MCQ Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 13

A body of mass

$M$ is attached to one end of a rod of mass

$M$ and length

$L$ . The entire system is rotated about an axis passing through the other end of the rod. The

$M.I.$ of the system about die given axis of rotation is:

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

Explanation

Moment of inertia of rod about its centre of mass is determined by the general form of moment of inertia which is given by:

$I = \int\limits_0^M {{r^2}dm}$

Now mass $m$ is termed into length element $dr$

$I = \int\limits_{ - L/2}^{L/2} {{r^2}} \times \dfrac{M}{L}dr$

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

Now moment of inertia of rod about its one end is given by

${I_R} = \dfrac{{M{L^2}}}{{12}} + M \times {\left( {\dfrac{L}{2}} \right)^2}$

${I_R} = \dfrac{{M{L^2}}}{{12}} + \dfrac{{M{L^2}}}{4}$

${I_R} = \dfrac{{M{L^2}}}{3}$

Moment of inertia of another mass M which is attached at the other end of the rod

$I = M{L^2}$

Total moment of inertia

${I_T} = {I_R} + I$

${I_T} = \dfrac{{4M{L^2}}}{3}$

Thus moment of inertia of a rod and mass about its one of the end is $\dfrac{{4M{L^2}}}{3}$ .

A light rod of length

$l$ has two masses

${m_1}$ and

${m_2}$ attached to its two ends . the moment of inertia of the system about an axis perpendicular to the rod and passing through the center of mass is

0%
$\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}{l^2}$
0%
$\frac{{{m_1} + {m_2}}}{{{m_1}{m_2}}}{l^2}$
0%
$({m_1} + {m_2}){l^2}$
0%
$\sqrt {{m_1}{m_2}} {l^2}$

A light rod of length

$l$ has two masses

$m_1$ and

$m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is

0%
$\dfrac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}{l^2}$
0%
$\dfrac{{{m_1} + {m_2}}}{{{m_1}{m_2}}}{l^2}$
0%
$\left( {{m_1} + {m_2}} \right){l^2}$
0%
$\sqrt {{m_1}{m_2}} {l^2}$

A solid disc has a radius

$R$ . Density of its material is given by

$\rho={\rho}_{0}(2-\cfrac{r}{R})$ where

${\rho}_{0}$ is a constant and

$r$ is the radial distance measured from the centre. Then the moment of inertia of the disc about an axis passing through its centre and perpendicular to the plane of the disc is

0%
$\cfrac{2}{7}M{R}^{2}$
0%
$\cfrac{5}{18}M{R}^{2}$
0%
$\cfrac{9}{20}M{R}^{2}$
0%
$\cfrac{7}{12}M{R}^{2}$

Three rods are placed in the form of equilateral triangle. Calculate the M.I. about axis passing through the centre and perpendicular to the plane (Assume mass and length of each rod is M and L respectively). number of sides n = 3

0%
$\frac{M L^{2}}{12}$
0%
$\frac{M L^{2}}{2}$
0%
$\frac{M L^{2}}{8}$
0%
$\frac{M L^{2}}{4}$

Location of centre of mass of uniform semi-circular plate of radius R from its centre is:

0%
$\dfrac{2R}{3\pi}$
0%
$\dfrac{R}{3\pi}$
0%
$\dfrac{3R}{4\pi}$
0%
$\dfrac{4R}{3\pi}$

Two objects of masses 200 gram and 500 gram possess
velocityies $10\mathop i\limits^ \wedge$ m/s and $3\mathop i\limits^ \wedge + 5\mathop j\limits^ \wedge$ m/s respectively. The velocity of their centre of mass
in m/s is:

0%
$5\mathop i\limits^ \wedge - 25\mathop j\limits^ \wedge$
0%
$\frac{5}{7}\mathop i\limits^ \wedge - 25\mathop j\limits^ \wedge \;$
0%
$5\mathop i\limits^ \wedge +\dfrac{{25}}{7}\mathop j\limits^ \wedge \;$
0%
$25\mathop i\limits^ \wedge - \dfrac{5}{7}\mathop j\limits^ \wedge$

A uniform rod of mass

$2\ kg$ and length

$1\ m$ lies on a smooth horizontal plane.A particle of mass 1 kg moving at a speed of

$2\, ms^{-1}$ perpendicular to the length of the rod strikes it at a distance

$\dfrac{1}{4}m$ from the center and stops. What is the angular velocity of the rod about its center just after the collision?

