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

For which of the following does the centre of mass lie outside the body?

0%
A pencil
0%
A shot put
0%
A dice
0%
A bangle

A metallic sphere having mass

$2 \,kg$ is moving with a velocity of

$10 \,m /s.$ The momentum of the sphere in kg metre / sec. will be-

0%
$1 / 5$
0%
$5$
0%
$12$
0%
$20$

Explanation

$P=m×v$

$P=2×10$

$P=20m/s$

In kinetic theory of gases , a molecule of mass

$m$ of an ideal gas collides with a wall of vessel with velocity

$V$ . The change in the linear momentum of the molecule is

0%
$2 \,mV$
0%
$mV$
0%
$-mV$
0%
Zero

The total momentum of the molecules of

$1\,gm\, mol$ of a gas in a container at rest of

$300\,K$ is

0%
$2 \times \sqrt{3 R \times 300} \, gm \times cm / sec$
0%
$2 \times 3 \times R \times 300 \, gm \times cm / sec$
0%
$1 \times \sqrt{3 \times R \times 300} \, gm \times cm / sec$
0%
Zero

$57.0-g$ tennis ball is traveling straight at a player at

$21.0\ m/s$ . The player volleys the ball straight back at

$25.0\ m/s$ . If the ball remains in contact with the racket for

$0.060\ 0\ s$ , what average force acts on the ball?

0%
$22.6\ N$
0%
$32.5\ N$
0%
$43.7\ N$
0%
$72.1\ N$
0%
$102\ N$

The ratio of the radii of gyration of a circular
disc to that of a circular ring, each of same mass
and the radius around their respective axes is -

0%
$\dfrac {\sqrt2}{1}$
0%
$\dfrac{\sqrt2}{\sqrt3}$
0%
$\dfrac{\sqrt3}{\sqrt2}$
0%
$\dfrac{1}{\sqrt{2}}$

Explanation

$MK_{1}^{2}=MR^{2}$

$K_{1} = R$

$MK_{2}^{2}=\frac{MR^{2}}{2}$

$K_{2}=\frac{R}{\sqrt{2}}$

$\frac{K_{2}}{K_{1}}=\frac{\sqrt{2}}{R}=\frac{1}{\sqrt{2}}$

Four thin metal rods, each of mass

$M$ and length

$L$ , are welded to form a square. The moment of inertia of the composite structure about a line which bisects any two opposite rods is:

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

Explanation

The MI of the rod AB(

$I_1$ )and CD(

$I_2$ ) about the axis PQ( as shown in the figure ) are same.

$I_1=I_2=\dfrac{ML^2}{12}$
The MI of the rod BC(

$I_3$ )and AD(

$I_4$ ) about the axis PQ( as shown in the figure ) are same..

$I_3=I_4={M}{(\dfrac{L}{2})^2}$

$AD \; \left | \right | \:PQ$ and

$BC \; \left | \right | \:PQ$ and all points on AD are at same distance from the axis PQ, all points on BC are at same distance from the axis PQ,

$\therefore I =I_1 + I_2 +I_3 +I_4= 2 \times \dfrac{ML^2}{12} + 2 \times {M}{(\dfrac{L}{2})^2}=\dfrac{2}{3}ML^2$

Three identical spheres each of mass m and radius R are placed touching each other so that their centres A, B and C lie on a straight line. The position of their centre of mass from the centre of A is

0%
$\displaystyle \frac{2R}{3}$
0%
2R
0%
$\displaystyle \frac{5R}{3}$
0%
$\displaystyle \frac{4R}{3}$

Explanation

[B] From Symmetry B is the centre of ABC combined.

Position of centre of mass of ABC (combined) from

centre of

$A = R + R = 2R$

An automobile is started from rest with one of its doors initially at right angles. lf the hinges of the door are toward the front of the car, the door will shut as the automobile picks up speed. The acceleration a is constant and the centre of mass is at a distance d from the hinges. The time

$T$ needed for the doors to close is given by:

$\left [ Given:\displaystyle\int_{0}^{\frac{\pi }{2}}\dfrac{d\theta }{\sqrt{sin\theta }}=N \right ]$ (Consider the door as a square plate)

0%
$\mathrm{T}=(\sqrt{\dfrac{3\mathrm{d}}{2\mathrm{a}}}) \mathrm{N}$
0%
$\mathrm{T}=(\sqrt{\dfrac{\mathrm{d}}{3\mathrm{a}}}) \mathrm{N}$
0%
$\mathrm{T}=(\sqrt{\dfrac{2\mathrm{d}}{\mathrm{a}}}) \mathrm{N}$
0%
$\mathrm{T}=(\sqrt{\dfrac{2\mathrm{d}}{3\mathrm{a}}}) \mathrm{N}$

Explanation

Equation of motion is

$I\alpha =Ma(dcos\theta )$

Now moment of inertia of the door about the hinges is given as

$I=\dfrac{Md^{2}}{3}+Md^{2}$

$=M\left ( \dfrac{4}{3}d^{2} \right )$

Hence,

$\dfrac{4}{3}Md^{2}\alpha =Ma dcos\theta$

$\Rightarrow \dfrac{4}{3}d\times\omega \dfrac{d\omega }{d\theta }=acos\theta \left ( \because \alpha =\omega \dfrac{d\omega }{d\theta } \right )$

$\Rightarrow \dfrac{4}{3}d\int \omega d\omega =\int acos\theta d\theta$

$\Rightarrow \dfrac{4}{3}d\dfrac{\omega ^{2}}{2}=asin\theta$

$\omega ^{2}=\dfrac{3a}{2d}sin\theta +C$

where, C

$=$ constant of integration

$\theta =0$ (initially), we have

$\theta =0$

$\Rightarrow C=0$

$\Rightarrow \omega^{2}=\dfrac{3a}{2d}sin\theta$

$\Rightarrow \omega =\sqrt{\dfrac{3a}{2d}sin\theta }$

$\dfrac{d\theta }{\sqrt{sin\theta }}=\sqrt{\dfrac{3a}{2d}}dt$ and integrating between

$t=0$ and

$t=T$

(corresponding to

$\theta =0$ and

$\theta =\dfrac{\pi }{2}$ )

$\int_{0}^{\dfrac{\pi }{2}}\dfrac{d\theta }{\sqrt{sin\theta }}=\sqrt{\dfrac{3a}{2d}}.T$

$\Rightarrow T=\left ( \sqrt{\dfrac{2d}{3a}} \right )N$

$\left ( since,given:\displaystyle N=\int_{0}^{\dfrac{\pi }{2}}\dfrac{d\theta }{\sqrt{sin\theta }} \right )$

A uniform sheet each of thickness 10 units is
cut into the shape as shown. Compute then x
and y-coordinates of the centre of mass of the
piece from point A.

