MCQ Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 6

All discs irrespective of their mass and radius move with same acceleration along an incline.

0%
True
0%
False

Explanation

This statement is true, since the acceleration of a disc along an incline is given by

$a =2g sin \theta / 3$ . Thus, any disc is independent of the radius

A ring of mass m and radius r rolls on an inclined plane. A torque is produced in the ring due to

0%
$mg sin \theta$ and the radius of the ring
0%
the perpendicular distance between the force and the point of contact and the force $mg$ acting on it
0%
the perpendicular distance between the force and the point of contact and the normal force acting on the object
0%
the perpendicular distance between the force and the point of contact and the force $mg sin \theta$ acting on it

Explanation

Torque is determined using the formula

$r\times F=rF\sin\theta$

Here

$r\sin\theta$ stands for perpendicular distance between the force and the point of contact as shown in figure

1016194_1006653_ans_4f437444d02141ab8bd269dc5b8b26c3.PNG

A vertical rod AC, with C at ground and A being the top most point is moving in such a way, that A moves with a speed of 8 m/s, mid point of the rod B moves with a speed of 4 m/s. along the same direction, with what velocity will the point C move

0%
4 m/s
0%
8 m/s
0%
0 m/s
0%
1 m/s

The angular momentum of a particle rotating with an angular velocity

$\omega$ distant r from a given origin is I. If the origin is shifted by 2r, then the new angular momentum of the particle with same angular velocity will be

0%
2I
0%
9I
0%
4I
0%
3I

Explanation

The angular momentum (L) of a particle of mass m distant r from a fixed origin, rotating with an angular velocity

$\omega$ is given by

$L=I\omega = mr^2 \omega$

If the distance is increased by 2r, then the total distance from the fixed orgin will be 3r

Thus, angular momentum increases by 9 times

Thus, the correct option is b

Statement related to center of gravity that is incorrect is

0%
The center of gravity of an object is defined as the point through which its whole weight appears to act.
0%
The center of gravity is sometimes confused with the center of mass.
0%
The center of gravity always lies inside the object.
0%
For an object placed in a uniform gravitational field, the center of gravity coincides with the center of mass.

In the image shown . AC is a rigid rod of length 12 cms rotating about point P, whose velocities are shown in the figure. Find the distance PC = x

0%
12 cms
0%
8 cms
0%
4 cms
0%
2 cms

Explanation

Since the rod is rotating about the point P, the angular velocities of all the points are same.

Thus, the linear velocities of point C and A are related to the angular velocity

$\omega$ as

$x \omega=3$ and

$(d+x)\omega$ =12.

Solving both the equations, we get

$x=4cm$

Which of the below statement is correct

0%
All points in an object move in a straight line in rectilinear motion, while moves in a curved path when rotated about a fixed point
0%
All points in an object move with the same linear speed when rotated about a fixed point
0%
All points in an object move in a straight line, while the object moves in a curved path pivoted about a fixed point
0%
All points in an object moves in a curved path, while the object is moving along a straight line

A disc rolling down the incline has lesser acceleration as compared to the same disc sliding down the incline

0%
True
0%
False

Explanation

A disc sliding will have an acceleration equal to

$g sin \theta$

A disc rolling on the other hand will have an acceleration less than the one sliding.This can be seen by solving the torque equations as shown below:

Torque produced due to CM at the contact point is

$mgR sin \theta= I \alpha$
Substituting the values, we get,

$mgR sin \theta= (mR^2/2) (a/R) \implies a= (2g/3) sin \theta$

Thus, acceleration while rolling is less than sliding

Thus the statement is correct

Only rotating bodies can have angular momentum.