0%
$3\,rad\,s^{-1}$
0%
$4\,rad\,s^{-1}$
0%
$1\,rad\,s^{-1}$
0%
$2\,rad\,s^{-1}$

A long, thin carpet is laid on a floor, One end of the carpet is bent back and then pulled backwards with constant unit velocity, just above the part of the carpet which is still at rest on the floor. The speed of centre of mass of the moving part is

0%
$1\ m/s$
0%
$\dfrac{3}{4}\ m/s$
0%
$\dfrac{1}{2}\ m/s$
0%
$\dfrac{2}{4}\ m/s$

Two cylinders of equal length and equal diameter, one of wood and the other of iron are placed on an inclined on an inclined plane. Which will roll down faster?

0%
Iron
0%
Wooden
0%
Both will roll down at the same time
0%
Can't say

A body of mass 'm' is dropped and another body of mass M is projected vertically up with speed 'u' simultaneously from the top of the tower of height H. If the body reaches the highest point before the dropped body reaches the ground , then maximum height raised by the centre of mass of the system from ground is

0%
$H + \frac{{{u^2}}}{{2g}}$
0%
$\frac{{{u^2}}}{{2g}}$
0%
$H + \frac{1}{{2g}}{\left( {\frac{{Mu}}{{m + M}}} \right)^2}$
0%
$H + \frac{1}{{2g}}{\left( {\frac{{mu}}{{m + M}}} \right)^2}$

For the figure given below, the speed of centre of mass is

0%
$\frac{2}{3} \hat{i} +\frac{8}{3}\hat{j}$
0%
$2{i} +4\hat{j}$
0%
$\frac{8}{3} \hat{i} +\frac{2}{3}\hat{j}$
0%
$\frac{2}{5} \hat{i} +\frac{8}{3}\hat{j}$

Explanation

We know that center of mass

$x_{cm} = \dfrac{m_1 \vec{x}_1 + m_2 \vec{x}_2}{m_1 + m_2}$

Then,

$V_{cm} = \dfrac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$

Given:-

$m_1 = 1 kg$

$m_2 = 2 kg$

$\vec{v}_1 = 2 \hat{j}$

$\vec{v}_2 = 4 \hat{i}$

$\vec{v}_{cm} = \dfrac{1 \times 2 \hat{j} + 2 \times 4 \hat{i}}{3} = \dfrac{8 \hat{i}}{3} + \dfrac{2}{3} \hat{j}$

Option C is correct.

The speed of centre of mass is

$\dfrac{8}{3} \hat{i} + \dfrac{2}{3} \hat{j}$

2057910_1162008_ans_fb90182eee394c9cb73028901478f127.png

A uniform rod is released from its vertical position with its bottom on a smooth surface. If v is the speed of centre of mass of the rod and

$\omega$ is its angular velocity when it makes an angle

$\theta$ with the horizontal, then

0%
$v= \cfrac{\omega l}{2} sin\theta$
0%
$v= \cfrac{\omega l}{2}$
0%
$v= \cfrac{\omega l}{2} cos\theta$
0%
$v= \cfrac{\omega l}{2} sin\theta cos\theta$

The handle of a water pump is

$90$ cm long from its piston rod. If the pivot handle is at distance of

$15$ cm from the piston rod, calculate (a) least effort required at its other end to overcome a resistance of

$60$ kgf and (b) mechanical of advantage of handle.