0%
17.5, 22.5
0%
22.5, 17.5
0%
10.5, 22.5
0%
190/7, 125/7

Three identical thin rods each of mass

$m$ and length

$L$ are joined together to form an equilateral triangular frame. The moment of inertia of frame about an axis perpendicular to the plane of frame and passing through a corner is:

0%
$\dfrac{2mL^{2}}{3}$
0%
$\dfrac{3mL^{2}}{2}$
0%
$\dfrac{4mL^{2}}{3}$
0%
$\dfrac{3mL^{2}}{4}$

Explanation

Sum of MI of the two rods, at vertex of which the axis passes, is

$2\times mL^2/3=2mL^2/3$

For the third rod, the distance between center of the rod and the axis is

$d=\sqrt3L/2$

Using parallel axis theorem, its MI is

$mL^2/12+3mL^2/4=5mL^2/6$

So total MI of the system is

$2mL^2/3+5mL^2/6=3mL^2/2$

A thin uniform metallic triangular sheet of mass M has sides

$AB=BC=L$ . Its moment of inertia about the axis AC lying in the plane of the sheet is:

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

Explanation

The given triangular sheet has an area equal to half of a square sheet having same side length as the length of the base of the triangle.

Let MI of the triangular sheet about AC be I.

I is half of the MI of the square plate about an axis passing through the centre O and being in the plane of the plate.

Let the MI of the square about an axis XY be

$I_{XY}$ .

Due to symmetry

$I_{AB} = I_{CD}$ AND

$I_{PQ} = I_{RS}$

Let

$I_{square}$ be the MI a bout an axis passing through the centre of the square and perpendicular to the plane of the square plate.

Then using Perpendicular Axis Theorem

$I_{square} = I_{AB} + I_{CD}=I_{PQ} + I_{RS}$

$\therefore I_{square} = 2 I_{AB} = I_{PQ}$

But,

$I_{AB}= 2I$

$\therefore I = \dfrac{1}{2}I_{AB} = \dfrac{1}{2} *(\dfrac{1}{2}\dfrac{ML^2}{6})$ ....(1) where M is the mass of the square plate.

But the triangular sheet has half the mass of the square plate. Writing I in terms of the mass of the plate.

$\therefore I =\dfrac{M}{2}*L^2/12=\dfrac{1}{12}M_{triangle}L^2$

In the system shown, a force of

$100$ N is applied at the end shown. What is the magnitude

$\tau$ (in N-m) of the torque which was produced on the drum for starting motion? (Given that the coefficient of static friction is

$0.1$ )

0%
$10$
0%
$5$
0%
$20$
0%
$25$

Explanation

For the motion to start, the 'moment-couple' on the drum must be equal to the product of frictional force of the rod against the drum and the drum radius. Since the rod is in equilibrium, we can take torque about the fulcrum to find the normal force of the drum on the rod (Normal force N acts at centre of rod)

$\sum \tau _{fulcrum}=0$
or

$-100(2+2)+N(2)=0$
or

$N=200\ N$
Friction force

$f=\mu N$

$= (0.1)(200) = 20 N$

$\tau =$ Torque produced by couple

$=$

$(20)1$
or

$\tau =20\ Nm$

A car (open at the top) of mass 9.75 kg is coasting along a level track at 1.36 m/s, when it begins to rain hard. The raindrops fall vertically with respect to the ground. When it has collected 0.5 kg of rain, the speed of the car is

0%
0.68 m/s
0%
1.29 m/s
0%
2.48 m/s
0%
9.8 m/s

Explanation

Initial

$m_{1} = 9.75$ kg
final

$m =9.75 + 0.5 = 10.25$

$m_{1}v_{1} = m_{2}v_{2}$

$(9.75)(1.36) = (10.25)v_{2}$

$v_{2} = 1.29$ m/s

The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is

$0_0$ . Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is :

0%
$0_0 + \dfrac{ML^2}{2}$
0%
$0_0 + \dfrac{ML^2}{4}$
0%
$0_0 + 2ML^2$
0%
$0_0 + ML^2$

Explanation

Let the end points of the rod are designated as A and B.
Applying parallel axis theorem,
Given,

$I_{cm}=O_o$

$I_A= I_{cm} + M(\frac{L}{2})^2= O_o + M\frac{L^2}{4}$

Find the momentum of each particle.

0%
$p=2\mu\sqrt{v_1^2+v_2^2}$
0%
$p=\mu\sqrt{v_1^2+v_2^2}$
0%
$p=3\mu\sqrt{v_1^2+v_2^2}$
0%
$p=4\mu\sqrt{v_1^2+v_2^2}$

Explanation

The particles

$m_1$ and

$m_2$ move with velocities

$\vec{v_1}$ and

$\vec{v_2}$ respectively.

$\vec{v_{cm}} = \dfrac{m_1\vec{v_1} + m_2\vec{v_2}}{m_1 + m_2}$

In the CM frame,

$\vec{v'_1} = \vec{v_1} - \vec{v_{cm}}$

$\vec{v'_2} = \vec{v_2} - \vec{v_{cm}}$

$\vec{v_{rel}} = \vec{v_1} - \vec{v_2}$

$\vec{p'_1}=m_1( \vec{v_1} - \vec{v_{cm}})=\dfrac{m_1m_2}{m_1 + m_2}(\vec{v_1} -\vec{v_2})=\dfrac{m_1m_2}{m_1 + m_2}\vec{v_{rel}}$

$\vec{p'_2}=m_2( \vec{v_2} - \vec{v_{cm}})=\dfrac{m_1m_2}{m_1 + m_2}(\vec{v_2} -\vec{v_1})=-\dfrac{m_1m_2}{m_1 + m_2}\vec{v_{rel}}$

$|\vec{p'_1}| = |\vec{p'_2}| =|\dfrac{m_1m_2}{m_1 + m_2}\vec{v_{rel}}|=\dfrac{m_1m_2}{m_1 + m_2}|\vec{v_{rel}}|= \dfrac{m_1m_2}{m_1 + m_2}\sqrt{(v_1^2 + v_2^2)}= \mu \sqrt{(v_1^2 + v_2^2)}$

since

$\vec{v_1} \perp \vec{v_2}$ and

$|\vec{v_1} -\vec{v_2}|=\sqrt{v_1^2 + v_2^2 -2v_1v_2 \cos \dfrac{\pi}{2}}=\sqrt{v_1^2 + v_2^2}$

where

$\mu =\dfrac{m_1m_2}{m_1 + m_2}$ is defined as the reduced mass.

