MCQ Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 7

In moment of inertia of a uniform rod of length

$'l'$ and mass

$"m"$ above an axis passing through one end of rod and inclined at angle

$\theta$ to rod is:

0%
$\dfrac {ml^{2}}{3}\cos^{2}\theta$
0%
$\dfrac {ml^{2}}{3}\sin^{2}\theta$
0%
$\dfrac {ml^{2}}{12}\cos^{2}\theta$
0%
$\dfrac {ml^{2}}{12}\sin^{2}\theta$

A thin uniform circular ring is rolling down an inclined plane of inclination

$30^0$ without slipping.Its linear acceleration along the inclined plane is

0%
$g$
0%
$\dfrac{9}{2}$
0%
$\dfrac{9}{3}$
0%
$\dfrac{9}{4}$

Explanation

Rolling of a body on inclined plane without slipping

$a=\dfrac {9\sin \theta}{1+\dfrac {K^2}{R^2}}$

$K=$ radius of gyration

$\theta =$ angle of inclination

For the circular ring

$K=R$

$\therefore a=\dfrac {9\sin 30}{1+\dfrac {R^2}{R^2}}$

$\boxed {a=\dfrac {9}{4}}$

1417394_1088544_ans_ca445f15f7054cadac2e7b34a9e90551.png

The block of mass

$m_{2}=10\ kg$ is given a sharp impulse so that it acquires a velocity

$v_{0}=30\ m/s$ towards right. Find the velocity of the centre of mass.(

$m_{2}=5\ kg$ and

$k=30\ N/m$ )

0%
$5\ m/s$
0%
$10\ m/s$
0%
$15\ m/s$
0%
$20\ m/s$

Explanation

Mass of block 1 & 2, ${{m}_{1}}=10\,kg\,\,and\,{{m}_{2}}=5\,kg$

Initial velocity 1 & 2, ${{v}_{1}}=30\,m{{s}^{-1}}\,\,and\,\,{{v}_{2}}=0$

From conservation of momentum

Initial momentum = momentum of center of mass

${{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=({{m}_{1}}+{{m}_{2}}){{V}_{cm}}$

$\Rightarrow {{V}_{cm}}=\dfrac{{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\dfrac{10\times 30+5\times 0}{10+0}=20\,m{{s}^{-1}}$

Hence, center of mass velocity is $20\,m{{s}^{-1}}$

Moment of inertia of a uniform rod of length L and mass M, about an axis passing through L/4 from one end and perpendicular to its length is

0%
$\dfrac{7}{36} ML^2$
0%
$\dfrac{7}{48} ML^2$
0%
$\dfrac{11}{48} ML^2$
0%
$\dfrac{ML^2}{12}$

Two like parallel forces

$20\ N$ and

$30\ N$ act at the ends

$A$ and

$B$ of a rod

$1.5\ m$ long. The resultant of the forces will act at the point:

0%
$90\ cm$ from A
0%
$75\ cm$ from B
0%
$20\ cm$ from B
0%
$85\ cm$ from A

Explanation

Consider the point at which resultant of the forces will act, is 'x' m from the point A.

Thus the point is (1.5 - x) m far from the point B.

Thus, using the formula Torque = Distance x Force acting

$20x = 30(1.5 - x)$

or, $2x = 4.5 - 3x$

or, $5x = 4.5$

or, $x = 0.9m or 90cm$

Thus the resultant point will be 0.9 m or 90 cm far from point A

Find out angular velocity of disc . If the disc is confined to roll without slipping at points of contact Radius of a disc is

$5$ cm:-

0%
$\frac { 3v }{10}$
0%
$\frac { 4v }{10}$
0%
$\frac { 3v }{14}$
0%
$\frac { 2v }{9}$

Explanation

$\frac { v }{ 2r-x } =\frac { 2v }{ x }$

$3x=4r$

$x=\frac { 4 }{ 3 } r$

$\omega =\frac { v }{ r } =\frac { 2v }{ \frac { 4 }{ 3 } r } =\frac { 3v }{ 2r } =\frac { 3v }{ 10 } (since r=5)$