0%
(a) $12$ kgf (b) $50$
0%
(a) $32$ kgf (b) $5$
0%
(a) $32$ kgf (b) $10$
0%
(a) $12$ kgf (b) $5$

In the figure shown, there is a conical shaft rotating on a bearing of very small clearance

$t$ . The space between the conical shaft and the bearing, is filled with a viscous fluid having coefficient of viscosity

$\eta$ . The shaft is having radius

$R$ and height

$h$ . If the external torque applied by the motor is

$\tau$ and the power delivered by the motor is

$P$ working in

$100\%$ efficiently to rotate the shaft with constant

$\omega$ . Then :-

0%
$P =\dfrac{\pi \omega^2 \eta R^3\sqrt{R^2 + h^2}}{2t}$
0%
$\tau =\dfrac{\pi \omega \eta R^3\sqrt{R^2 + h^2}}{2t}$
0%
$P =\dfrac{\pi \omega^2 \eta R^3 h}{2t}$
0%
$\tau =\dfrac{\pi \omega \eta R^3h^2}{2t \sqrt{R^2 + h^2}}$

Two blocks of masses

$10$ kg and

$6$ kg are connected by a spring and placed on a frictionless horizontal surface. A hammer strikes the block of mass

$10$ kg as a result of which this block is given a velocity

$3.2$ m/s towards the lighter block immediately on being struck. Velocity imparted to the centre of mass of their system is:

0%
zero
0%
$3$ m/s
0%
$2$ m/s
0%
$1.4$ m/s

A body of mass 'm' falls from a height 'h' and rebounds. If 'e' is the coefficient of restitution between the ground and the body, the change in linear momentum of the body is

0%
$m(1+e) \sqrt{2gh}$
0%
$(me) \sqrt{2gh}$
0%
$3(1 + e) \sqrt{2gh}$
0%
$m(1 - e) \sqrt{2gh}$

Figure shows a thin metallic triangular sheet ABC. The mass of the sheet is M. The moment of inertia of the sheet about side AC is :

0%
$\frac{Ml^2 }{18}$
0%
$\frac{Ml^2 }{12}$
0%
$\frac{Ml^2 }{6}$
0%
$\frac{Ml^2 }{4}$

A thin rod of mass

$m$ and length

$l$ is suspended from one of its ends. It is set into oscillation about a horizontal axis. Its angular speed is

$\omega$ while passing through its mean position. How high will its centre of mass rise from its lowest position?

0%
$\dfrac{\omega^{2}l^{2}}{2g}$
0%
$\dfrac{\omega^{2}l^{2}}{3g}$
0%
$\dfrac{\omega^{2}l^{2}}{g}$
0%
$\dfrac{\omega^{2}l^{2}}{8g}$

A scalene triangular lamina of uniform mass density and negligible thickness has one of its vertices at the origin. The position vectors of its other two vertices are

$\vec a$ and

$\vec b$ . The location of its centre of mass will be

0%
$\frac{\vec a + \vec b}{2}$
0%
$\frac{\vec a + \vec b}{3}$
0%
$\frac{2(\vec a + \vec b)}{3}$
0%
$\frac{3(\vec a + \vec b)}{4}$

Find the position of center of mass of a disc of radius

$R$ from which a hole of radius r is cut out. The center of the hole is at a distance

$R/2$ from the center of the disc.

0%
$\dfrac{Rr^2}{2(R^2-r^2)}$ towards right of $O$
0%
$\dfrac{Rr^2}{2(R^2-r^2)}$ towards left of $O$
0%
$\dfrac{2Rr^2}{(R^2+r^2)}$ towards right of $O$
0%
$\dfrac{2Rr^2}{(R^2+r^2)}$ towards left of $O$

Explanation

Center of mass of bigger disc

$=O(0,0)$ Let

$O$ be the origin.

Center of mass of smaller disc

$=C(R/2,0)$

Let

$x\ cm=x_1$

$x\ cm=R/2=x_2$

Now,

$m_1+m_2=m$

$x\ cm=0=x$

By using the formula

$\cfrac{m_1x_1+m_2x_2}{m_1+m_2}=x$

We get center of mass

$x_1=\cfrac{Rr^2}{2(R^2-r^2)}$ towards left of

$O$ .

1221037_1210128_ans_a77ad93cf9494173bddbb4e87a986395.png

A thin disc of radius R and mass M has charge q uniformly distributed on it. It rotates with angular velocity

$\omega$ . The ratio of magnetic moment and angular momentum for the disc is?