The arrangement shown in figure above consists of two identical uniform solid cylinders, each of mass

$m$ , on which two light threads are wound symmetrically. The friction in the axle of the upper cylinder is assumed to be absent. If the tension of each thread in the process of motion is

$\displaystyle T=\frac{mg}{x}$ , then the value of

$x$ is :

0%
$10$
0%
$20$
0%
$15$
0%
$5$

Explanation

For the cylinder from the equation

$N_z=I\beta_z$ about its stationary axis of rotation.

$\displaystyle2Tr=\frac{mr^2}{2}\beta$ or

$\displaystyle\beta=\frac{4T}{mr}$ ......(1)
For the rotation of the lower cylinder from the equation

$N_{cz}=I_c\beta_z$

$\displaystyle2Tr=\frac{mr^2}{2}\beta^\prime$ or,

$\displaystyle\beta^\prime=\frac{4T}{mr}=\beta$
Now for the translational motion of lower cylinder from the equation

$F_x=mw_{cx}$ ,

$mg-2T=mw_c$ ......(2)
As there is no slipping of threads on the cylinders,

$w_c=\beta^\prime r+\beta r=2\beta r$ (3)
Simultaneous solution of (1), (2) and (3) yields

$\displaystyle T=\frac{mg}{10}$

A train of mass

$M$ is moving on a circular track of radius

$R$ with a constant speed

$v$ . The length of the train is half of the perimeter of the track. The linear momentum of the train will be

0%
$zero$
0%
$\dfrac {2Mv}{\pi}$
0%
$MvR$
0%
$Mv$

Explanation

Correct answer: Option B

Hint: In this case, the train's length is half the circumference of the circular track.

Step 1:Train’s linear momentum

$\rho = mv$ is the formula for calculating a body's linear momentum, where m is the mass and v is the velocity.

The linear density of an item is equal to where $m$ is the object's mass and $l$ is the object's length.

To get an equation for the train's linear momentum, consider a small part of it.

Consider a small element that is at an angle $\theta$ to the X-axis. An angle $d\theta$ is subtended by this element. This is seen in the diagram.

The small element's length will be $dl =Rd\theta$

The small element's mass can therefore be represented as dm = $\lambda dl$ -------(1)

We obtain by substituting for $\lambda = \dfrac{M}{{\pi R}}$ and $dl = Rd\theta$ in equation (1):

$dm = \dfrac{M}{{\pi R}} \times Rd\theta$ $= \dfrac{M}{\pi }d\theta$

Because the vertical component of the velocity $vcos\theta$ at A and B is oriented in the positive and negative Y directions, they will cancel each other out. As a result, only the horizontal component $v\sin \theta$ is important.

We may now write $dp = dm\left( {vsin\theta } \right)$ for the small element's linear momentum. --------(2)

When we replace $dm = \dfrac{M}{\pi }d\theta$ in equation (2), we obtain

$dp = \dfrac{M}{\pi }(v\sin \theta )d\theta$ ---------(3)

The linear momentum of the small element is expressed in equation (3).

Step 2: Finding total linear momentum

Integrating equation (3) will give us the train's total linear momentum.

$P = \mathop \smallint \limits_0^\pi d{p_x} = \mathop \smallint \limits_0^\pi \dfrac{M}{\pi }\left( {v\sin \theta } \right)d\theta$

$P = \dfrac{{Mv}}{\pi }\int\limits_0^\pi {sin\theta d\theta }$ $= - \dfrac{{Mv}}{\pi }[cos\theta ]_0^\pi$

Applying limits,

$P = - \dfrac{{Mv}}{\pi }[ - 1 - ( + 1)]$ $= \dfrac{{2Mv}}{\pi }$

As a result, the train's linear momentum is calculated as $P = \dfrac{{2Mv}}{\pi }$

The figure shows a uniform rod lying along the x-axis. The locus of all the points lying on the x-y plane, about which the moment of inertia of the rod is same as that about O, is:

0%
an ellipse
0%
a circle
0%
a parabola
0%
a straight line

A chain

$AB$ of mass

$m$ and length

$L$ is hanging on a smooth horizontal table as shown in the figure. If it is released from the position shown then the displacement of centre of mass of chain in magnitude, when end

$A$ moves a distance

$\dfrac{L}{2}\ is\ X\sqrt{2}m$ .
Find

$X$ . (

$L = 32\ m$ )

0%
$8$
0%
$9$
0%
$10$
0%
$16$

Explanation

In the first case, we consider two parts one of mass

$3m/4$ and length

$3L/4$ and other of mass

$m/4$ and length

$L/4$ .
The center of mass of these masses is calculated as

$x_1=\displaystyle\dfrac{(3m/4)(-3L/8)+(m/4)(0)}{m}=-9L/32=-9$
and

$y_1=\displaystyle\dfrac{(3m/4)(0)+(m/4)(-L/8)}{m}=-L/32=-1$

Thus the coordinates of the center of mass is

$(-9,-1)$

Similarly, when the chain moves by

$L/2$ distance we get the new position of the center of mass as:

$x_2=\displaystyle\dfrac{(m/4)(-L/8)+(3m/4)(0)}{m}=-9L/32=-1$
and

$y_2=\displaystyle\dfrac{(m/4)(0)+(3m/4)(-3L/8)}{m}=-L/32=-9$

Thus the coordinates of the center of mass in this case is

$(-1,-9)$
Thus the distance between the two position of the center of mass is calculated as

$8\sqrt 2$ . Thus the value of

$X$ is

$8$ .

Two identical bricks of length

$L$ are piled one on top of the other on a table as shown in the figure. The maximum distance

$S$ the top brick can overhang the table with the system still balanced is :

0%
$\dfrac{1}{2}L$
0%
$\dfrac{2}{3}L$
0%
$\dfrac{3}{4}L$
0%
$\dfrac{7}{8}L$

Explanation

The center of mass of the set of bricks must be above or to the left of the point of support for the system of be balanced.