0%
True
0%
False

Moment of inertia of a thin rod of mass

$m$ and length

$l$ about at axis passing though a point

$l/4$ from one and perpendicular to the rod is:

0%
$\frac { { ml }^{ 2 } }{ 12 }$
0%
$\frac { { ml }^{ 2 } }{ 13 }$
0%
$\frac { { 7ml }^{ 2 } }{ 48 }$
0%
$\frac { { ml }^{ 2 } }{ 9 }$

A constant torque is applied to a rigid body, whose moment of inertia is

$4 kg-m^2$ around the axis of rotation. If the wheel starts from rest and attains an angular velocity of 20 rad/s in 10 s, what is the applied torque

0%
$18 N-m$
0%
$8 N-m$
0%
$4 N-m$
0%
$6 N-m$

Explanation

Torque =

$I \alpha=4(20-0)/10=8 N-m$

The correct option is (b)

Three objects of same mass but different geometries, capable of rotating have radius of gyration as 0.2, 0.5 and 0.A torque is applied to these objects when they are rotating with constant angular velocity. Which object will have a larger response time for showing the change in their angular velocity

0%
Object with a radius of gyration of 0.7 will take a long time to respond to the torque
0%
Object with a radius of gyration of 0.5 will take a long time to respond to the torque
0%
Object with a radius of gyration of 0.2 will take a long time to respond to the torque
0%
All objects with have same response time

A ball moving with a momentum of

$5kgm/s$ strikes against a wall at an angle of

${45}^{o}$ and is reflected at the same angle. The change in momentum will be (kg-m/s) :

0%
$7.07$
0%
$14.14$
0%
$8.00$
0%
$7.5$

Explanation

$\Delta p=2mv\cos { \theta } =2p\cos { \theta } =2\times 5\times \cfrac { 1 }{ \sqrt { 2 } } =7.07$

Two blocks of masses

$10\ kg$ and

$4\ kg$ are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of

$14\ m/s$ to the heavier block in the direction of the lighter block, The velocity of the centre of mass is:

0%
$30\ m/s$
0%
$20\ m/s$
0%
$10\ m/s$
0%
$5\ m/s$

Explanation

Formula,

$V_{cm}=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$

$=\dfrac{10\times 14+4\times 0}{10+4}$

$=10m/s$

A jar filled with two non mixing liquids

$1$ and

$2$ having densities

$\rho_{1}$ and

$\rho_{2}$ respectively. A solid ball, made of a material of density

$\rho_{3}$ , is dropped in the jar. It comes to equilibrium in the position shown in the figure

0%
$\rho_{3} < \rho_{1} < \rho_{2}$
0%
$\rho_{1} < \rho_{3} < \rho_{2}$
0%
$\rho_{1} < \rho_{2} < \rho_{3}$
0%
$\rho_{3} < \rho_{2} < \rho_{1}$

Explanation

Basic principal of density is that higher density material settle down and lower density floats. So here we can see that

$\rho_1<\rho_3<\rho_2$

$\rho_3$ density of solid base

$\rho_1 \&\rho_2$ density of liquids

Using the parallel axes theorem, find the

$M.l.$ of a sphere of mass

$m$ about an axis that touches it tangentially. Give that

$I_{cm}=\dfrac {2}{5}mr^{2}$ .

0%
$\dfrac {7}{5}\ mr^{2}$
0%
$\dfrac {7}{15}\ mr^{2}$
0%
$\dfrac {8}{15}\ mr^{2}$
0%
$\dfrac {9}{15}\ mr^{2}$

Find the angular momentum of a particle of mass

$m$ describing a circle of radius

$r$ with angular speed

$\omega$ .

0%
$2m{ \omega }r^{ 2 }$ .
0%
$m{ \omega }r^{ 2 }$ .
0%
$3m{ \omega }r^{ 2 }$ .
0%
$4m{ \omega }r^{ 2 }$ .

Explanation

We know,

Angular momentum

$(L)=I\omega$

For a particle of mass

$m$ revolving around Radius

$R$

$I=mR^2$ (about centre of revolution)

$L=mR^2\omega$

Hence option

$\textbf B$ s correct answer.

A wheel having moment of inertia

$4\ kg\ {m}^{2}$ about its symmetrical axis, rotates at rate of

$240$ rpm about it. The torque which can stop the rotation of the wheel in one minute is:

0%
$\dfrac {5\pi}{7}Nm$
0%
$\dfrac {8\pi}{15}Nm$
0%
$\dfrac {2\pi}{9}Nm$
0%
$\dfrac {3\pi}{7}Nm$

Explanation

$\alpha =\dfrac{240\times 2\pi}{60\times (60)}$

$\alpha =\dfrac{8\pi}{60}$
now

$T=I\alpha$

$=4\times \dfrac{8\pi}{60}$

$=\dfrac{8\pi}{15}$

The linear momentum

$P$ of a particle varies with time as follow

$P=a+{bt}^{2}$ . Where

$a$ and

$b$ are constants. The net force acting on the particle is:

0%
Proportional to $t$
0%
Proportional to ${t}^{2}$
0%
Zero
0%
Constant

Explanation

Given Momentum

$P=a+bt^{2}$ .