0%
$\dfrac{q}{2M}$
0%
$\dfrac{R}{2M}$
0%
$\dfrac{q^2}{2M}$
0%
$\dfrac{2M}{q}$

Explanation

$\gamma =\frac { \mu }{ L } =\frac { i\ast \pi \ast { r }^{ 2 } }{ m\ast { r }^{ 2 }\ast w } =\frac { q\ast w\ast \pi }{ 2\pi m } =\frac { q }{ 2m }$

A rod of steel is fixed at

$(1,0)$ and toy is fixed at

$(2,1)$ on it. The rod is rotated through an angle

$15^{o}$ in clockwise direction, the new position of the toy will be

0%
$\left(1+\sqrt {6},\dfrac {1}{\sqrt {2}}\right)$
0%
$\left(1-\sqrt {6},\sqrt {2}\right)$
0%
$\left(1+\sqrt {6},\sqrt {2}\right)$
0%
$none\ of\ these$

The angular velocity of a body changes from

${\omega _1}to\;{\omega _2}$ without applying torque.The ratio of initial radius of gyration to the final of gyration is-

0%
$\sqrt {{\omega _2}} :\sqrt {{\omega _1}}$
0%
$\sqrt {{\omega _1}} :\sqrt {{\omega _2}}$
0%
${\omega _2}:{\omega _1}$
0%
${\omega _1}:{\omega _2}$

A spool ( initially at rest ) is pulled on a smooth horizontal surface by constant horizontal forceF. Such that the spool comes into a state of sustained rolling motion. Kinetic energy of the spool, after centre of mass of the spool has moved a distance d, is

0%
$Fd$
0%
$\cfrac {Fdr_1}{r_2}$
0%
$\cfrac { Fd(r_{ 1 }+r_{ 2 }) }{ r_2 }$
0%
$\cfrac {Fdr_1}{({r_1}+{r_2})}$

When a force of

$1$ N acts on

$1$ kg mass at rest for

$1$ s, its final momentum is

$P$ . When

$1$ N force acts on

$1$ kg mass at rest through a distance

$1$ m, its final momentum is

$P'$ . The ratio of

$P$ to

$P'$ is:

0%
$1:1$
0%
$1:\sqrt{2}$
0%
$1:2$
0%
$2:1$

Explanation

$F=1N,t=1S$

$\triangle P=F\times t=1N\times 1S=1kgm/s$

${ P }_{ f }-{ P }_{ i }=1kgm/s$

${ P }_{ f }-0=1kgm/s$

${ P }_{ f }=1kgm/s$

$F=1N,M=1kgm,S=1m$

$a=\dfrac { F }{ M } ,\dfrac { 1N }{ 1Kg } =1m/{ s }^{ 2 }$

${ V }_{ f }^{ 2 }-{ V }_{ i }^{ 2 }=2as$

${ V }_{ f }^{ 2 }-0=2\left( 1m/{ s }^{ 2 } \right) \left( 1m \right)$

${ V }_{ f }^{ }=\sqrt { 2ms }$

multiply both sides

${ mv }_{ f }=\sqrt { 2m/s } \times 1kg$

${ P }^{ \prime }=\sqrt { 2 } kgm/s$

$\dfrac { p }{ { p }^{ \prime } } =\dfrac { 1 }{ \sqrt { 2 } } =\dfrac { 1 }{ \sqrt { 2 } } =1:\sqrt { 2 }$

The centre of mass of a system of three particles of masses 1 g, 2 g and 3 g is taken as the origin of a coordinate system. The position vector of a fourth particle of madd 4 g such that the centre of mass of the four particle system lies at the point (1,2,3) is

$\alpha$

$(\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } )$ , where

$\alpha$ is a constant. The value of

$\alpha$ is

0%
$\frac { 10 }{ 3 }$
0%
$\frac { 5 }{ 2 }$
0%
$\frac { 1 }{ 2 }$
0%
$\frac { 2 }{ 5 }$

ABC is a right - angle triangular plate of uniform thickness. The sides are such that AB > BC as shown in figure.

$I_{1}, I_{2} and I_{3}$ are moments of inertia about AB, BC and AC respectively. Then which of the following relations is correct?

