MCQ Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 12

Moment of inertia of a straight wire about an axis perpendicular to the wire and passing through one of its end is

$I$ . This wire is now framed into a circle (a ring) of single turn. The moment of inertia of this ring about an axis passing through centre and perpendicular to its plane would be:

0%
$\left( \cfrac { 3 }{ { \pi }^{ 2 } } \right) I$
0%
$\left( \cfrac { 3 }{ { 4\pi }^{ 2 } } \right) I$
0%
$\left( \cfrac { { \pi }^{ 2 } }{ 3 } \right) I$
0%
$\left( \cfrac { {4 \pi }^{ 2 } }{ 3 } \right) I$

Explanation

For a wire of length l and mass m its moment of inertia about an axis passing through its one end and perpendicular to its length is

$ml^2/3$ so

$I=ml^2/3$ or

$ml^2 = 3I$

Now when this wire is changed in a ring then the circumference of the circle is

$2\pi r =length = l$ so radius

$r=l/2\pi$

Now Moment of Inertia of a ring of radius r and mass m about an axis passing through its centre and perpendicular to its plane is

$mr^2$

putting value of r in terms of l we get new M.O.I as

$(m)l^2/4\pi^2 = ml^2/4\pi^2$

$3I/4\pi^2$

Option B is correct.

Inside a smooth spherical shell of radius R a ball of the same mass is released from the shown position ( fig. ). Find the distance travelled by the shell on the horizontal floor when the ball comes to the just opposite position of itself with respect to its position in the shell.

0%
$\dfrac{3R}{5}$
0%
$\dfrac{R}{4}$
0%
$\dfrac{3R}{4}$
0%
$\dfrac{5R}{4}$

Explanation

The correct option is C.

Given,

A spherical shell having radius

$R$ . And a ball is placed inside the spherical shell.

So,

Let the shell travel

$x$ distance on the floor.

The ball travels a distance of

$(x+0)$ in x-direction with respect to the floor.

So, conserving center of mass is:

$\Rightarrow m(\dfrac{3R}{4})+mx=0$

$\Rightarrow m(\dfrac{3R}{4})=-mx$

$\Rightarrow x=(-\dfrac{3R}{4})$

Thus, the Ball travels

$\dfrac{3R}{4}$ in the opposite direction.

969153_1017479_ans_73f4004332a04bbc93c3d953d3f0a22a.png

An annular ring with inner and outer radii

${R}_{1}$ and

${R}_{2}$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring

${F}_{1}/{F}_{2}$ is:

0%
$\cfrac { { R }_{ 2 } }{ { R }_{ 1 } }$
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${ \left( \cfrac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 2 }$
0%
$1$
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$\cfrac { { R }_{ 1 } }{ { R }_{ 2 } }$

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

Explanation

Moment of inertia through centre of rod=

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

Moment of inertia through a point at a distance of

$\dfrac{l}{4}$ from an end is I=

$\dfrac{Ml^2}{12}+\dfrac{Ml^2}{16}=\dfrac{7Ml^2}{48}$

A small ball strikes a stationary uniform rod, which is free to rotate, in gravity-free space. The ball does not stick to the rod The rod will rotate about:

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its centre of mass
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the centre of mass of 'rod plus ball'
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the point of impact on the ball on the rod
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the point about which the moment of inertia of the rod plus ball' is minimum

A ball is made of a material of density

$\rho$ when

${ \rho }_{ oil }<{ \rho }<{ \rho }_{ water }$ with

${ \rho }_{ oil }$ and

${ \rho }_{ water }$ representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?

Explanation

$\rho_)<\rho<\rho_\omega$

$\rho_\omega\rightarrow$ Density of water

$\rho_0\rightarrow$ Density of oil

$\rho\rightarrow$ Density of ball

The principal is thathigher density materials settels down and lower density floats over it So

$\rho_0>\rho$ ball will go down the oil and

$\rho_\omega>\rho$ so ball will float above the water.

The radius of gyration of a body about an axis passing through its centre of mass is

$24\ cm$ . Calculate the radius of gyration of the body about parallel axis passing through a point at a distance

$7\ cm$ from its centre of mass.

0%
$16\ cm$
0%
$20\ cm$
0%
$30\ cm$
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$25\ cm$

The radius of gyration of a body about an axis at a distance of

$4\ cm$ from its centre of mass is

$5\ cm$ . The radius of gyration about a parallel axis through centre of mass is:

0%
$2\ cm$
0%
$5\ cm$
0%
$4\ cm$
0%
$3\ cm$

Explanation

$mr_{gCM}^2+mr^2=mr_{g}^2$

$r_{gCM}^2+r^2=r_{g}^2$

$r_{gCM}^2+4^2=5^2$

$r_{gCM}=3cms$

The ratio of the radii of gyration of a spherical shell and a solid sphere of the same mass and radius about a tangential axis is:

0%
$\sqrt { 3 } :\sqrt { 7 }$
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$\surd 5:\surd 6$
0%
$\sqrt { 25 } :\sqrt { 21 }$
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$\sqrt { 21 } :\sqrt { 25 }$