The center of mass of the upper brick needs to be above the right end of lower brick. The center of upper brick is at a distance of

$L/2$ from its right end.

The distance center of mass of both the books from the right edge of upper brick

$=\dfrac{(mL/2)+(mL)}{2m}=\dfrac{3L}{4}=S$ .

Three point like equal masses

$m_1, m_2$ and

$m_3$ are connected to the ends of a mass-less rod of length

$L$ which lies at rest on a smooth horizontal plane. At

$t = 0$ , an explosion occurs between

$m_2$ and

$m_3$ , and as a result, mass

$m_3$ is detached from the rod, and moves with a known velocity

$v$ at an angle of

$30^o$ with the y-axis. Assume that the masses

$m_2$ and

$m_3$ are unchanged during the explosion.
What is the velocity of the centre of mass of the system consisting of three masses after the expulsion?

0%
$\displaystyle \dfrac{v}{4} (\widehat i - 3 \widehat j)$
0%
$\displaystyle \dfrac{v}{4} (- \widehat i + \sqrt{3} \widehat j)$
0%
$-v$
0%
none of the above

Two wheels

$A$ and

$B$ are released from rest from points

$X$ and

$Y$ respectively on an inclined plane as shown in figure. Which of the following statement(s) is/are incorrect?

0%
Wheel $B$ takes twice as much time to roll from $Y$ to $Z$ than that of wheel $A$ from $X$ to $Z$ .
0%
At point $Z$ velocity of wheel $A$ is four times that of wheel $B$
0%
Acceleration of the wheel $A$ is four times that of wheel $B$
0%
Both wheel take same time to arrive at point $Z$ .

Explanation

The wheels will start pure rolling after being released from rest.

Let

$v$ and

$w$ be the linear and angular velocities respectively.

$v=rw$ (pure rolling)

Work-energy theorem:

$W = \Delta K.E$

$M gH = \dfrac{1}{2}M v^2 + \dfrac{1}{2}I w^2$

$M gH = \dfrac{1}{2}M v^2 + \dfrac{1}{2}\times M R^2 w^2$

$M gH= \dfrac{1}{2}M v^2 + \dfrac{1}{2}\times M v^2$

$\implies v= \sqrt{g H}$

Thus

$\dfrac{v_B}{v_A} = \dfrac{\sqrt{g (4h)}}{\sqrt{g (h)}} \implies v_B = 2 v_A$

Also acceleration of the body rolling without slipping on the inclined plane,

$a = \dfrac{g sin \theta}{1+ \frac{I}{MR^2}}$ where

$I = K MR^2$ and

$K = 1$

Thus

$a_A = a_B$

Now

$v = 0 + at \implies t = \dfrac{v}{a}$

$\dfrac{t_B}{t_A} = \dfrac{v_B}{a_B} \times \dfrac{a_A}{v_A} = \dfrac{2v}{v}$

$\implies t_B = 2 t_A$

A particle of mass

$'m'$ is rigidly attached at

$'A'$ to a ring of mass

$'3m'$ and radius

$'r.'$ The system is released from rest and rolls without sliding. The angular acceleration of ring just after release is

0%
$\cfrac{g}{4r}$
0%
$\cfrac{g}{6r}$
0%
$\cfrac{g}{8r}$
0%
$\cfrac{g}{2r}$

Three particles of masses 1 kg, 2 kg and 3 kg are situated at the corners of the equilateral triangle move at speed 6 ms

$^{-1}$ , 3 ms

$^{-1}$ and 2 ms

$^{-1}$ respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find velocity of COM of the system at this instant :

0%
$3 ms^{-1}$
0%
$5 ms^{-1}$
0%
$6 ms^{-1}$
0%
$zero$

Explanation

Given :

$m_1 = 1kg$

$m_2 = 2kg$

$m_3 = 3kg$

$v_1 = 6$

$m/s$

$v_2 = 3$

$m/s$

$v_3 = 2$

$m/s$

Let velocity of their centre of mass be

$\vec{V_{CM}}$

Using

$\vec{V_{CM} } = \dfrac{m_1 \vec{v_1} + m_2 \vec{v_2} + m_3 \vec{v_3}}{m_1+m_2+m_3}$

$\vec{V_{CM} } = \dfrac{1 (-6sin30^o \hat{i} - 6cos30^o\hat{j}) + 2 (3\hat{i}) + 3 (-2cos60^o\hat{i} + 2sin60^o \hat{j})}{1+2+3}$

$\vec{V_{CM} } = \dfrac{1 (-6 \times \frac{1}{2} \hat{i} - 6\times \frac{\sqrt{3}}{2}\hat{j}) + 2 (3\hat{i}) + 3 (-2\times \frac{1}{2}\hat{i} + 2\times \frac{\sqrt{3}}{2} \hat{j})}{6}$

$\implies \vec{V_{CM}} = 0$

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 square about the median line is

0%
$\dfrac{Ml^{2}}{3}$
0%
$\dfrac{Ml^{2}}{4}$
0%
$\dfrac{Ml^{2}}{6}$
0%
none of these

Explanation

Consider the frame made using four rods as shown.
We have the moment of inertia of rod AB about the center of mass as

$\dfrac{Ml^2}{12}+M(\dfrac{l}{2})^2=\dfrac{Ml^2}{3}$
Thus we get the moment of inertia of all the four rods as

$\dfrac{4}{3}Ml^2$
Now using perpendicular axis theorem we see the the moment of inertia about perpendicular medians is given as

$2I=\dfrac{4}{3}Ml^2$
Thus we get

$I=\dfrac{2}{3}Ml^2$
Thus none of the above is the correct answer.

A steel plate of thickness

$h$ has the shape of a square whose side equals

$l$ , with

$h\ll l$ . The plate is rigidly fixed to a vertical axle

$OO$ which is rotated with a constant angular acceleration

$\beta$ (figure shown above). Find the deflection

$\lambda$ assuming the sagging to be small.

0%
$\displaystyle\lambda=\frac{9\rho\beta l^5}{5Eh^2}$
0%
$\displaystyle\lambda=\frac{7\rho\beta l^5}{3Eh^2}$
0%
$\displaystyle\lambda=\frac{7\rho\beta l^5}{5Eh^2}$
0%
None of these

Explanation

The deflection of the plate can be noticed by going to a co- rotating frame. In this frame each element of the plate experiences a pseudo force proportional to its mass. These forces have a moment which constitutes the bending moment of the problem. To calculate this moment we note that the acceleration of an element at a distance $\xi$ from the axis is $a=\xi\beta$ and the moment of the forces exerted by the section between $x$ and $l$ is

$\displaystyle N=\rho lh\beta\int_x^l{\xi^2d\xi}=\frac{1}{3}\rho lh\beta(l^3-x^3)$ .

