid cylinder of mass

$M$ and radius

$R$ about its own axis is

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$\dfrac { R }{ \sqrt { 2 } }$
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$\dfrac { R }{ 2 }$
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$\dfrac { R }{ \sqrt { 3 } }$
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$\dfrac { R }{ 3 }$
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$\dfrac { R }{ 4 }$

Explanation

The radius of gyration of a solid cylinder

$K=\sqrt { \dfrac { I }{ M } } =\sqrt { \dfrac { M{ R }^{ 2 } }{ 2 } }$

$K=\dfrac { R }{ \sqrt { 2 } }$

State whether true or false :
The pendulum is said to be in mean position or stable equilibrium when the centre of mass of the bob lies directly below the the point of suspension.

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True
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False

A thin rod of length L is suspended from one end and rotated with n rotations per second. The rotational kinetic energy of the rod will be :

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$2mL^2\pi^2n^2$
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$\frac{1}{2}mL^2\pi^2n^2$
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$\frac{2}{3}mL^2\pi^2n^2$
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$\frac{1}{6}mL^2\pi^2n^2$

Explanation

Moment of inertia of the rod about the axis passing through its end

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

Frequency of rotation

$f = n$

Angular velocity of rod

$w = 2\pi f = 2\pi n$

Kinetic energy of the rod

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

$\implies \ K.E. = \dfrac{1}{2}\times \dfrac{mL^2}{3}\times (2\pi n)^2 = \dfrac{2}{3}mL^2 \pi^2 n^2$

A rod of mass

$m$ and length

$L$ , lying horizontally, is free to rotate about a vertical axis through its centre. A horizontal force of constant magnitude

$F$ acts on the rod at a distance of

$L/4$ from the centre. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time

$t$ after the motion starts.

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$\dfrac{3Ft^2}{5mL}$
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$\dfrac{5Ft^2}{2mL}$
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$\dfrac{3Ft^2}{2mL}$
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$\dfrac{2Ft^2}{3mL}$

Explanation

We have

torque $=\tau=rF\sin { 90° } =\cfrac{ LF }{ 4 }$

And perpendicular axis theorem gives us

$I={ I }_{ 2 }={ I }_{ x }+{ I }_{ y }=\cfrac{ M{ L }^{ 2 } }{ 12 }$

Angular acceleration,

$\alpha =\cfrac{ \tau }{ I } =\cfrac{ LF }{ 4 } \times \left( \cfrac{ 12 }{ M{ L }^{ 2 } } \right) =\cfrac{ 3F }{ ML }$

If $\theta$ is the angle rotated in time $t$ , and initial angular velocity ${ w }_{ 0 }$ being zero we have

$\theta ={ w }_{ 0 }t+\cfrac{ 1 }{ 2 } \alpha { t }^{ 2 }=\cfrac{ 3F{ t }^{ 2 } }{ 2ML }$

Consider a ring rolling down a smooth inclined plane of vertical height 'h' and inclination

$\theta$ . Then the true statement in the following is?

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Acceleration along the plane is g $\sin\theta$ and the potential energy at the topmost point is mgh
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Acceleration along the plane is g and the potential energy at the top most point is mgh
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Acceleration along the plane is g $\sin\theta$ and the potential energy at the top most point as mgh $\sin\theta$
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None of these

Explanation

Acceleration along the plane is

$g\sin \theta$ and the potential energy at the topmost point is

$mgh$ .

Two bodies of mass

$10kg$ and

$2kg$ are moving with velocities

$2i-7j+3k$ and

$-10i+35j-3k$

$m/s$ respectively. The velocity of their CM is

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$2i$ m/s
0%
$2k$ m/s
0%
$(2i+2k)$ m/s
0%
$(2i+2j+2k)$ m/s

Explanation

Velocity of CM =

$\dfrac{m_1v_1+m_2v_2}{m_1+m_2}=\dfrac{10(2i-7j+3k)+2(-10+35-3k)}{10+2}=2k$

$m/s$

The radius of gyration of a solid sphere of radius

$R$ about a certain axis is also equal to

$R$ . If

$r$ is the distance between the axis and the centre of the sphere, then

$r$ is equal to:

0%
$R$
0%
$0.5R$
0%
$\sqrt { 0.6 } R$
0%
$\sqrt { 0.3 } R$

Explanation

The radius of gyration of a solid sphere of radius

$R$ about a certain axis is equal to

$R$ .

$k=R$

$Mk^2=MR^2$

$I=I_{cm}+mr^2$

$MR^2=\dfrac{2}{5}MR^2+Mr^2$

$r^2=\dfrac{3}{5}R^2$

$r=\sqrt{0.6}R$

The correct option is C.