0%
$I_{1}= I_{2} = I_{3}$
0%
$I_{2} > I_{1} > I_{3}$
0%
$I_{3} < I_{2} < I_{1}$
0%
$I_{3} > I_{1} > I_{2}$

Explanation

We know that

$I=mr^2$

$\therefore I\propto r^2$

So moment of inertia depend only on man and perpendicular distance between it and axis of rotation it does not depend

$m$ thickness of plane

$\therefore I_1=I_2=I_3$

1349456_1222565_ans_f3595e4a3cc449c4a5c2947f4472f019.png

A solid sphere is thrown on a horizontal rough surface with initial velocity of center of mass

$v_o$ without rolling. Velocity of its center of mass when it starts pure rolling is

0%
$\dfrac {2v_o}{7}$
0%
$\dfrac {2v_o}{5}$
0%
$\dfrac {5v_o}{3}$
0%
$\dfrac {5v_o}{7}$

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The centre of mass of the new disc is

$\alpha R$ fromthe centre of the bigger disc. The value of

$\alpha$ is

0%
$\cfrac{1}{2}$
0%
$\cfrac{1}{6}$
0%
$\cfrac{1}{4}$
0%
$\cfrac{1}{3}$

Explanation

$\textbf{Step 1 - Draw the given situation Refer Figure}$

Let

$\sigma =$ Surface mass density.

So,

$M_{1} = \sigma A_{1} = \sigma \cdot 4\pi R^{2}$

$M_{2} = \sigma A_{2} = \sigma \pi R^{2}$

$\textbf{Step 2 - Calculation of centre of mass}$

$X_{cm} = \dfrac {M_{1}x_{2} + M_{2}x_{2}}{M_{1} + M_{2}}$

$X_{cm} = \dfrac {(\sigma \cdot 4\pi R^{2}) \times 0 - (\sigma \cdot \pi R^{2})\cdot R}{(\sigma \cdot 4\pi R^{2}) - (\sigma \cdot \pi R^{2})}$

$X_{cm} = \dfrac {-\sigma \cdot \pi R^{3}}{3\sigma \cdot \pi R^{2}}$

$X_{cm} = \dfrac {-R}{3}$

Due to symmetry about x-axis

So,

$Y_{cm} = 0$

Thus,

$\vec {r}_{cm} = X_{cm} \hat i + Y_{cm} \hat {j}$

$\vec {r}_{cm} = \dfrac {-R}{3} \hat i + 0$

$|\vec {r}_{cm}| = \dfrac {R}{3}$

$....(1)$

as a given -

$|\vec {r}_{cm}| = \alpha R$

$....(2)$

So, comparing

$(1)$ and

$(2)$

$\alpha = \dfrac {1}{3}$

${\textbf{Correct option: D}}$

If moment of inertia of a body of mass

$m$ is double than moment of inertia of rod of same mass

$m$ about transverse axis passing through its centre then radius of gyration of given body is

0%
Six times than that of rod
0%
Double than that of rod
0%
$\sqrt {2}$ times than that of rod
0%
$12\sqrt {2}$ times than that of rod

Four solid rigid balls each of mass

$m$ and radius

$r$ are fixed on a rigid-ring of radius

$2r$ and mass

$2m$ . The system is whirled about

$'O'$ as shown. The radius of gyration of the system is

$128\,m{r^2}/30.$ how

$?$

0%
$r\sqrt{\cfrac{128}{5}}$
0%
$r\sqrt{\cfrac{88}{5}}$
0%
$r\sqrt{\cfrac{128}{30}}$
0%
$r\sqrt{\cfrac{88}{30}}$

Explanation

about

$O$

$M.I.$ of ring

$=2m{\left( {2r} \right)^2}$

$M.I.$ of a ball

$=2m{r^2}/5 + m{\left( {2r} \right)^2} = 22m{r^2}$

So total

$= 8m{r^2} + 4 \times 22m{r^2}/5=r\sqrt {\frac{{88}}{5}}$

Hence,

option

$(B)$ is correct answer.