We know force

$F=\cfrac{dp}{dt}=rate\,of\,change\,of\,momentum$ .

$\Rightarrow F=2bt$

$Force \propto time$ .

A block of mass

$2\ kg$ is moving with a velocity of

$2\hat { i } -\hat { j } +3\hat { k }\ m/s$ . Find the magnitude and direction of momentum of the block with the

$x-$ axis.

0%
$2\sqrt {14}\ kg\ m/s$ , $\tan ^{ -1 }{ \left( \sqrt { \dfrac { 5 }{ 7 } } \right) }$
0%
$2\sqrt {7}\ kg\ m/s$ , $\tan ^{ -1 }{ \left( \sqrt { \dfrac { 2 }{ 7 } } \right) }$
0%
$2\sqrt {14}\ kg\ m/s$ , $\tan ^{ -1 }{ \left( \sqrt { \dfrac { 2 }{ 7 } } \right) }$
0%
$2\sqrt {9}\ kg\ m/s$ , $\tan ^{ -1 }{ \left( \sqrt { \dfrac { 3 }{ 7 } } \right) }$

Explanation

Given,

$m=2 kg$

$\vec v=2\hat i- \hat j+3 \hat k$

$m/s$

Momentum ,

$\vec P=m \vec v$

$\vec P=2(2\hat i-\hat j+3\hat k)=4\hat i-2\hat j+6\hat k kg$

$m/s$

It's magnitude ,

$|P|=\sqrt{(4)^2+(-2)^2+(6)^2}$

$|P|=2\sqrt{14} kg$

$m/s$

The direction of momentum with the x-axis is,

$tan\theta=\dfrac{4 }{\sqrt{(4)^2}+(-2)^2+(6)^2}$

$\theta = tan^{-1} (\sqrt{\dfrac{2}{7}})$

The correct option is C.

The density of rod

$AB$ increases linearly from

$A$ to

$B$ . Its midpoint is

$O$ and its centre of mass is at

$C$ . Four axes pass through

$A, B, O$ and

$C$ , all perpendicular to the length of the rod. The moments of inertia of the rod about these axes are

${I}_{A}$ ,

${I}_{B}$ ,

${I}_{C}$ respectively.

0%
${I}_{A}>{I}_{B}$
0%
${I}_{A}<{I}_{B}$
0%
${I}_{B}>{I}_{C}$
0%
${I}_{O}<{I}_{C}$

Explanation

$I_C$ is least and

$C$ lies between

$O$ and

$B$ as density is increasing from

$A$ to

$B$

$I_A > I_B$ more mass is concentrate towards

$B$ .

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

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

0%
$I_{0}+ML^{2}/2$
0%
$I_{0}+ML^{2}/4$
0%
$I_{0}+2ML^{2}$
0%
$I_{0}+ML^{2}$

Explanation

Here, we apply theorem of parallel axis

${I_{ 0}}$ moment of inertia of Red length

$L$ about its axis

${I_B}$ moment of inertia of Rod at its end

$\begin{array}{l} { I_{ B } }={ I_{ 0 } }+M{ \left( { \frac { L }{ 2 } } \right) ^{ 2 } } \\ { I_{ B } }={ I_{ 0 } }+\frac { { M{ L^{ 2 } } } }{ 4 } \end{array}$

1103487_1023714_ans_a897f31fa7d34a939243d3c29f6be1ea.PNG

The

$M.I.$ of a thin rod of length

$\ell$ about the perpendicular axis through its centre is

$I$ . The

$M.I.$ of the square structure made by four such rods about a perpendicular axis to the plane and through the centre will be :

0%
$4\ I$
0%
$8\ I$
0%
$12\ I$
0%
$16\ I$

Explanation

Given Moment of inertia (MOI) ,

$I=\dfrac{1}{12}ml^2$

$\Rightarrow ml^2= 12I$

Using parallel axis theorem of moment,

$I=I_{cm}+ mx^2$ , Where

$I_{cm}$ is MOI about centre of mass and

$x$ is the distance between centre of mass and axis of rotation.