Two particles A and B initially at rest, move towards each other under a mutual force of attraction. At the instant when the speed of A is v and speed of B is 2v, the speed of centre of mass of the system is :

0%
zero
0%
v
0%
1.5 v
0%
3 v

An automobile engine develops

$100$

$kW$ when rotating at a speed of

$1800\ rev/min$ . The torque it delivers is

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$3.33\ N-m$
0%
$200\ N-m$
0%
$530.5\ N-m$
0%
$2487\ N-m$

Explanation

$\textbf{Step 1: Equation of power}$

Work done by torque is given by

$;$

$W=\tau.\theta$

Differentiating with respect to time

$\Rightarrow\ \ \text{Power, } P = \dfrac{dW}{dt}=\tau.\dfrac{d\theta}{dt}=\tau.w$

$....(1)$

where

$w$ is angular velocity

$\textbf{Step 2: Calculation of angular velocity}$

$1$ RPM

$\Rightarrow \dfrac{2\pi}{60}\,rad/s$

$\Rightarrow 1800$ RPM

$=\dfrac{2\pi}{60}\times 1800=60\,\pi \,rad/s=w$

$\textbf{Step 3: Calculation of power}$

From eq

$....(1):$

$P=\tau.w$

$\Rightarrow\ \ 100\,KW=\tau(60\pi)$

$\Rightarrow\ \ \tau =\dfrac{100000}{60\pi}=530.5\,Nm$

Hence option

$C$ is correct.

A rigid body rotates about a fixed axis with variable angular velocity equal to (a-bt) at time t where a and b are constants. The angle through which it rotates before it comes to rest is___?

0%
$\dfrac{a^2}{b}$
0%
$\dfrac{a^2}{2b}$
0%
$\dfrac{a^2}{4b}$
0%
$\dfrac{a^2}{2b^2}$

Explanation

Let the angle rotated be

$\theta$

Given,

$\omega=a-bt$

$\dfrac{d\theta}{dt}=a-bt$

Also,

It will come to rest when,

$\omega=a-bt=0$

$t_f=\dfrac{a}{b}$

$\displaystyle \int_0^\theta d\theta=\int_0^{\dfrac{a}{b}} a-bt$

$\theta=\dfrac{a^2}{b}-\dfrac{a^2}{2b}=\dfrac{a^2}{2b}$

Option

$\textbf B$ is the correct answer

A particle of mass

$m$ describes a circle of radius

$r$ . The centripetal acceleration of the particle is

$\dfrac{4}{r^2}$ . Then the momentum of the particle is

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$2\dfrac {m}{r}$
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$2\dfrac {m}{\sqrt {r}}$
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$4\dfrac {m}{\sqrt {r}}$
0%
None of these

Explanation

For a particle of mass m performing uniform circular motion with radius r centripetal acceleration

$a_{c}$ is given as $a_{c}=\frac{v^{2}}{r}$ where v is the linear velocity

$\because a_{c}=\frac{v^{2}}{r}$

$\frac{4}{r^{2}}=\frac{v^{2}}{r} \; \; \; ...\: \: a_{c}=\frac{4}{r^{2}}\; \; (given)$

$\therefore v=\frac{2}{r^{\frac{1}{2}}} \; \; ... (i)$

Linear momentum $p=mv$

using (i)

$\therefore p=mv=m*\frac{2}{r^{1/2}}$

Therefore the momentum of the particle is $2*\frac{m}{r^{1/2}}$

The centre of mass of a body :

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lies always at the geometrical centre
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lies always inside the body
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lies always outside the body
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may lie within or outside the body

Four identical rods are joined end to end to form a square. The mass of each rod is

$M$ . The moment of inertia of the system about one of the diagonals is:

0%
$\cfrac { 2M{ l }^{ 2 } }{ 3 }$
0%
$\cfrac { 13M{ l }^{ 2 } }{ 3 }$
0%
$\cfrac { M{ l }^{ 2 } }{ 6 }$
0%
$\cfrac { 13M{ l }^{ 2 } }{ 6 }$

The moment of inertia of a rod about its perpendicular bisector is I. When the temperature of the rod is increased by

$\triangle T$ , the increase in the moment of inertia of the rod about the same axis is (Here ,

$\alpha$ is the coefficient of linear expansion of the rod)

0%
$\alpha I\triangle T$
0%
$2\alpha I\triangle T$
0%
$\dfrac{\alpha I \triangle T}{2}$
0%
$\dfrac{2I\triangle T}{\alpha}$