The moment of inertia of a thin rod of length L and mass M about an axis passing through a point at a distance

$\dfrac{L}{3}$ from one of its ends and perpendicular to the rod is :-

0%
$\dfrac{7}{48}ML^2$
0%
$\dfrac{ML^2}{9}$
0%
$\dfrac{ML^2}{12}$
0%
$\dfrac{ML^2}{2}$

The moment of inertia of a uniform rod of length

$2l$ and mass

$m$ about an axis

$xx'$ passing through its centre and inclined at an angle

$\alpha$ is

0%
$\cfrac{ml^{2}}{3}\sin^{2}\alpha$
0%
$\cfrac{ml^{2}}{12}\sin^{2}\alpha$
0%
$\cfrac{ml^{2}}{6}\cos^{2}\alpha$
0%
$\cfrac{ml^{2}}{2}\cos^{2}\alpha$

$Wb/m^{-2}$ making an angle

$30^\circ$ with the field. Find the couple acting on it.

0%
$2.5 \ Nm$
0%
$5.5 \ Nm$
0%
$7.5 \ Nm$
0%
$9.0 \ Nm$

A torque of

$2$ newton

$-m$ produced an angular acceleration of

$2\ rad/sec^{2}$ a body. If its radius of gyration is

$2m$ , mass will be-

0%
$2\ kg$
0%
$4\ kg$
0%
$1/2\ kg$
0%
$1/4\ kg$

If linear mass density of a rod of length

$2$ m varies as

$\lambda =3x+2$ , then the position of the centre of gravity of the rod is?

0%
$1$ m
0%
$1.5$ m
0%
$1.2$ m
0%
$2$ m

Explanation

Given length l=2 m

Considering an elemental section dx at a distance x from any one free end

Mass of the section dx=

$\lambda dx$ =(3x+2) dx

Now the position of center of gravity of rod

${ X }_{ C.G }=\dfrac { \int _{ 0 }^{ L }{ xdm } }{ \int _{ 0 }^{ L }{ dm } } \\ =\dfrac { \int _{ 0 }^{ 2 }{ x(3x+2)dx } }{ \int _{ 0 }^{ 2 }{ (3x+2)dx } } \\ =\dfrac { { 2 }^{ 3 }+{ 2 }^{ 2 } }{ 3(\dfrac { { 2 }^{ 2 } }{ 2 } )+{ 2 }^{ 2 } } =1.2\quad m$

Two discs of same mass and same thickness have densities as

$17\ g/{cm}^{3}$ and

$51\ g/{cm}^{3}$ . The ratio of their moment of inertia are in the ratio

0%
$1:3$
0%
$3:1$
0%
$\dfrac{1}{\sqrt [ 3 ]{ 9 } }$
0%
$\sqrt [ 3 ]{ 9 }$

Explanation

$I=\dfrac{1}{2}M{R}^{2}$ ......

$(1)$

Now,

$M=\dfrac{4}{3}\pi{R}^{3}d\rho$ where

$m$ is the mass,

$d$ is the thickness and

$\rho$ is the density

$\Rightarrow R={\left(\dfrac{3M}{4\pi d\rho}\right)}^{\frac{1}{3}}$

$\Rightarrow {R}^{2}={\left(\dfrac{3M}{4\pi d\rho}\right)}^{\frac{2}{3}}$

Hence eqn

$(1)$ becomes,

$I=\dfrac{1}{2}M\times {\left(\dfrac{3M}{4\pi d\rho}\right)}^{\frac{2}{3}}$ .......

$(2)$

Now if

$M$ and

$d$ are remaining constant as per the given condition then

$I\propto\dfrac{I}{{\rho}^{\frac{2}{3}}}$ ......