From the fundamental equation

$\displaystyle EI\frac{d^2y}{dx^2}=\frac{1}{3}\rho lh\beta(l^3-x^3)$ .

The moment of inertia $\displaystyle I=\int_{\displaystyle-\frac{h}{2}}^{\displaystyle+\frac{h}{2}}{z^2ldz}=\frac{lh^3}{12}$

Note that the neutral surface (i.e. the surface which contains lines which are neither stretched nor compressed) is a vertical plane here and $z$ is perpendicular to it.

$\displaystyle\frac{d^2y}{dx^2}=\frac{4\rho\beta}{Eh^2}(l^3-x^3)$ . Integrating

$\displaystyle\frac{dy}{dx}=\frac{4\rho\beta}{Eh^2}\left(l^3x-\frac{x^4}{4}\right)+c_1$

Since $\displaystyle\frac{dy}{dx}=0$ , for $x=0$ , $c_1=0$ . Integrating again,

$\displaystyle y=\frac{4\rho\beta}{Eh^2}\left(\frac{l^3x^2}{2}-\frac{x^5}{20}\right)+c_2$

$c_2=0$ because $y=0$ for $x=0$

Thus $\displaystyle\lambda=y(x=l)=\frac{9\rho\beta l^5}{5Eh^2}$

A particle of mass m comes down on a smooth inclined plane from point B at a height of h from rest. The magnitude of change in momentum of the particle between position A (just before arriving on horizontal surface) and C (assuming the angle of inclination of the plane as

$\theta$ with respect to the horizontal) is :

0%
0
0%
$2m \sqrt{(2gh)} sin \theta$
0%
$2m \sqrt{2gh} sin \displaystyle \left ( \frac{\theta}{2} \right )$
0%
$2m \sqrt{(2gh)}$

Explanation

Let velocity of the particle at the base of the inclined plane be

$v$

Work-energy theorem :

$W_{g} = \Delta K.E$

$mgh = \dfrac{1}{2} mv^2 - 0$

$\implies v = \sqrt{2gh}$

Momentum of the particle at point C,

$\vec{P_c} = mv \hat{i}$

Momentum of the particle at point A,

$\vec{P_A} = mvcos\theta \hat{i} - mvsin\theta \hat{j}$

Change in momentum between A and C,

$\Delta \vec{ P} = \vec{P_c} - \vec{P_A}$

$\Delta \vec{ P} = mv (1-cos\theta) \hat{i} - mvsin\theta \hat{j}$

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

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

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

$\bigg(\because 1-cos\theta = 2sin^2 \dfrac{\theta}{2}\bigg)$

$\implies$

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

$=2 m\sqrt{2gh} sin \bigg(\dfrac{\theta }{2}\bigg)$

The radius of gyration of a plane lamina of mass $M$ , length $L$ and breadth $B$ about an axis passing through its center of gravity and perpendicular to its plane will be

0%
$\displaystyle{\sqrt{\dfrac{(L^2+B^2)}{12}}}$
0%
$\displaystyle{\sqrt{\dfrac{(L^2+B^2)}{8}}}$
0%
$\displaystyle{\sqrt{\dfrac{(L^2+B^2)}{2}}}$
0%
$\displaystyle{\sqrt{\dfrac{(L^3+B^3)}{12}}}$

Explanation

The moment of inertia about a rectangular lamina with area density

$\rho$ is

$\int \int \rho r^2 dA$ . Here r is the distance from each point to the axis of rotation and

$dA$ is the differential of area.
Take the center of the rectangle to be at

$(0,0)$ so the vertices are

$(L/2, B/2)$ , etc. Then the distance from (x,y) to the axis of rotation (passing through

$(0,0)$ perpendicular to the plate) is

$x^2+ y^2$ and the moment of inertia is

$\displaystyle\int^{L/2}_{-L/2}\int^{B/2}_{B/2}\rho(x^2+y^2)dxdy$
or

$\displaystyle\int^{L/2}_{-L/2}\rho(Bx^2+\frac{1}{12}B^3)$
or

$\displaystyle\frac{1}{12}\rho(L^3B+LB^3)$
Now as

$M=\rho LB$ , thus we get moment of inertia as

$\displaystyle\frac{M}{12}(L^2+B^2)$

so radius of gyration is

$K=\sqrt{\dfrac{I}{M}}$

$\implies K= \sqrt{\displaystyle\frac{1}{12}(L^2+B^2)}$

(i) Centre of gravity (C.G.) of a body is the point at which the weight of the body acts
(ii) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius
(iii) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be considered to be concentrated at its C.G.
(iv) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the C.G. of the body to the axis
Which one of the following pairs of statements is correct ?

0%
(iv) and (i)
0%
(i) and (ii)
0%
(ii) and (iii)
0%
(iii) and (iv)

Two particles of equal mass have initial velocities

$2\hat{i}\, ms^{-1}\, and\, 2\hat{j}\, ms^{-1}$ . First particle has a constant acceleration

$(\hat{i}\,+\,\hat{j})\, ms^{-1}$ while the acceleration of the second particle is always zero. The centre of mass of the two particles moves in :

0%
Circle
0%
Parabola
0%
Ellipse
0%
Straight line

Explanation

Velocity of center of mass is given by:

$v = \dfrac{v_{1} m_{1} + v_{2} m_{2}}{m_{1} + m_{2}} = i + j$
Acceleration of center of mass is given by:

$a = \dfrac{a_{1} m_{1} + a_{2} m_{2}}{m_{1} + m_{2}} = \dfrac{1}{2}(i + j)$
Acceleration and velocity both have the same direction, hence the body moves along the same line (i+j).
Hence it moves in a straight line.
Option D is correct.