Explanation

Moment of inertia a of a uniform rod of mass M, length L about its perpendicular bisector is

$I=\dfrac{1}{12}ML^2$ ...................(i)
When the temperature of the rod is increased by

$\triangle T,$ there is increase in length of the road

$\triangle L$ is given by

$\triangle L=L\alpha \triangle T$ ................ (ii)

$\therefore$ New moment of inertia about the same axis is

$I'=\dfrac{1}{12}m(l+\triangle L)^2=\dfrac{1}{12}M[L^2+(\triangle L)^2+2L\triangle L]$
Since

$\triangle L$ is a small quantity , the term

$\triangle L)^2$ being very small can be neglected.

$\therefore I'=\dfrac{1}{12} M[L^2+2L\triangle L]$

$=\dfrac{1}{12}ML^2+\dfrac{1}{12}ML^22\alpha \triangle T$ ......Using (ii)

$=I+2\alpha I \triangle T$ ......Using (i)

$\therefore$ Increase in moment of inertia,

$=I'-I=2\alpha I\triangle T$

The moment of inertia of a solid sphere about an axis passing through the centre of gravity is

$1/2M{R}^{2}$ , then its radius of gyration about a parallel axis at a distance

$2R$ from first axis is:

0%
$5R$
0%
$\sqrt { \cfrac { 22 }{ 5 } } R$
0%
$\cfrac { 5 }{ 2 } R\quad$
0%
$\sqrt { \cfrac { 12 }{ 5 } } R$

The cylinders P and Q are of equal of mas and length but made of metals with densities

${\rho _P}$ and

${\rho _Q}\,\,\left( {{\rho _P} > {\rho _Q}} \right).$ If their moment of inertia about an axis passing through centre and normal to the circular face be

${I_P}$ and

${I_Q}$ , then:

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${I_P} = {I_Q}$
0%
${I_P} > {I_Q}$
0%
${I_P} < {I_Q}$
0%
${I_P} \leqslant {I_Q}$

Explanation

Given: The cylinders P and Q are of equal of mas and length but made of metals with densities

$\rho_P$ and

$\rho_Q(\rho_P>\rho_Q)$ . If their moment of inertia about an axis passing through centre and normal to the circular face be

$I_P$ and

$I_Q$ ,

To find the relation between

$I_P$ and

$I_Q$

Solution:

Let mass be m,

We know,

$m=\rho V$ (as mass=density

$\times$ volume)

$\implies m=\rho\times \pi r^2l\\\implies r^2=\dfrac {m}{\pi \rho l}.......(i)$

Moment of inertia about given axis is,

$I=\dfrac 12mr^2=\dfrac 12m\times \dfrac {m}{\pi \rho l}\\\implies I=\dfrac {m^2}{2\pi \rho l}\\\implies I\propto \dfrac 1\rho\\\implies \dfrac {I_P}{I_Q}=\dfrac {\rho_Q}{\rho_P}$

Given,

$\rho_P>\rho_Q\\\implies I_P<I_Q$

Four identical rods are joined end to end to form a square. The mass of each rod is

$M$ . The moment of inertia of the system about an axis passing through the point of intersecion of diagonals and perpendicular to the plane of the square is:

0%
$\cfrac { 4M{ l }^{ 2 } }{ 3 }$
0%
$\cfrac { 13M{ l }^{ 2 } }{ 3 }$
0%
$\cfrac { M{ l }^{ 2 } }{ 6 }$
0%
$\cfrac { 13M{ l }^{ 2 } }{ 6 }$

Explanation

To the moment of inertia about axis passing through c and perpendicular to the plane of rods we use perpendicular axis theorum 4 times

$I_c= 4(\dfrac{ml^2}{12}+\dfrac{ml^2}{4})=\dfrac{4ml^2}{3}$

hence Option A is correct

1721650_984286_ans_7538e6bac471497d837f2255def60b79.JPG

The moment of inertia of a body rotating about a given axis is

$12.0 kg m^2$ in the SI system. What is the value of the moment of inertia in a system of units in which the unit of length is

$5$ cm and the unit of mass is

$10g$ ?

0%
$2.4 \times 10^3$
0%
$6.0 \times 10^3$
0%
$5.4 \times 10^3$
0%
$4.8 \times 10^5$

Explanation

$I=12Kg{ m }^{ 2 }$

New system

$\Rightarrow$ unit length

$=5cm$ and unit mass

$=10g$ .