$\therefore$ Net MOI

$= 4\left ( \dfrac{1}{12}ml^2+m\left (\dfrac{l}{2}\right )^2\right) =\dfrac{4}{3}ml^2=16I$

A disc rolls down a plane of length

$L$ and inclined at angle q, without slipping. its velocity on reaching the bottom will be :-

0%
$\sqrt{\dfrac{4gL \sin \theta}{3}}$
0%
$\sqrt{\dfrac{2gL \sin \theta}{3}}$
0%
$\sqrt{\dfrac{10gL \sin \theta}{7}}$
0%
$\sqrt{4gL \sin \theta}$

A rubber ball of mass

$10 gm$ and volume

$15 cm^3$ dipped in water to a depth of

$10 m$ . Assuming density of water uniform throughout the depth, find
(a) the acceleration of the ball, and
(b) the time taken by it to the surface if it is released from rest. take

$g = 980 cm/s^2$ .

0%
(a) $5.9 m/s^2$ (b) $2.02$ sec.
0%
(a) $4.9 m/s^2$ (b) $2.02$ sec.
0%
(a) $4.9 m/s^2$ (b) $12.02$ sec.
0%
(a) $8.9 m/s^2$ (b) $2.02$ sec.

Explanation

Given,

$m=10gm$

$V=15cm^3$

$H=10m$

$g=980 cm/s^2$

$\rho=1kg/cm^2$

The effective upward force on the ball,

$F=F_B-mg =ma$

$a=\dfrac{1}{m}(F_B-mg)$

$a=\dfrac{1}{2}(\rho gh-mg)$

$a=\dfrac{1}{10}(15\times 1\times 980-10\times 980)$

$a=490cm/s^2=4.9 m/s^2$

From the kinematics,

$s=ut+\dfrac{1}{2}at^2$ initial speed of ball is zero.

$10=\dfrac{1}{2}\times 4.9\times t^2$

$t=2.02sec$

Thus, the correct option is B.

A spherical shell and a solid cylinder of same radius rolls down an inclined plane. The ratio of their accelerations will be :-

0%
$15 : 14$
0%
$9 : 10$
0%
$2 : 3$
0%
$3 : 5$

Explanation

The formula for

$ac{c^n}$ of an object

$i.e,$ sphere cylender etc

$.)$ of mass

$,$ radius and

$MOI$ given respectively is given by

$a = \dfrac{{g\sin \theta }}{{1 + \dfrac{I}{{M{R^2}}}}}$

$I$ of spherical shell

$,$

${I_{shell}} = \dfrac{{2M{R^2}}}{3}$

$I$ of solid cylinder

$,$

${I_{cyl}} = \dfrac{{M{R^2}}}{2}$

Therefore

$,$

${a_{shell}} = \dfrac{{g\,\sin \theta }}{{1 + \dfrac{{2M{R^2}}}{{3M{R^2}}}}} = \dfrac{{g\,\sin \theta }}{{\dfrac{5}{3}}}$

${a_{cylinder}} = \dfrac{{g\,\sin \theta }}{{1 + \dfrac{{M{R^2}}}{{2M{R^2}}}}}$

Therefore

$, =$

$\dfrac{3}{5} \times \dfrac{3}{2} = \dfrac{9}{{10}} = 9:10$

hence

$,$

option (B) is correct

1148191_1024460_ans_212e88e3edb949488343a189ac56b329.PNG

A thin rod of mass

$6m$ and length

$6L$ is bent into regular hexagon. The

$M.I$ of the hexagon about a normal axis to its plane and through centre of system is

0%
$m{ L }^{ 2 }$
0%
$3m{ L }^{ 2 }$
0%
$5m{ L }^{ 2 }$
0%
$11m{ L }^{ 2 }$

Explanation

Each rod has mass

$m$ and length

$L$ and distance of their C.O.M from center of hexagon

$(d) = L\sin 60^{o}=\dfrac{L\sqrt{3}}{2}$ (Internal angle of regular hexagon

$= 120^{o})$

Moment of inertia of one rod about axis through its C.O.M

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

So its moment of inertia about axis through center of hexagon and perpendicular to the plan is:

$I_{centre} = I+md^{2}$ (Parallel axis theorem)

$=\displaystyle \frac{mL^{2}}{12} + \frac{3mL^{2}}{4}=\frac{5mL^{2}}{6}$

So for

$6$ rods the net moment of inertia this axis will be:

$6 \times \dfrac{5mL^2}{6} = 5mL^2$

A spherically symmetric gravitational system of particles has a mass density

$\rho =\left\{\begin{matrix} \rho_0,\ \ r\leq R \\ 0,\ \ r > R\end{matrix}\right.$ where

$\rho_0$ is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed V as a function of distance

$r(0 < r < \infty)$ from the centre of the system is represented by.

Explanation

Given mass density

$P=\left\{ { P }_{ 0 }\quad for\quad r\le R\\ O\quad for\quad r>R \right\}$

Since gravitational depends on mass, which is zero for

$r>R$ , hence it only acts for

$r\le R$ .

${ \vec { F } }_{ graw }={ \vec { F } }_{ circular }$

$\Rightarrow \dfrac { { GM }_{ sphere }\times M }{ { R }^{ 2 } } =\dfrac { M{ \left( e \right) }^{ 2 } }{ R }$

$\Rightarrow { V }^{ 2 }\alpha \dfrac { { M }_{ sphere } }{ R }$ since

$r\le R$

$\Rightarrow { V }^{ 2 }\alpha \dfrac { { R }^{ 3 } }{ R } \alpha { R }^{ 2 }\Rightarrow V\alpha R$ straight line

$+ve$ slope

Beyond

$r=R$ , mass density

$=0\Rightarrow$ acceleration due to gravity

$=0\Rightarrow$ velocity stays constant.

974704_1040481_ans_736c071a12b247c3a35746d162ca399c.jpg

Three identical masses each of mass

$1kg$ , are placed at the corners of an equilateral triangle of side

$l$ . Then the moment of inertia of this system about an axis along one side of the triangle is

0%
3 $Ml^2$
0%
$l^2$
0%
$\dfrac{3}{4}$ $Ml^2$
0%
$\dfrac{3}{2}$ $Ml^2$

A thin rod of linear pass density

$\lambda$ at right angle at its mid point

$(C)$ and fixed to points

$A$ and

$B$ such that it can rotate about an axis passing through

$AB$ .The moment of inertia about an axis passing through

$AB$ is :

0%
$\dfrac { \lambda l^3} {6 \sqrt2}$
0%
$\dfrac { \lambda l^3} {2 \sqrt2}$
0%
$\dfrac { \lambda l^3}{4}$
0%
$\dfrac { \lambda l^3}{ \sqrt2}$

Explanation

We have length of each rod

$=\dfrac{1}{\sqrt 2}$

Let a small part

$dx$ be at a distance

$x$ from the of first rod it's small mass

$dm=\lambda dx$ (Linear density of rod

$= \lambda$ )

distance of dm from axis

$=x \cos 45^o=\dfrac{x}{\sqrt 2}$

So moment of interia of

$dm$ about the axis is :

$dI=dm \left(\dfrac{x}{\sqrt 2}\right)^2=\dfrac{\lambda x^2dx}{2}$

$I=\displaystyle \int^{\dfrac{L}{\sqrt {2}}}_{0}\dfrac{\lambda x^2dx}{2}=\dfrac{\lambda}{2}\dfrac{x^3}{3}^{\dfrac{L}{\sqrt 2}}_0=\dfrac{\lambda L ^3}{3 \times 4\sqrt{2}}$

Now the second rod will also have same moment of inertia about the axis

Hence net moment of inertia

$=2 \dfrac{\lambda L^3}{3 \times 4 \sqrt 2}=\dfrac{\lambda L^3}{6 \sqrt 2}$

Initially two stable particles

$x$ and

$y$ start moving towards each other under mutual attraction. If at one time the velocities of

$x$ and

$y$ are

$V$ and

$2V$ respectively, what will be the velocity of centre of mass of the system?