$(3)$

Given: moment of inertia of first disk of density

${\rho}_{1}=17$ g/

${cm}^{3}$ and

moment of inertia of second disk of density

${\rho}_{2}=51$ g/

${cm}^{3}$

$\dfrac{{I}_{1}}{{I}_{2}}={\left(\dfrac{{\rho}_{2}}{{\rho}_{1}}\right)}^{\frac{2}{3}}$

$={\left(\dfrac{51}{17}\right)}^{\frac{2}{3}}$

$={\left(\dfrac{3}{1}\right)}^{\frac{2}{3}}$

$={9}^{\frac{1}{3}}$

A wheel having moment of inertia

$2\ kgm^{-2}$ about its axis, rotates at

$50\ rpm$ abut this axis. The angular retardation that can stop the wheel in one minute is

0%
$\dfrac {\pi}{36}rad\ s^{-2}$
0%
$\dfrac {\pi}{18}rad\ s^{-2}$
0%
$\dfrac {\pi}{72}rad\ s^{-2}$
0%
$\dfrac {\pi}{9}rad\ s^{-2}$

An object comprises of a uniform ring of radius

$R$ and its uniform chord

$AB$ (not necessarily made of same material) as shown in figure. Which of the following can be the centre of mass of the object.

0%
$\left(\dfrac{R}{3}, \dfrac{R}{3}\right)$
0%
$\left(\dfrac{R}{2}, \dfrac{R}{2}\right)$
0%
$\left(\dfrac{R}{4}, \dfrac{R}{4}\right)$
0%
$\left(\dfrac{R}{\sqrt{2}}, \dfrac{R}{\sqrt{2}}\right)$

Four identical thin rods each of mass M and length l, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is?

0%
$\dfrac{1}{3} Ml^2$
0%
$\dfrac{8}{3} Ml^2$
0%
$\dfrac{2}{3} Ml^2$
0%
$\dfrac{13}{3} Ml^2$

Explanation

Moment of inertia about centre of rod is

$I_0=\dfrac {ml^2}{12}$

By parallel axis theorem for rod

$AB, MI$ about

$0$ ,

$I=I_0 +md^2\quad (d=l/2)$

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

$MI$ due to all four rods,

$I=4I=\dfrac {4ml^2}{3}=\dfrac {8}{3}I_0$

1407159_1098274_ans_b00f465e50384bbd858f517175cde3c8.png

The moment of inertia of a thin uniform rod of mass

$M$ and length

$L$ about an axis perpendicular to its length is

$\dfrac{ML^{2}}{g}$ . The distance of the axis from the centre of the rod is:-

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

Four thin uniform rods of length

$L$ and mass

$m$ are joined to form a square. The moment of inertia of square about an axis along its one diagonal is :

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

Explanation

From Perpendicular axis theorem

$I_z=I_y +I_x\quad \left\{I_x =I_y \right\}$

$\boxed {I_z =2I_x}$

from parallel axis theorem

$I_z =I_{com}+md^2$

$\Rightarrow \ I_z=\dfrac {ML^2}{12}+M\left (\dfrac {L}{2}\right)^2$

$\Rightarrow \ I_z =\dfrac {ML^2}{12}+\dfrac {ML^2}{4}\ \Rightarrow \boxed {I_z =\dfrac {1}{3} ML^2}\to$ For single Ros

for

$4$ rod

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

$\boxed {I_x =\dfrac {I_z}{2}=\dfrac {\dfrac {4}{3}ML^2}{2}=\dfrac {2}{3}ML^2}$

1457812_1112279_ans_51a00d2ea3424230ac543a117f78252a.png

The linear density of a rod of length

$L$ varies as

$\rho=A+Bx$ where

$x$ is the distance from the left end. The distance of centre of mass from

$O$ is

0%
$\cfrac { 3AL+2{ BL }^{ 2 } }{ 3(2A+BL) }$
0%
$\cfrac { 2AL+2{ BL }^{ 2 } }{ 3(2A+BL) }$
0%
$\cfrac { AL+2{ BL }^{ 2 } }{ (2A+BL) }$
0%
$\cfrac { 3AL+2{ BL }^{ 2 } }{(2A+BL) }$

Four identical thin rods each of mass

$M$ and length

$l$ , form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

0%
$\cfrac{2}{3}M{l}^{2}$
0%
$\cfrac{13}{3}M{l}^{2}$
0%
$\cfrac{1}{3}M{l}^{2}$
0%
$\cfrac{4}{3}M{l}^{2}$

Two point like spheres of mass

$5\ kg$ are joined by

$1m$ long weightless rod. The

$M.I.$ in

$kg$ .