A solid cylinder at rest at the top of an inclined plane of height 2.7 m rolls down without slipping. If the same cylinder has to slide down a frictionless inclined plane and acquire the same velocity as that acquired by the centre of mass of the rolling cylinder at the bottom of the inclined plane, the height of the inclined plane in meters should be

0%
2.2
0%
1.2
0%
1.6
0%
1.8

The moment of inertia of a straight thin rod of mass M and length L about an axis perpendicular to its length and passing through its one end is -

0%
$\frac { { ML }^{ 2 } }{ 12 }$
0%
$\frac { { ML }^{ 2 } }{ 3 }$
0%
$\frac { { ML }^{ 2 } }{ 2 }$
0%
${ ML }^{ 2 }$

Two particles

$A$ 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 that of

$B$ is

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

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

$AB$ and

$CD$ are two identical rods each of length

$l$ and mass

$m$ joined to from a cross is fixed inside a ring of mass

$m$ and radius

$l/2$ . Moment of intertia of the system about a bisector of the angle between the rods (xy) is

0%
$\dfrac{{7m{l^2}}}{6}$
0%
$\dfrac{{13m{l^2}}}{12}$
0%
$\dfrac{{m{l^2}}}{12}$
0%
$\dfrac{{5m{l^2}}}{24}$

In the figure shown a hole of radius 2 cm is made in a semi-circular disc of the radius $6\pi$ at a distance 8 cm from center C of the disc. The distance of the center of mass of this system from C is :

0%
4 cm
0%
8 cm
0%
6 cm
0%
12 cm

Figure shows a cubical box that has been constructed from uniform metal plat of negligible thickness. The box is open at the top and has edge length

$40 /cm$ . The

$z$ co-ordinate of the centre of mass of the box in

$cm$ , is

0%
$12$
0%
$16$
0%
$20$
0%
$22$

Four identical rods of mass

$M=6 kg$ each are welded at their ends to form a square and then welded to a massive ring having mass

$m = 4 k g$ having radius

$R = 1 m$ . If the system is allowed to roll down the incline of inclination

$\theta = 30 ^ { \circ }$ , assume there is no slipping. The acceleration of the system will be

0%
$\mathrm { g } / 2$
0%
$\mathrm { g } / 4$
0%
$\mathrm {7 g } / 24$
0%
$\mathrm { g } / 8$

A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height

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

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

A pulley of radius

$2m$ is rotated about its axis by a force

$F=(20t-5{t}^{2})$ newton (where

$t$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its aixs of rotation is

$10kg{m}^{2}$ the number of rotations made by the pulley before its direction of motion if reversed, is:

0%
less than $3$
0%
more than $3$ but less than $6$
0%
more than $6$ but less than $9$
0%
more than $9$

A square metal frame is made up of four identical thin rods each of mass

$6\ kg$ and length

$1\ m$ . The moment of inertia of the frame in

$kg-m^{2}$ about an axis passing through point

$O$ and perpendicular to the plane of the frame is:

0%
$4$
0%
$8$
0%
$12$
0%
$1$

A sphere of mass m and radius r rolls without slipping over a tunnel of width

$d= \frac {8r}{5}$ as shown in the figure (in the figure,the sphere rolls perpendicular to the plane of the page). The velocity of centre of mass of sphere is v directed into the plane of the page. The maximum speed of a point on the sphere is

0%
$\frac {4v}{3}$
0%
$\frac {8v}{3}$
0%
$\frac {6v}{3}$
0%
$\frac {3v}{2}$

A Three particles of masses 1 kg, 2 kg, 1 kg are at the points whose position vectors are

$\hat{i}+\hat{j}, 2\hat{i}-\hat{j}, 3\hat{i}+\hat{j}$ . The position vector of their centre of mass is :

0%
$\frac{1}{4}\left ( 6\hat{i}+\hat{j} \right )$
0%
$2\hat{i}$
0%
$\frac{1}{3}\left ( 6\hat{i}+\hat{j} \right )$
0%
$8\hat{i}$

A solid sphere of radius

$R$ is rolled by a foce

$F$ acting at the top of the sphere as shown in the figure. The sphere rolls without slipping on rough stationary surface. Initially sphere is in the rest poition then:

0%
work done by friction when the centre of mass moves a distance $S$ is $2FS$
0%
speed of the $CM$ when $CM$ moves a distance $S$ is $\sqrt{\dfrac{20}{7} \dfrac{FS}{M}}$
0%
work done by the force $F$ when Centre of Mass moves a distance $S$ is $FS$
0%
speed of the $CM$ when moves a distance $S$ is $\sqrt{\dfrac{$RS}{M}}$