0%
$V$
0%
Zero
0%
$\cfrac{V}{3}$
0%
$\cfrac{V}{5}$

Explanation

Since they are moving under mutual force at attraction their internal forces cancel each other and

$M\overrightarrow { { a }_{ cm } } ={ F }_{ external }\\ \overrightarrow { { F }_{ external } } =0\quad \\ \overrightarrow { { a }_{ cm } } =0\\ Since\quad initial\quad velocity\quad =0\\ Final\quad velocity\quad =0\quad \\ Since\quad \overrightarrow { a } =0$

A smooth horizontal rod passes through two identical rings, each of mass

$m$ . Rings are connected to a block of same mass through two strings of length

$2l$ and

$l$ as shown in figure. Block is released, when block is crossing lowest point its velocity is

$2\ m/s$ , velocity of left ring is

0%
$4 \ m/s$
0%
$2 \ m/s$
0%
$3 \ m/s$
0%
$None\ of\ these$

Consider the following two statements:
(A) The linear momentum of a particle is independent of the frame of reference
(B) The kinetic energy of a particle is independent of the frame of reference

0%
Both A and B are true
0%
A is true but B is false
0%
A is false but B is true
0%
both A and B are false

If is the angular momentum is

$L$ and

$I$ is the moment of inertia of a rotating body, then

$\dfrac { L^ 2 }{ 2I }$ represents its ____________

0%
rotational PE
0%
total energy
0%
rotational KE
0%
translation KE

Explanation

For rotating body,

given that:

Moment of inertia = $I$

We know that,

Angular momentum $L= I\omega$

Now,

Rotational kinetic energy = $\dfrac{\left( I{{\omega }^{2}} \right)}{2}$

So, rotational kinetic energy = $\left( \dfrac{{{L}^{2}}}{2I} \right)$

Hence, $\left( \dfrac{{{L}^{2}}}{2I} \right)$ is the rotational kinetic energy.

Find the moment of inertia of the triangular lamina of mass M about the axis of rotation AA' shown in the figure:

0%
$\frac{Ml^2}{3}$
0%
$\frac{Ml^2}{6}$
0%
$\frac{2}{3} Ml^2$
0%
$\frac{1}{3} Ml^2$

Explanation

mass per unit area =

$M \times \dfrac{al}{2}$ where a= base which is AA' =l(given)

M.O.I

$I = \dfrac{1}{2}\times \dfrac{M}{2}al\times \dfrac{1}{2}axdx$

where

$x$ is te small strip along 'l' from the base a(AA')

$M.O.I = \dfrac{1}{2}\times \dfrac{M}{l} \int x^2dx$ =

$\dfrac{M}{2l}[\dfrac{x^3}{3}] = \dfrac{M}{2l}[\dfrac{l^3}{3}] = \dfrac{Ml^2}{6}$

A uniform metal disc of radius R is taken and out of it a disc of diameter

$\dfrac{R}{2}$ is cut off from the end.The centre of mass of the remaining part will be :

0%
$\dfrac{R}{28}$ from the centre
0%
$\dfrac{R}{3}$ from the centre
0%
$\dfrac{R}{5}$ form the centre
0%
$\dfrac{R}{6}$ from the centre

Explanation

No correct option.

Given,

$Radius =R,D=\dfrac{R}{2}$

Let origin be the center

$0=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}$

Mass of the diameter

$\dfrac{R}{2}=\dfrac{m}{8}$

$0=\dfrac{\dfrac{7m}{8}X_1+\dfrac{m}{8}(\dfrac{R}{4})}{\dfrac{7m}{8}+\dfrac{m}{8}}=\dfrac{7m}{8}x_1+\dfrac{m}{8}\times\dfrac{R}{4}\Rightarrow x_1=\dfrac{-R}{28}$

Th moment of inertia of a rod of mass

$M$ and length

$L$ about an axis passing through on edge of perpendicular to its length will be :-

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

Explanation

Given,

Mass of rod $=M$

Length of Rod $=L$

Distance of rods edge from center $=\dfrac{L}{2}$

Moment of inertia of rod at center ${{I}_{c}}=\dfrac{1}{12}M{{L}^{2}}$

For moment of inertia at edge apply parallel axis theorem. ${{I}_{e}}={{I}_{c}}+M{{\left( \dfrac{L}{2} \right)}^{2}}$