$m^{2}$ unit about an axis perpendicular to the rod and through the center of any of the sphere will be (in kg-

$m^{2})$ -

0%
10
0%
2.5
0%
5.0
0%
1.0

Explanation

REF.Image.

Point line spheres

$\Rightarrow$ moment or inertia

of sphere -1 will be zero because its

point line so where mass will be

at the axis of rotation so

$r = 0 \Rightarrow I_1 = 0$ for sphere -2

That of sphere -2 about axis passing

through sphere -1 will be

$I_2 = m_2 r^2 = 5 \times 1^2 = 5 kgm^2$

Net

$\boxed{I = I_1 + I_2 = 0+5 = 5 kgm^2}$

1194624_1116673_ans_b7c65ad3e3074afc8342fed792209b1b.jpg

Two spheres of masses

$4kg$ and

$8kg$ are moving with velocities

$2{ms}^{-1}$ and

$3{ms}^{-1}$ away from each other along the same line. The velocity of centre of mass is

0%
$\cfrac{8}{3}{ms}^{-1}$ towards second sphere
0%
$\cfrac{8}{3}{ms}^{-1}$ towards first sphere
0%
$\cfrac{4}{3}{ms}^{-1}$ towards second sphere
0%
$\cfrac{4}{3}{ms}^{-1}$ towards first sphere

Explanation

Away from each other so

$opposite$ velocities.

Now taking

$second$ sphere as

$positive$ velocity,

we have

$V_{cm}=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}=\dfrac{4(-2)+8(3)}{4 +8}=\dfrac{4}{3}m/s$ positive so towards the second sphere.

A spherical body of radius

$'R'$ rolls on a horizontal surface with linear velocity

$'v'$ . Let

$L_1$ and

$L_2$ be the magnitudes of angular momenta of the body about centre of mass and point of contact

$P$ . Then,

0%
$L_2 = 2 L_1$ ; if radius of gyration $K = R$
0%
$L_2 = 2 L_1$ ; for all cases
0%
$L_2 > 2 L_1$ ; if radius of gyration $K < R$
0%
None of these

Explanation

Angular momentum about center of mass:

$L_{1}=I\omega = MK^2\omega$

angular momentum at the point of contact:

$L_{2}=I\omega + MRv = MK^2\omega + MR(R\omega)$

$\because\, (v=R\omega)$

$L_{2} = MK^2\omega + MR^2\omega$

$L_{2} = M\omega(K^2 + R^2)$

$\therefore\, L_{2}= 2L_{1}\, for K=R$

$L_{2}>L_{1}\, for K>R$

A particle is moving along a straight line parallel to x-axis with constant velocity . Find angular momentum about the origin in vector form

0%
$+ m{v^2}b\hat k$
0%
$mb\hat k$
0%
$2mvb\hat k$
0%
$mvb\hat k$

Explanation

A particle is moving along a straight line

$\parallel$ to

$x$ -axis.

Position vector

$(\hat r)=b\hat j$

Mass of body=

$m$

Velocity of body

$\hat V=V\hat i$

Angular momentum=

$m(\hat r\times \hat V)=mbV\hat k$

The mass per unit length of a non-uniform rod of length

$L$ varies as

$m = \lambda x$ where

$\lambda$ is constant. The centre of mass of the rod will be at :

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

Explanation

$m=\lambda x$

taking differentiable on both side

$dm=\lambda xdx$

$x_{cm}=\dfrac{\displaystyle\int^L_0xdm}{\displaystyle\int^L_0dm}$

$=\dfrac{\displaystyle\int^L_0xx\lambda dx}{\displaystyle\int^L_0\lambda xdx}$

$=\dfrac{\lambda\displaystyle\int^L_0x^2dx}{\lambda\displaystyle\int^L_0xdx}$

$=\left[\dfrac{x^3/3}{x^2/2}\right]^L_0$

$=\dfrac{L^3}{3}\times \dfrac{2}{L^2}$

$x_{cm}=\dfrac{2}{3}L$ .