Three rings each of mass m and radius r are so placed that they touch each other. The radius or gyration of the system about the axis as shown in the figure is:

0%
$\displaystyle \sqrt{\frac{6}{5}}r$
0%
$\displaystyle \sqrt{\frac{5}{6}}r$
0%
$\displaystyle \sqrt{\frac{6}{7}}r$
0%
$\displaystyle \sqrt{\frac{7}{6}}r$

Explanation

correct option is (D)

$\textbf{HINT:}$ Moment of inertia is defined as the product of mass of section and the square of the distance between the reference axis and the centroid of the section . Spinning figure skaters can reduce their moment of inertia by pulling in their arms, allowing them to spin faster due to conservation of angular momentum

$\textbf{Step1:}$ finding momentum of inertia of upper ring about axis

Moment of inertia of the upper ring about the axis $\mathrm{I}_{1}=\dfrac{1}{2} \mathrm{mr}^{2}$

Now, Momentum of inertia of one the lower ring

Moment of inertia of the one of the lower ring about the axis $\mathrm{I}_{2}=\dfrac{1}{2} \mathrm{mr}^{2}+$ $\mathrm{mr}^{2}=\dfrac{3}{2} \mathrm{mr}^{2}$

$\textbf{Step 2:}$ finding momentum of inertia of of second lower ring

Similarly, moment of inertia of the second lower ring about the axis $\mathrm{I}_{3}=$ $\dfrac{3}{2} \mathrm{mr}^{2}$

$\textbf{Step 3:}$ : fining total momentum

Total moment of inertia of the system about the axis $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}=\dfrac{7}{2} \mathrm{mr}^{2}$

Radius of gyration $\mathrm{k}=\sqrt{\dfrac{\mathrm{I}}{3 \mathrm{~m}}}=\sqrt{\dfrac{7}{6}} \mathrm{r}$

Consider the following two statement-
[A] Linear momentum of a system of particle is zero
[B] kinetic energy of a system of particles is zero. Then

0%
A does not imply B but B implies A
0%
A implies B and B implies A
0%
A does not imply B & B does not imply A
0%
A implies B but B does not imply A

Linear mass density of the two rods system,

$AC$ and

$CB$ is

$x$ . Moment of inertia of two rods about an axis passing through

$AB$ is

0%
$\displaystyle \frac{xl^{3}}{4\sqrt{3}}$
0%
$\displaystyle \frac{xl^{3}}{\sqrt{3}}$
0%
$\displaystyle \frac{xl^{3}}{4}$
0%
$\displaystyle \frac{xl^{3}}{6\sqrt{2}}$

Explanation

$\displaystyle L=\frac{l}{2} \sec 45^{\circ} =\frac{l}{\sqrt{2}}$

$\displaystyle m=Lx =\frac {lx}{\sqrt{2}}$

$\displaystyle I=2 \left[ \frac{mL^{2}}{3} \sin^{2} 45^{\circ} \right] =2 \left[ \left( \frac{lx}{3\sqrt{2}}\right) \left( \frac{l}{\sqrt{2}}\right)^{2} \left( \frac{1}{2} \right) \right]$

$\displaystyle =\frac{xl^{3}}{6\sqrt{2}}$

The moment of inertia of the plate about z-axis is

0%
$\displaystyle\frac{ML^2}{12}$
0%
$\displaystyle\frac{ML^2}{24}$
0%
$\displaystyle\frac{ML^2}{6}$
0%
None of these

Explanation

By unfolding, we can treat the given scenario as the square sheet of side

$L$ rotating about O.

Now as MOI of rod of length

$a$ about its centre of mass is

$\dfrac{Ma^2}{12}$

Thus

$I_{x'} = I_{y'} = \dfrac{M(L)^2}{12}$

Now using perpendicular axis theorem:

$I_z = I_{x'}+I_{y'}$

$I_z =2I_{x'} = \dfrac{ML^2}{6}$

In the diagram shown below all three rods are of equal length L and equal mass M. The system is rotated such that rod B is the axis. What is the moment of inertia of the system?

0%
$\dfrac {ML^{2}}{6}$
0%
$\dfrac {4}{3} ML^{2}$
0%
$\dfrac {ML^{2}}{3}$
0%
$\dfrac {2}{3} ML^{2}$

Explanation

Moment of inertia of system

$= M.I\ of\; A + M.I.\ of\; B + M.I\ of\ C$

$M.I\ of\ A = M. T$ through centre and perpendicular to length

$= \dfrac {1}{12} ML^{2}$
M.I. of

$C = M.I.$ of

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

$M.I$ of

$B = 0$
(moment of mass about an axis passing through its own position is zero)

$\therefore Total M.I. = \dfrac {1}{12} ML^{2} + \dfrac {1}{12} ML^{2} = \dfrac {1}{16} ML^{2}$ .

There are two particles of same mass. If one of the particles is at rest always and the other has an acceleration

$\bar{a}$ . Acceleration of center of mass is :

0%
zero
0%
$\displaystyle\frac{1}{2}\, \bar{a}$
0%
$\bar{a}$
0%
centre of mass for such a system can not be defined.

Explanation

${a}_{c.o.m}$ =

$\dfrac{{m}_{1}\times{a}_{1}+{m}_{2}\times{a}_{2}}{{m}_{1}+{m}_{2}}$
given both body having same mass so

${a}_{c.o.m}$ =

$\dfrac{{m}\times{a}_{1}+{m}\times{a}_{2}}{{2m}}$

${a}_{1}$ =0 given that body 1 is in rest
so

${\bar{a}}_{c.o.m}$ =

$\dfrac{1}{2}$

$\times\bar{a}$

Fill in the blanks.
The effect of rotation of the earth is ......... at the equator and .............. at the poles. If the earth stops rotating. the weight of a body would .......... due to the absence of .............. force.

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Minimum, maximum, decreases, gravitional
0%
Maximum, minimum, increases, centrifugal
0%
Maximum, minimum, decreases, centrifugal
0%
Minimum, maximum, increases, gravitional

If torques of equal magnitudes are applied to a hollow cylinder and a solid sphere both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry and the sphere is free to rotate about an axis passing through its center. Which of the two will acquire a greater angular speed after a given time?