$I=12\times 1000\times \left( { 10 }^{ 4 } \right) g{ cm }^{ 2 }$

$I$ unit length

$=5cm$ . unit mass

$=10g$ .

$I=12\times \dfrac { 10000\times { 10 }^{ 3 } }{ 25\times 10 } \left( 10g \right) \left( 5{ cm }^{ 2 } \right)$

$I=48000\left( 10g \right) { \left( 5cm \right) }^{ 2 }$

$=4.8\times { 10 }^{ 5 }$

A ball of mass m moving with a constant velocity u strikes against a ball of same mass at rest. If e is the coefficient of restitution, then what will be the ration of velocity of two balls after collision?

0%
$\dfrac { 1-e }{ 1+e }$
0%
$\dfrac { e-1 }{ e+1 }$
0%
$\dfrac { 1+e }{ 1-e }$
0%
$\dfrac { e+1 }{ e-1 }$

Explanation

momentum should be conserved

$mu+0=mv_1 +mv_2$

$\Rightarrow \ v_1 +v_2 =u.....(1)$

restitution coefficient should be given

$e=\dfrac {v_2 -v_1}{u_1 -u_2}=\dfrac {v_2 -v_1}{u-0}$

$\Rightarrow \ v_2 -v_1 =e\ u.....(2)$

adding

$(1)$ &

$(2)$ we get

$2v_2 =u(e+1).....(3)$

subtracting

$(1)$ &

$(2)$ we get

$2v_1=u(1-e)....(4)$

dividing

$(4)$ by

$(3)$

we get

$\boxed {\dfrac {v_1}{v_2}=\dfrac {1-e}{1+e}}$

The radius of gyration of a ring of mass 80 g and diameter 6 cm, about an axis passing through its center of gravity and perpendicular to the plane is:

0%
3 cm
0%
6 cm
0%
9 cm
0%
4 cm

Two particles of mass

$1\ kg$ and

$3\ kg$ move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is

$2\ m/s$ , their centre of mass has a velocity of

$0.5\ m/s$ , then the velocity of the centre of mass is

$0.75\ m/s$ .

0%
True
0%
False

Explanation

As given in the question that there is no external force acting on the system

$\implies a_{sys}=0 \implies \Delta v_{com}=0$

But here we find that

$\Delta v_{com} = v_{com2}-v_{com1}=2.5ms^{-2} \ne0$

hence it's false and Option B is correct.

The figure below shows a pattern of two fishes. Write the mapping rule for the rotation of Image A to Image B.

0%
Image B has to be rotated by 90 degrees anticlockwise to map with image A
0%
Image B should first have a mirror reflection and then be rotated to map with image A
0%
The width of image B should be reduced by 1 unit (1 square in the graph shown) and then rotated by 90 degrees to map with image A
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The width of image B should be increased by 1 unit (1 square in the graph shown) and then rotated by 90 degrees to map with image A

A long slender rod of mass

$2\ kg$ and length

$4\ m$ is placed on a smooth horizontal table. Two particles of masses

$2\ kg$ and

$1\ kg$ strike the rod simultaneously and stick to the rod after collision as shown in Figure.
Velocity of the centre of mass of the rod after collision is

0%
$12\ m/s$
0%
$9\ m/s$
0%
$6\ m/s$
0%
$3\ m/s$

Explanation

By conservation of linear momentum,

$2\times 10+1\times 10=\left( 2+12 \right) v$

$v=\cfrac { 30 }{ 5 }$

$v=6㎧$

A string is wrapped around a cylinder of mass

$M$ and radius

$R$ . The string is pulled vertically upwards to prevent the centre of mass from falling as the cylinder unwinds the string, The work done on the cylinder for reaching an angular speed

$\omega$ is:

0%
$\cfrac { 2M{ R }^{ 2 }{ \omega }^{ 2 } }{ 3 }$
0%
$\cfrac { M{ R }^{ 2 }{ \omega }^{ 2 } }{ 3 }$
0%
$\cfrac { M{ R }^{ 2 }{ \omega }^{ 2 } }{ 2 }$
0%
$\cfrac { M{ R }^{ 2 }{ \omega }^{ 2 } }{ 4 }$

Explanation

Work done is the rotational KE acquired be cylinder,

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

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

$=\dfrac{MR^2}{4}\omega ^2.$

Hence, the answer is

$\dfrac{MR^2}{4}\omega ^2.$

Three particles of equal masses are placed at the corners of an equilateral triangle as shown in the figure. Now particle A starts with a velocity

$\nu_1$ towards line

$AB$ , particle

$B$ starts with a velocity

$\nu_2$ towards line

$BC$ and particle

$C$ starts with velocity

$\nu_3$ towards line

$CA$ . The displacement of

$CM$ of three particle A, B and C after time

$t$ will be (given if

$\nu_1 = \nu_2 = \nu_3$ )

0%
zero
0%
$\dfrac{\nu_1 + \nu_2 + \nu_3}{3}t$
0%
$\dfrac{\nu_1 + \dfrac{\sqrt{3}}{2}\nu_2 + \dfrac{\nu_3}{2}}{3}t$
0%
$\dfrac{\nu_1 + \nu_2 + \nu_3}{4}t$

Explanation

REF.Image.