A shell fired from a gun at an angle to the horizontal explodes in mid air. Then the centre of mass of the shell fragments will move

0%
vertically down
0%
horizontally
0%
along the same parabolic path along which the 'intact' shell was moving
0%
along the tangent to the parabolic path of the 'intact' shell, at the point of explosion.

Force-time graph for the motion of a body is shown in figure. Change is linear momentum between 0 s to 8 s is :

0%
Zero
0%
4 N-s
0%
8 N-s
0%
None of the above

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%
$5\ m/s$

The moment of inertia of a uniform rod of length

$2l$ and mass '

$m$ ' about an axis through centre and inclined at an angle

$\theta$ to rod is:

0%
$\dfrac{ml^2}{3}\sin^2\theta$
0%
$\dfrac{ml^2}{6}\cos^2\theta$
0%
$\dfrac{ml^2}{2}\cos^2\theta$
0%
$\dfrac{ml^2}{12}\cos^2\theta$

Explanation

$\lambda =\dfrac{m}{2l}$ mass per unit length

$dm=\lambda dx$
The perpendicular distance of element from given axis

$=x\sin \theta$

$\therefore dI=(dm)(x\sin \theta)^2$

$=\dfrac{m}{2l}x^2dx \sin^2\theta$

$\displaystyle \int_0^I dI=\dfrac{m}{2l}\sin^2\theta \int_{-l}^{l}x^2dx$

$I=\dfrac{m}{2l}\sin^2\theta \left(\dfrac{l^3}{3}+\dfrac{l^3}{3}\right)$

$=\dfrac{ml^2}{3}\sin^2\theta$
1010649_1136596_ans_525528e261cf464b9fce5145d0c92ef2.PNG

A man stands at the centre of a turn table it extended horizontally, with a

$5 kg$ mass hand. He is set into rotation with an angular of one revolution in

$2s$ . His new angular is he drops his hands to his sides is (Assume moment of inertia of the man is

$6 \ kgm^2$ . The distance of the wavelength from the axis is

$1 m$ and final distance is

$0.2 m$ )

0%
$2.5 \ rev/s$
0%
$1.25 \ rev/s$
0%
$5 \ rev/s$
0%
None of these

A person sitting firmly over a rotating stool has his arms stretched. If he folds his arms, his angular momentum about the axis of rotation

0%
increases
0%
decreases
0%
remains unchanged
0%
doubles

Two persons A and B of weight 80 kg and 50 kg respectively are standing at opposite ends of a boat of mass 70 kg and length 2m, at rest. When they interchange their positions then the displacement of the center of mass of the boat will be

0%
60 cm towards left
0%
30 cm towards right
0%
30 cm towards left
0%
stationary

A wire of length

$l$ and mass

$m$ is bent in the form of a rectangle

$ABCD$ with

$\dfrac {AB}{BC}=2$ . The moment of inertia of this wire frame about the side

$BC$ is:

0%
$\dfrac {11}{252}ml^{2}$
0%
$\dfrac {8}{203}ml^{2}$
0%
$\dfrac {5}{136}ml^{2}$
0%
$\dfrac {7}{162}ml^{2}$

The radius of gyration of a uniform rod of length

$l$ and mass

$M$ about an axis passing through its centre and perpendicular to its length is:

0%
$\dfrac{l^2}{12}$
0%
$\dfrac{l}{2\sqrt{3}}$
0%
$\dfrac{l}{2}$
0%
$\dfrac{l}{\sqrt{2}}$

Explanation

Radius of gyration =k

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

$\because mk^2=I$

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

$k=\dfrac{l}{\sqrt{12}}$

$k=\dfrac{l}{2\sqrt{3}}$
option (B)