0%
${ \omega }_{ 1 } > { \omega }_{ 2 }$
0%
${ \omega }_{ 1 } = { \omega }_{ 2 }$
0%
${ \omega }_{ 2 } > { \omega }_{ 1 }$
0%
None of these

Explanation

Consider

${ I }_{ 1 }$ and

${ I }_{ 2 }$ be the moments of inertia of the hollow cylinder and solid sphere about its axis through its center respectively.
Then,

${ I }_{ 1 }=M{ R }^{ 2 }$ ....(i)
and

${ I }_{ 2 }=\dfrac { 2 }{ 5 } M{ R }^{ 2 }$ ....(ii)
Let

$\tau$ be the magnitude of the torque applied on each of them. If

${ \alpha }_{ 1 }$ and

${ \alpha }_{ 2 }$ are the angular accelerations produced in the cylinder and sphere respectively, then

$\tau ={ I }_{ 1 }{ \alpha }_{ 1 }$
and

$\tau ={ I }_{ 2 }{ \alpha }_{ 2 }$

$\therefore { I }_{ 1 }{ \alpha }_{ 1 }={ I }_{ 2 }{ \alpha }_{ 2 }\Rightarrow \dfrac { { \alpha }_{ 1 } }{ { \alpha }_{ 2 } } =\dfrac { { I }_{ 2 } }{ { I }_{ 1 } }$

$=\dfrac { \dfrac { 2 }{ 5 } M{ R }^{ 2 } }{ M{ R }^{ 2 } } =\dfrac { 2 }{ 5 } \Rightarrow { \alpha }_{ 2 }=\dfrac { 5 }{ 2 } { \alpha }_{ 1 }$

$\Rightarrow { \alpha }_{ 2 }=2.5{ \alpha }_{ 1 }$ ....(iii)
If

${ \omega }_{ 1 }$ and

${ \omega }_{ 2 }$ be the angular speed of the cylinder and sphere after time

$t$ , then

${ \omega }_{ 1 }={ \omega }_{ 0 }+{ \alpha }_{ 1 }t$ ....(iv)
and

${ \omega }_{ 2 }={ \omega }_{ 0 }+{ \alpha }_{ 2 }t$

$={ \omega }_{ 0 }+2.5{ \alpha }_{ 1 }t$ .....(v)
where,

${ \omega }_{ 0 }=$ initial angular speed

$\therefore$ From equation (iv) and (v), it is clear that

${ \omega }_{ 2 } > { \omega }_{ 1 }$

$\therefore$ The sphere will acquire more angular speed as compared to that of the cylinder after a given time.

A cord of negligible mass is wound round the rim of fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in the figure. The fly wheel is mounted on a horizontal axle with frictionless bearings. Compute the angular acceleration of the wheel.

0%
$8.25 kgm^2$
0%
$6.25 kgm^2$
0%
$4.8 kgm^2$
0%
$1.0 kgm^2$

Explanation

Given :

$M =20$ kg

$R = 0.2$ m

$F= 25$ N

Using

$\tau = I\alpha$

$\therefore$

$FR = (MR^2)\alpha$

$\implies \alpha = \dfrac{F}{MR}$

Thus, angular acceleration

$\alpha =\dfrac{25}{20\times 0.2} = 6.25$

$rad/s^2$

A bar magnet of moment of inertia

$I$ is vibrated in a magnetic field of induction

$\displaystyle 0.4\times { 10 }^{ -4 }T$ . The time period of vibration is

$12\ s$ . The magnetic moment of the magnet is

$\displaystyle 120\ { Am }^{ 2 }$ . The moment of inertia of the magnet is (in

$\displaystyle kg{ m }^{ 2 }$ ) approximately:

0%
$\displaystyle 2.1{ \pi }^{ 2 }\times{ 10 }^{ -2 }$
0%
$\displaystyle 1.57\times { 10 }^{ -2 }$
0%
$\displaystyle 1728\times { 10 }^{ -2 }$
0%
$\displaystyle 172.8\times { 10 }^{ -4 }$

Explanation

An SHM takes place due to torque acting on magnet due to presence of magnetic field.

$\tau=\mu Bsin\theta\approx \mu B\theta=I\omega^2\theta$

$\implies I=\cfrac{\mu B}{\omega ^2}=\cfrac{\mu BT^2}{4\pi ^2}$

$I=\cfrac{120 \times 0.4 \times 10^{-4} \times 12^2}{4 \times \pi^2}$

$=172.8\times 10^{-4} {kgm}^2$

A wheel of radius

$r$ and moment of inertia

$I$ about its axis is fixed at the top of an inclined plane of inclination

$\theta$ as shown in the figure. A string is wrapped round the wheel and its free end supports a block of mass

$M$ which can slide on the plane. Initially, the wheel is rotating at a speed of

$\omega$ in a direction such that the block slides up the plane. How far will the block move before stopping?

0%
$\dfrac{(I+ Mr^3) {\omega}^2}{4 Mg \sin \theta}$
0%
$\dfrac{(I+ Mr^2) {\omega}^2}{2 Mg \sin \theta}$
0%
$\dfrac{(I+ Mr^2) {\omega}^2}{5 Mg \sin \theta}$
0%
$\dfrac{(I+ Mr^2) {\omega}^2}{7 Mg \sin \theta}$

Explanation

$\triangle u=mglsin\theta$

MOI of system

$=(I+mr^{ 2 })$

Also initial kinetic energy =

$\triangle u$

$\therefore \cfrac { 1 }{ 2 } (I+mr^{ 2 }){ \omega }^{ 2 }=mglsin\theta \\ \Rightarrow l=\cfrac { (I+mr^{ 2 }){ \omega }^{ 2 } }{ 2mgsin\theta }$

A rod is hinged at centre and rotated by applying the torque starting from rest ,the power developed by the torque with respect to time is of

0%
cubic nature
0%
quadratic nature
0%
linear nature
0%
sinusoidal nature

Explanation

$I\alpha =\tau \\ \Rightarrow \alpha =\cfrac { \tau }{ I } \\ \omega =\alpha t=\cfrac { \tau t }{ I } \\ \rho =\cfrac { dKE }{ dt } =\cfrac { d\left( \cfrac { 1 }{ 2 } I{ \omega }^{ 2 } \right) }{ dt } =\cfrac { d\left( \cfrac { 1 }{ 2 } \cfrac { { \tau }^{ 2 }{ t }^{ 2 } }{ t } \right) }{ dt } =\cfrac { { \tau }^{ 2 } }{ I } t$

Thus we see that

$\rho \propto t$

Hence the power devlopedby the torque with respect to time is of linear nature.