$I=\dfrac{1}{12}M{{L}^{2}}+M{{\left( \dfrac{L}{2} \right)}^{2}}=\dfrac{1}{3}M{{L}^{2}}$

moment of inertia at edge is $\dfrac{1}{3}M{{L}^{2}}$

A small block slides with velocity

$0.5 \sqrt gr$ on the horizontal friction less surface as shown in the figure. The block leaves the surface at point

$C$ . The angle

$\theta$ in the figure is:

0%
$\cos^{-1} (4/9)$
0%
$\cos^{-1} (3/4)$
0%
$\cos^{-1} (1/2)$
0%
$none\ of\ the\ above$

Explanation

Vertical displacement, $s=r(1-\cos \theta )$

Apply kinematic equation,

${{V}^{2}}={{u}^{2}}+2as=\left[ {{\left( 0.5\sqrt{gr} \right)}^{2}}+2gr(1-\cos \theta ) \right]$

Kinetic energy = Potential Energy

$\dfrac{1}{2}m{{v}^{2}}=mgr\left( 1-\cos \theta \right)$

${{v}^{2}}=2gr(1-\cos \theta )\,\,.........\,\,(1)$

Normal reaction on block

$N=mg\cos \theta -\dfrac{m{{v}^{2}}}{r}$

$0=mg\cos \theta -\dfrac{m{{v}^{2}}}{r}$

$0=mg\cos \theta -\dfrac{m{{\left( 0.5\sqrt{gr} \right)}^{2}}+2gr(1-\cos \theta )}{r}$

$0=mg\cos \theta -2.25mg+2mg\cos \theta$

$\cos \theta =\dfrac{3}{4}$

$\theta ={{\cos }^{-1}}\left( \dfrac{3}{4} \right)$

Hence, at angle ${{\cos }^{-1}}\left( \dfrac{3}{4} \right)$ , block will leave the surface.

Two blocks of masses

$2kg$ and

$1kg$ respectively are tied to the ends of a strings which passes over a light frictionless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by centre of mass in

$2$ seconds is:(

$g=10m/s^2$ )

0%
$1.42m$
0%
$2.22m$
0%
$3.12m$
0%
$3.33m$

Explanation

Given,

Two blocks having masses

$2kg,1kg,t=2s$

So,

$a=\dfrac{(2-1)mg}{(2+1)m}=\dfrac{g}{3} a_{CM}=\dfrac{2\times\dfrac{g}{3}-1\dfrac{g}{3}}{2+1}=\dfrac{g}{9}$

So,

$S=ut +\dfrac{1}{2}at^2=\dfrac{1}{2}\times\dfrac{g}{2}\times4=\dfrac{20}{9}=2.22m$

1000415_1061653_ans_5638af77e21640b1b566123cbd51164c.png

In applying the equation for motion with uniform angular acceleration

$\omega =\omega _0+\alpha t$ the radian measure

0%
Must be used for both $\omega$ and $\alpha$
0%
May be used for both $\omega$ and $\alpha$ `
0%
may be used for $\omega$ but not $\alpha$
0%
cannot be used for both $\omega$ and $\alpha$

Two particle A and B initially at rest move towards each other under a mutual force of attraction. At the instant when velocity of A is v and thata of B is

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

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

The linear and angular acceleration of a particle are 10 m/

$sec^2$ and 5 rad/

$sec^2$ respectively. It will be at a distance from the axis of rotation

0%
50 m
0%
$\frac{1}{2}$ m
0%
1 m
0%
2 m

A solid sphere and a hollow sphere have same mass and radius. Ratio of radius of gyration of the solid sphere to that of the hollow sphere about a tangent axis is:

0%
$\sqrt {\frac{7}{3}}$
0%
$\sqrt {\frac{3}{5}}$
0%
$\sqrt {\frac{5}{3}}$
0%
$\sqrt {\frac{{21}}{5}}$

Explanation

For hollow sphere, movement of inertia

$= \frac{2}{3}M{R^2}$

$M{K^{2\,}}\, = \frac{2}{3}\,M{K^{2\,}}$

or

$K = \,\sqrt {\frac{2}{3}R}$

For solid sphere, movement of inertia

$= \frac{2}{5}M{R^2}$

$M{K_1}^{2\,}\, = \frac{2}{5}\,M{K^{2\,}}$

or

${K_1} = \,\sqrt {\frac{2}{5}R}$

Now,

$\frac{{{K_1}}}{K}\, = \sqrt {\frac{3}{5}}$

Two rods equal mass

$m$ and length

$\ell$ lie along the

$x$ axis and

$y$ axis with their centres origin. What is the moment of inertia of both about the line