Two objects masses 200 g and 500 g posses velocities

$10i ms^{-1}$ and

$3i + 5 j ms^{-1}$ respectively. The velocity of their centre of mass

$ms^{-1}$ is

0%
$5 i - 25 j$
0%
$\dfrac{5}{7} i - 25 j$
0%
$5 i + \dfrac{25}{7} j$
0%
$25 i - \dfrac{5}{7} j$

If momenta of two particles of a system are given by

$\overrightarrow { { p }_{ 1 } } =2\hat { i } -\hat { j } +3\hat { k }$ and

$\overrightarrow { { p }_{ 2 } } =$

$-\hat { i } +2\hat { j } +3\hat { k }$ , then the angle made by the direction of motion of the system with

$x$ -axis is

0%
$\cos ^{ -1 }{ 1 }$
0%
$\cos ^{ -1 }{ \sqrt { 36/38 } }$
0%
$45^0$
0%
$\cos ^{ -1 }{ \sqrt { 1/38 } }$

$2$ kg body and a

$3$ kg body are moving along the x-axis. At a particular instant the

$2$ kg body has a velocity of

$3$ m/s and the

$3$ kg body has the velocity of

$2$ m/s. The velocity of the centre of mass at that instant is

0%
$5$ m/s
0%
$1$ m/s
0%
Zero
0%
$2.4$ m/s

Two bodies of 6 kg and 4 kg masses have their velocities

$5\hat{i} -2\hat{j} +10\hat{k}$ and

$10\hat{i} -2\hat{j} +5\hat{k}$ respectively. Then the velocity of their centre of mass is

0%
$5\hat{i}+2\hat{j}-8\hat{k}$
0%
$7\hat{i}+2\hat{j}-8\hat{k}$
0%
$7\hat{i}-2\hat{j}+8\hat{k}$
0%
$7\hat{i}+2\hat{j}+8\hat{k}$

A T joint is formed by two identical rods A and B each of mass m and length L in the X -Y plane as shown. Its moment of inertia about axis coinciding with A is

0%
$\frac{2mL^2}{3}$
0%
$\frac{mL^2}{12}$
0%
$\frac{mL^2}{6}$
0%
None of these

Explanation

$\textbf{Step 1 - Expression for net moment of inertia Refer Figure}$

$I_{tot.} = I_{1} + I_{2}$

$\textbf{Step 2 - Calculate net moment of inertia}$

For vertical rod,

$I_{1} = 0$ because axis passes through the rod.

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

$\therefore I_{tot.} = 0 + \dfrac {ML^{2}}{12}$

$I_{tot.} = \dfrac {ML^{2}}{12}$

Hence option B correct.

2180177_1162477_ans_53bababa75d64aaf8f04cee8d06b09cb.jpg

Two particles of equal mass are moving along the same straight line with the same speed in opposite direction. What is the speed of the centre of mass of the system?

0%
$\dfrac{v}{2}$
0%
$2v$
0%
$v$
0%
$0$

In free space, a shell moving with velocity 60 m/s along the positive x-axis of an inertial frame, when passes the origin, explodes into two pieces of masses ratio 1 :Velocity of the mass center after the explosion is

0%
20 m/s
0%
60 m/s
0%
90 m/s
0%
None of the above

Mass of thin long metal rod is

$2$ kg and its moment of inertia about an axis perpendicular to the length of the rod and passing through its one end is

$0.5kg{m^2}$ . Its radius of gyration is:

0%
$25$ cm
0%
$40$ cm
0%
$50$ cm
0%
$1$ m

Explanation

Solution we know that,

$I_{end} = \dfrac{ML^{2}}{3} = \dfrac{1}{2}$ [given]

$I_{com} = \dfrac{ML^{2}}{12} = \dfrac{1}{4}(\dfrac{ML^{2}}{3}) = \dfrac{1}{8}$

Also,

radius of gyration

$= \sqrt{\dfrac{I_{com}}{m}}$

$= \sqrt{\dfrac{1}{8\times 2}} = \dfrac{1}{\sqrt{16}} = 25cm$

1128856_1144666_ans_28f6ed279fbd4d488abbc0aec0c780cf.jpg

The linear density of a thin rod of length

$1.0$ m varies as

$\lambda =2$ kg/m

$+\left(\dfrac{2kg}{m^2}\right)x$ , where x is the distance from its one end. The distance of its centre of mass from its end is?