A particle of mass

$m$ is subjected to an attractive central force of magnitude

$k/r^2$ ,

$k$ being a constant. If at the instant when the particle is at an extreme position in its closed orbit, at a distance

$a$ from the centre of force, its speed is

$(k/2ma)$ , if the distance of other extreme position is

$b$ . Find

$a/b.$

0%
4
0%
3
0%
5
0%
6

Explanation

The attractive force in orbit,

$k_r=-\dfrac{k}{r^2}$ . . . . .(1)

Potential energy,

$U=-\dfrac{k}{r}$ . . . . . .(2)

By the conservation of at energy at extreme position

$a$ and

$b$

$K_a+U_a=K_b+U_b$

$\dfrac{1}{2}mv_1^2-\dfrac{k}{a}=\dfrac{1}{2}mv_2^2-\dfrac{k}{b}$ . . . . .(3)

At the instant when the particle is at an extreme position

$a$ in the a closed orbit then it's speed is,

$v_1=\sqrt{\dfrac{k}{2ma}}$ . . . .(4)

By the conservation of angular momentum,

$mv_1a=mv_2b$

$v_2=\dfrac{a}{b}v_1=\dfrac{a}{b}\sqrt{\dfrac{k}{2ma}}$ . . . . . .(5)

Substitute

$v_1$ and

$v_2$ in equation (3), we get

$a/b=3$

The correct option is B.

Find the moment of inertia of the rod AC about an axis BD as shown in figure. Mass of the rod is m and length is L

0%
$\dfrac{3mL^2 sin^2 \alpha}{2}$
0%
$\dfrac{mL^2 sin^2 \alpha}{3}$
0%
$\dfrac{3mL^2 sin^2 \alpha}{4}$
0%
$\dfrac{3mL^2 sin^2 \alpha}{5}$

A hoop of mass

$m$ is projected on a floor with linear velocity

$v_{0}$ and reverse spin

$\omega_{0}$ . The coefficient of friction between the hoop and the ground is

$\mu$ .
a. Under what condition will the hoop return back?
b. How far will it go?
c. How long will it continue to slip when its centre of mass becomes stationary?
d. What is the velocity of return?

0%
a. $\omega_{0} > \dfrac {v_{0}}{R}$ .
b. $\dfrac {v_{0}^{2}}{R}$ ,
c. $\dfrac {R}{3\mu_{g}}\left (\omega_{0} - \dfrac {v_{0}}{R}\right )$ ,
d. $\dfrac {\omega_{0}R}{2} - \dfrac {v_{0}}{2}$ .
0%
a. $\omega_{0} > \dfrac {v_{0}}{R}$ .
b. $\dfrac {v_{0}^{3}}{R}$ ,
c. $\dfrac {R}{2\mu_{g}}\left (\omega_{0} - \dfrac {v_{0}}{R}\right )$ ,
d. $\dfrac {\omega_{0}R}{4} - \dfrac {v_{0}}{2}$ .
0%
a. $\omega_{0} > \dfrac {v_{0}}{R}$ .
b. $\dfrac {v_{0}^{2}}{R}$ ,
c. $\dfrac {R}{2\mu_{g}}\left (\omega_{0} - \dfrac {v_{0}}{R}\right )$ ,
d. $\dfrac {\omega_{0}R}{3} - \dfrac {v_{0}}{2}$ .
0%
a. $\omega_{0} > \dfrac {v_{0}}{R}$ .
b. $\dfrac {v_{0}^{2}}{R}$ ,
c. $\dfrac {R}{2\mu_{g}}\left (\omega_{0} - \dfrac {v_{0}}{R}\right )$ ,
d. $\dfrac {\omega_{0}R}{2} - \dfrac {v_{0}}{2}$ .

Three identical rods, each of mass

$m$ and length

$l$ , form an equilateral triangle. Moment of inertia about one of the sides is

0%
$\cfrac { m{ l }^{ 2 } }{ 6 }$
0%
$m{ l }^{ 2 }\quad$
0%
$\cfrac { 3m{ l }^{ 2 } }{ 4 }$
0%
$\cfrac { 2m{ l }^{ 2 } }{ 3 }$

Explanation

$M.I$ of

$AB=0$

$M.I$ of

$BC=\dfrac{ml^2}{3}\sin ^230$

$=\dfrac {ml^2}{4}\times \dfrac{1}{3}$

$M.I$ of

$AC=\dfrac{ml^2}{12}$

$M.I=\dfrac{2ml^2}{12}$

$=\dfrac{ml^2}{6}$

Putting in first equation,

$\Rightarrow mg-T=m\dfrac{2T}{M}$

$\Rightarrow T+\dfrac{2Tmn}{M}=mg$

$\Rightarrow T\left(\dfrac{M+2m}{M}\right)=mg$

$\Rightarrow T=\dfrac{mgM}{M+2m}$

Hence, the answer is not given in options.

Figure below shows the variation of the moment of inertia of a uniform rod about an axis normal to its length with the distance of the axis from the end of the rod. The moment of inertia of the rod about an axis passing through its centre and perpendicular to the its length is then

0%
$0.5 kg-m^2$
0%
$0.15 kg-m^2$
0%
$0.10 kg-m^2$
0%
$0.05 kg-m^2$

Explanation

$\dfrac{mL^2}{3} = 0.2$

$\dfrac{mL^2}{12} = I$

$\dfrac{I}{0.2} = \dfrac{1}{4} \Rightarrow I = 0.05$

A plot of angular momentum of a rigid body (along y axis) about an axis with time (along x axis) gives rise to a straight line, whose equation is given by y = 3x+Find the torque acting on the body

0%
8 N-m
0%
3 N-m
0%
30 N-m
0%
0.3 N-m

Explanation

We know that torque= dL/dt. Integrating, we get

$L=L_0+\tau t$ .

This equation resembles the equation of a straight line y= mx +C. Comparing with y=3x+1, we get the torque = 3 N-m

The correct option is (b)

A square frame ABCD is formed by four identical rods each of mass 'm' and length 'T'. This frame is in X - Y plane such that side AB coincides with X - axis and side AD along Y - axis. The moment of inertia of the frame about X - axis is

0%
$\dfrac{5ml^2}{3}$
0%
$\dfrac{2ml^2}{3}$
0%
$\dfrac{4ml^2}{3}$
0%
$\dfrac{ml^2}{12}$

Explanation

Moment of inertia of rod AD and BC are each

$\dfrac{1}{3}mT^2$

For rod AB, moment of inertia is zero since it is on X-axis and hence r=0.

For, rod CD, every element is at distance

$T$ from X-axis and hence its moment of inertia is

$mT^2$ .

Hence,

$I=2\times \dfrac{1}{3}mT^2+mT^2$

$\implies I=\dfrac{5}{3}mT^2$

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

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 11 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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