Two bodies of masses

$5kg$ and

$1kg$ are moving with velocities

$2\hat { i } -7\hat { j } +3\hat { k }$ and

$-10\hat { i } +35\hat { j } -3\hat { k }$

$m/s$ respectively. The velocity of the centre of mass of the two-body system is :

0%
along z-axis
0%
in the xy-plane
0%
in the xyz-space
0%
in the yz-plane

A square of side 4 cm and uniform thickness is divided into four equal squares. If one of them is cut off find the position of the center of mass of the remaining portion from its geometric center.

0%
$3\sqrt2 cm$
0%
$\cfrac{3}{\sqrt2}cm$
0%
$\cfrac{\sqrt2}{3}cm$
0%
$\cfrac{1}{\sqrt2}cm$

A small particle of mass m is projected at an angle $\theta$ with the x-axis, with initial velocity ${{v_o}}$ in the x-y plane. Just before time $\frac{{2{v_o}\sin \theta }}{g}$ , the angular momentum of the particle about the point of projection is

0%
$\dfrac{{2mv_0^3\sin \theta \cdot {{\cos }^2}\theta }}{g}$
0%
$\dfrac{{2mv_0^3{{\sin }^2}\theta \cdot \cos \theta }}{g}$
0%
$\dfrac{{mv_0^3\sin \theta \cdot \cos \theta }}{g}$
0%
$\dfrac{{2mv_0^3{{\sin }^2}\theta \cdot {{\cos }^2}\theta }}{g}$

Particle

$A$ of mass

$m$ is moving with a velocity

$- 6 i + 8 j \mathrm { m } / \mathrm { s }$ experiencing a uniform acceleration of

$4 i - 2 \hat { j } \mathrm { m } / \mathrm { s } ^ { 2 } .$ At that moment particle

$B$ of mass

$m$ is moving with a velocity of

$10\hat { i } - 4 \hat { j } \mathrm { m } / \mathrm { s }$ and experiencing uniform acceleration of

$- 2 i + 4 j \mathrm { m } / \mathrm { s } ^ { 2 }$ . The path traced by the center of mass of the

$A, B$ system is

0%
Parabola
0%
Circle
0%
A straight line
0%
Rectangular hyperbola

$\begin{array} { l } { \text { Three identical uniform rods each of length } 1 \mathrm { m } \text { and } } \\ { \text { mass } 2 \mathrm { kg } \text { are arranged to form an equilateral triangle. } } \\ { \text { What is the moment of inertia of the system about an to } } \\ { \text { axis passing through one corner and perpendicular to } } \\ { \text { the plane of the triangle: } } \end{array}$

0%
$4 \mathrm { kg } - \mathrm { m } ^ { 2 }$
0%
$3 \mathrm { kg } - \mathrm { m } ^ { 2 }$
0%
$2 k g - m ^ { 2 }$
0%
$3 / 2 \mathrm { kg } - m ^ { 2 }$

Explanation

$\begin{array}{l} I=\dfrac { { m{ l^{ 2 } } } }{ 3 } +\dfrac { { m{ l^{ 2 } } } }{ 3 } +\left[ { \dfrac { { m{ l^{ 2 } } } }{ { 12 } } +m{ { \left( { \dfrac { { \sqrt { 3 } l } }{ 2 } } \right) }^{ 2 } } } \right] \\ =\dfrac { { 2m{ c^{ 2 } } } }{ 3 } +\left[ { \dfrac { { m{ c^{ 2 } } } }{ { 12 } } +\dfrac { { 3m{ l^{ 2 } } } }{ 4 } } \right] \\ =\dfrac { { 2m{ l^{ 2 } } } }{ 3 } +\dfrac { { 10m{ l^{ 2 } } } }{ { 12 } } \\ =\dfrac { { 18m{ l^{ 2 } } } }{ { 12 } } \\ =\dfrac { { 3m{ l^{ 2 } } } }{ 2 } \\ =3\, kg\, -{ m^{ 2 } } \\ Hence,\, the\, option\, B\, is\, the\; correct\, answer. \end{array}$

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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