0%
$\dfrac{2}{3}$ m
0%
$\dfrac{5}{9}$ m
0%
$\dfrac{4}{3}$ m
0%
$\dfrac{1}{2}$ m

Explanation

REF.Image

$\lambda =2\dfrac{kg}{meter}+\dfrac{2kg}{meter^{2}}.x$

$\lambda =(2+2x) kg/m$ ; x is in meter

suppose there is an element of length dx

& at a distance x from left must end.

mass of this element

$dm = \lambda dx = (2+2x)dx$

$\Rightarrow$ mass of rod = sum of marvels of these elements

$\displaystyle M= \int_{x=0}^{1}dm = \int_{0}^{1}(2+2x)dx= [2x+\frac{2x}{2}]_{0}^{1}$

$m= 3 kg$

$\displaystyle x_{ c.m} = \frac{\displaystyle \int_{0}^{1}x\, dm}{\displaystyle \int dm}= \frac{\displaystyle \int_{0}^{1}(2+2x)xdx}{\displaystyle \int_{0}^{1}(2+2x)dx}$

$\displaystyle x_{ cm}= \frac{\displaystyle \int_{0}^{1}(2x+2x^{2})}{3}dx=\frac{[\dfrac{2x^{2}}{2}+\dfrac{2x^{3}}{3}]_{0}^{1}}{3}$

$x_{cm}=\dfrac{1+\dfrac{2}{3}}{3}$ meter =

$\dfrac{5}{9}$ meter

from are end (having less density)

1185139_1166048_ans_cdb087188cd940648fac514d00eaecce.png

Three identical uniform rods each of length

$1cm$ and Mass

$2kg$ are arranged to form an equilateral triangle what is the moment of inertia of the system about an axis passing through one corner ans perpendicular to the plane of the triangle?

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$4kg-m^2$
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$3kg-m^2$
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$2kg-m^2$
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None of these

Explanation

$I=I_{AB}+I_{AC}+I_{BC}$

$=\dfrac{ml^2}{3}+\dfrac{ml^2}{3}+I_{cm}+mh^2$

$=\dfrac{2ml^2}{3}+\dfrac{ml^2}{12}+m\left(\dfrac{1\sqrt{3}}{2}\right)^2 =\dfrac{3}{2}ml^2$

$=3kgm^2$

Moment of a inertia of a sphere about its diameter is

$2/5\ MR^{2}$ . What its moment of inertia about an axis perpendicular to its two diameter and passing through their point of intersection.?

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$I=\dfrac{2}{5}Mr^{2}$
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$I=\dfrac{3}{5}Mr^{2}$
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$I=\dfrac{4}{5}Mr^{2}$
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$I=\dfrac{5}{5}Mr^{2}$

Explanation

moment of inertia of sphere about its diameter:

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

two diameter always inetersect at the centre.

however,a sphere in symmetric from all directions,

so taking any other diameter axis will have moment,

$I=\dfrac{2}{5}MR^{2}$ (i.e. same)

1372055_1169073_ans_5867ca969d6e43b5990064389eb8cf0f.bmp

The wheel of radius

$r= 300 mm$ rolls to the right without slipping and has a velocity

$v_0= 3 m/s$ of its center O. The speed of the point A on the wheel for the instant represented in the figure is:-

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$4.36$ m/s
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$5$ m/s
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$3$ m/s
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$1.5$ m/s