$v_{1x} = \dfrac{-v_{1}}{2}, v_{1y} = \dfrac{-v_{1}\sqrt{3}}{2}$

$v_{2x} = v_{2} , v_{2y} = 0$

$v_{3x} = \dfrac{-v_{3}\sqrt{3}}{2} , v_{3y} = v_{3}\dfrac{\sqrt{3}}{2}$

using ,

$\vec{v}_{cm} = \hat{i}v_{x} + \hat{j}v_{y}$

where,

$V_{x} = \dfrac{-mv_{1/2}+ v_{2}m - mv_{3/2}}{3m} = 0$

$v_{y} = \dfrac{-m v_{1}\sqrt{3}/2 + mv_{3}\sqrt{3}/2}{3m} = 0$

$\therefore$ Net velocity = 0 (using vector theory)

i.e. CM is at rest.

So, displacement of CM is zero.

so, (a) is correct.

1400061_1002649_ans_067374b2560249f6a96fd7566894d7f4.png

If a car is moving forward, what is the direction of the moment of the moment caused by the rotation of the tires

0%
It is heading inwards, i.e. the direction is towards inside of the car
0%
It is heading outwards, i.e. the direction is towards outside of the car
0%
It is heading forward, i.e. the direction is towards the forward direction of the motion of the car
0%
It is heading backward, i.e. the direction is towards back side of the motion of the car

A gun of mass

$M$ , frees a shell of mass

$m$ horizontally and the energy of explosion is such as would be sufficient to project the shell vertically to a height '

$h$ '. The recoil velocity of the gun is:

0%
$\left(\dfrac{2m^2gh}{M(m + M)}\right)^{\dfrac{1}{2}}$
0%
$\left(\dfrac{2m^2gh}{M(m - M)}\right)^{\dfrac{1}{2}}$
0%
$\left(\dfrac{2m^2gh}{2M(m - M)}\right)^{\dfrac{1}{2}}$
0%
$\left(\dfrac{2m^2gh}{2M(m + M)}\right)^{\dfrac{1}{2}}$

Explanation

Applying linear momentum conservation,

$p_i=p_f$

$\Rightarrow 0=mV_g=mV_b$

$\Rightarrow V_b=\dfrac{mV_g}{m}$ ......(1)

According to question, total explosion energy is equal to

$mgh$

$\Rightarrow \dfrac{1}{2}mV_g^2+\dfrac{1}{2}mV_b^2=mgh$

$mv_g^2+m\left(\dfrac{M}{m}\right)^2V62g=2mgh$ ( By (1))

$v_g^2\left(M+\dfrac{M^2}{m}\right)=2mgh$

$v_g^2=2mgh\times \dfrac{m}{m(m+M)}$

$v_g=\sqrt{\dfrac{2m^2gh}{M(m+M)}}$ (A)

1438112_999628_ans_ea00428b5e0c4c5bba577aca27839267.png

Four particles of equal masses are placed on the vertices of a square and are rotated with a uniform angular velocity about one of the edges (A) as shown in the figure. Which particle will have a larger angular momentum

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

Explanation

Angular momentum L =

$I\omega$ . I is the moment of inertia. For discrete particles, I =

$mr^2$ .; r is the distance of the particle from the axis of rotation.

The distance of the mass at C is at a larger distance from the origin as compared to other masses. Thus, moment of inertia for mass C is larger and hence angular momentum for C is the largest as compared to other masses

The ratio of the radii of gyration of a hollow sphere and a solid sphere of the same masses and radii about an axis passing through their centre is

0%
$\sqrt{2/3}$
0%
$\sqrt{2/5}$
0%
$\sqrt{5/3}$
0%
$\sqrt{5/2}$

Explanation

The radius of gyration of a solid sphere is

$\sqrt{2/5}$ and that of a hollow sphere is

$\sqrt{2/3}$ . Dividing one by the other, we get, the ratio =

$\sqrt{5/3}$

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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