$x = y$ :

0%
$\dfrac{m\ell^2}{3}$
0%
$\dfrac{m\ell^2}{4}$
0%
$\dfrac{m\ell^2}{12}$
0%
$\dfrac{m\ell^2}{6}$

Explanation

Moment of inertial of a rod of about an axis passing through centre and perpendicular to length :

$I=\dfrac{M{l}^{2}}{12}$

moment of inertia of the rod about an axis inclined an angle

$\theta$ to the original axis:

${I}^{\prime}=I{\sin}^{2}{\theta}$

The line

$x=y$ makes an angle of

${45}^{\circ}$ about

$x$ and

$y$ axes.

Hence

${I}^{\prime}=I{\sin}^{2}{{45}^{\circ}}=\dfrac{I}{2}$

moment of inertia of both rods about the line passing through

$x=y$

${I}_{two/, rods}=2{I}^{\prime}$

$=2\times\dfrac{I}{2}=I$

$=\dfrac{M{l}^{2}}{12}$

A wheel of mass

$10kg$ has a moment of inertia of

$160kg -m^2$ about its iwn axis. The radius of gyration is

0%
$10m$
0%
$4m$
0%
$5m$
0%
$6m$

A body of mass

$5\ kg$ is acted on by a net force

$F$ which varies with time

$t$ as shown in graph. Then the net momentum in

$SI$ units gained by the body at the end of

$10$ seconds is

0%
$0$
0%
$100$
0%
$140$
0%
$200$

Explanation

We know that the Newton's second law of motion is

$F=\dfrac{\Delta p}{\Delta t}$ or change in momentum

or the

$gain$ in momentum is

$\Delta p=F\Delta t=\text{Area under the F-t graph}$

The net area till the end of

$t=10s$ is consists of a trapezium with parallel side as

$10second$ and

$8-4=4second$

and the separation being

$20N$ so area or momentum is

$\dfrac{sum\times separation}{2}=\dfrac{(10+4)second\times 20N}{2}=140Newtonsecond$

Option c is correct.

A triangular loop of side

$l$ carries a current

$I$ . It is placed in a magnetic field

$B$ such that the plane of the loop in the direction of

$B$ . The torque on the loop is:

0%
Zero
0%
$IBl$
0%
$\dfrac{\sqrt 3}{2} Il^2B^2$
0%
$\dfrac{\sqrt 3}{4} B l^2$

Explanation

Since

$q = {90^0}$

$,$

$\tau = NIAB = 1 \times I \times \left( {\frac{{\sqrt 3 }}{4}{l^2}} \right)B = \frac{{\sqrt 3 }}{4}{l^2}B$

Hence,

option

$(D)$ is correct answer.

A ballet dancer, dancing on a smooth floor is spinning about a vertical axis with her arms folded with angular velocity of

$20\ rad/s$ . When the stretches her arms fully, the spinning speed decreases in

$10\ rad/s$ /. If

$I$ is the initial moment of inertia of the dancer, the new moment of inertia is

0%
$21$
0%
$31$
0%
$I/2$
0%
$I/3$

Explanation

Here, angular momentum is conserved.

Initial angular momentum = Final angular momentum

$I\times 20= I'\times 10$

where,

$I'$ is the new moment of inertia.

Then,

$I'=2I$

The velocity of centre of mass of system of block

$A,B$ &

$C$ is-

0%
$\dfrac {v_{0}}{2}$
0%
$\dfrac {3v_{0}}{5}$
0%
$\dfrac {2v_{0}}{5}$
0%
$\dfrac {v_{0}}{5}$

Explanation

We know that even after collision net momentum of system in horizontal direction will be conserved ,

By definition

$v_{cm}= \dfrac{\int vdm}{\int dm}=\dfrac{mv_0+2mv_0}{2m+2m+m}=\dfrac{3v_0}{5}$

two identical particles move towards each other with velocity

$2v$ and

$v$ respectively. The velocity of centre of mass is-

0%
$v$
0%
$v/3$
0%
$v/2$
0%
$zero$

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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