Explanation

$V_0=\omega r_0$

$3m/s=\omega\times 300\,mm$

$3m/s=\omega\times 0.3m$

$10\,rad/s=\omega$

$v=\omega r=2m/s$

$v_0=3m/s$

$V_{net}=\sqrt{v^2+v^2_0+2v\,v_0\cos\theta}$

$=\sqrt{4+9+2\times 2\times 3\times \dfrac{1}{2}}$

$=\sqrt{13+6}$

$=\sqrt{19}\approx 4.36$

1298979_1164266_ans_8dcc4cafb1004c10a731ff4d2c8445c7.png

Find the M.I of rod about (i) an axis perpendicular to the rod and passing through left end (ii)An axis through its centre of mass and perpendicular to the length whose linear density varies as

$\lambda=ax$ where a is a positive constant and

$'x'$ is the position of an element of the rod relative to its left end.The length of the rod is

$l$

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$\dfrac{al^4}{4}$
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$\dfrac{al^3}{4}$
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$\dfrac{al^4}{2}$
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$\dfrac{al^3}{6}$

Find the torque of a force

$\vec{F}=-3\hat{i}+\hat{j}+5 \hat{k}$ acting at the point

$\vec{r}=7\hat{i}+3\hat{j}+\hat{k}$ with respect to origin:-

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$14 \hat{i}-32 \hat{j}-2 \hat{k}$
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$4 \hat{i}+4 \hat{j}+6 \hat{k}$
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$-14 \hat{i}+38 \hat{j}-16 \hat{k}$
0%
$-21\hat{i}+3 \hat{j}+5\hat{k}$

Explanation

$\vec{\tau}=\vec{r}\times \vec{F}=(7\hat{i}+3\hat{j}+\hat{k})(-3\hat{i}+\hat{j}+5\hat{k})$

$\vec{\tau}=\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix} =14\hat{i}-38\hat{j}+16\hat{k}$

Four uniform thin rods each of mass 1 kg and length 1 m are joined in the form of a square. If the square is rotated about axis AB, then it moment of inertia is

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$\frac{5}{3}{\text{kg}}\;{{\text{m}}^{\text{2}}}$
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$\frac{3}{4}{\text{kg}}\;{{\text{m}}^{\text{2}}}$
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$\frac{5}{4}{\text{kg}}\;{{\text{m}}^{\text{2}}}$
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$\frac{7}{4}{\text{kg}}\;{{\text{m}}^{\text{2}}}$

A bullet of mass m is fired at a certain angle

$\theta$ b with the horizontal. The bullet returns to ground after time t. Then the change in momentum of the bullet is

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zero
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$2mu sin \theta$
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$\dfrac{mgt cos \theta}{2}$
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mgt

Explanation

$\Delta D=mV_y-mv_y\\ \quad=m(\mu\sin\theta-(-u\sin\theta))\\ \quad=m(2u\sin\theta)\\ \quad=2mu\sin\theta$

A circular disc of M.I. 10

$kg-m^2$ rotates about its own axis at a constant speed of 60 r.p.m. under the action of an electric motor of power 31.4 W. If the motor is switched off, how many rotations will it cover before coming to rest?

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$3.14$
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$31.4$
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$314$
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$6.28$

Explanation

The moment of inetia of the disc is

$10\ kg-m^2$

The speedof the disc is

$60\ rpm$ i.e.

$60\times\dfrac{2\pi}{60}\ rad/s$

The power supplied by the motor is

$31.4\ W$

The relation between the power and the torque developed is:

$P=\tau\omega$

$\tau=\dfrac{P}{\omega}=\dfrac{31.4}{2\pi}$

$\tau=5\ Nm$

The angular acceleration of motor due to torque is :

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

$\alpha=\dfrac{5}{10}=0.5\ rad/s^2$

Now, when the motor is switched off, the angular displacement of the disc to come to rest is given as:

$\omega^2_f=\omega^2_0+2\alpha\theta$

$0-(2\pi)^2=2\times0.5\times\theta$

$\theta=4\pi^2$

In one rotation, displacement is

$2\pi$ . SO, total number of rotations is:

$n=\dfrac{\theta}{2\pi}=\dfrac{4\pi^2}{2\pi}$

$n=2\pi=6.28\ rotations$

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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