Explanation

Moment of inertia of a solid sphere about tangential axis is

$\dfrac{2}{3}mr^2+mr^2=\dfrac{5}{3}mr^2$

Radius of gyration=

$\sqrt{\dfrac{5}{3}}r$

Moment of inertia of a spherical shell about tangential axis is

$\dfrac{2}{5}mr^2+mr^2=\dfrac{7}{5}mr^2$

Radius of gyration=

$\sqrt{\dfrac{7}{5}}r$

Ratio of their radii of gyration=

$\sqrt{\dfrac{3}{7}}$

Four particles each of mass

$'m'$ are placed at the corners of a square of side

$'L''$ . The radius of gyration of the system about an axis normal to the square and passing through its centre.

0%
$\dfrac { L }{ 2 }$
0%
$\dfrac { L }{ \sqrt { 2 } }$
0%
$L$
0%
$\sqrt { 2 }\ L$

Explanation

I=m

$(\dfrac{l}{\sqrt{2}})^2\times 4=2ml^2$

$2ml^2=4mr^2$

Radius of gyration r=

$\dfrac{l}{\sqrt{2}}$

Two identical rods each of moment of inertia

$'I'$ about a normal axis through centre are arranged in the from of a cross. The

$M.I$ . of the system about an axis through centre and perpendicular to the plane of system is:

0%
$2I$
0%
$I$
0%
$2.5\ I$
0%
$6I$

Explanation

$I_z=I_x=I_y$

Where x and y and z are mutually perpendicular axes

$I_0=I+I=2I$

A thin rod of length

$0.6m$ is vertically straight on the horizontal floor. This falls freely to one side without slipping of its bottom. The angular velocity of the centre of a rod when its top end touches the floor is?

0%
$5\sqrt { 2} rad$
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$\sqrt { \dfrac { 3}{ 2 } }rad$
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$\sqrt { 3 } rad$
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$\sqrt { \dfrac { 3 }{ 4 } }rad$

A wire of length

$l$ is bent into the shape of an

$n$ sided regular polygon of mass

${m}_{1}$ then

0%
Its moment of inertia about its natural axis $\dfrac { { ml }^{ 2 } }{ { 4n }^{ 3 } } \left( \dfrac { 1 }{ 3 } +\cot ^{ 2 }{ \left( \dfrac { \pi }{ n } \right) } \right)$
0%
Its moment of inertia about its natural axis $\dfrac { { ml }^{ 2 } }{ { 4n }^{ 2 } } \left( \dfrac { 1 }{ 3 } +\cot ^{ 2 }{ \left( \dfrac { \pi }{ n } \right) } \right)$
0%
The moment of inertia about its natural axis in the limiting case $(n\rightarrow \alpha)$ is $\dfrac { { ml }^{ 2 } }{ { 4\pi }^{ 2 } }$
0%
The moment of inertia about its natural axis in the limiting case $(n\rightarrow \alpha)$ is $\dfrac { { ml }^{ 2 } }{ { 8\pi }^{ 2 } }$

Four rods of equal length l and mass m each form a square as shown in figure. The moment of inertia of the system about the axis 1 and axis 2 is

0%
$\dfrac { 2 }{ 3 } { ml }^{ 2 }$ , $\dfrac { 5 }{ 3 } { ml }^{ 2 }$ ,
0%
$\dfrac { 5 }{ 3 } { ml }^{ 2 }$ , $\dfrac { 10 }{ 3 } { ml }^{ 2 }$ ,
0%
$\dfrac { 1 }{ 3 } { ml }^{ 2 }$ , $\dfrac { 5 }{ 3 } { ml }^{ 2 }$ ,
0%
$\dfrac { 10 }{ 3 } { ml }^{ 2 }$ , $\dfrac { 2 }{ 3 } { ml }^{ 2 }$ ,

Two identical discs of mass

$M$ and radius

$R$ are joined to form a figure of eight [see Figure]. The radius of gyration about an axis through their point of contact and perpendicular to the plane:

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$\sqrt { 5 }\ R$
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$\sqrt { 3 }\ R$
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$\sqrt { 7 }\ R$
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$\sqrt { \dfrac { 3 }{ 2 } }\ R$

Two uniform thin rods each of mass

$'m'$ and length

$'L'$ are arranged to form a cross. The moment of inertia of the system about an angular bisector is

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

Explanation

Moment of inertia of the rod about its centre is I=

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

Moment of inertia of the system of rods which form a cross about an axis passing through the centre and perpendicular to the rods is I=

$I_x+I_y=2I$ =

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

Moment of inertia of the system through angular bisectors of rods =

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

Momentum of inertia of a rod of mass

$2\ kg$ and length

$1\ m$ about an axis passing through a point

$25\ cm$ from the center and normal to the length is:

0%
$2.9\ kg\ {m}^{2}$
0%
$1.9\ kg\ {m}^{2}$
0%
$0.29\ kg\ {m}^{2}$
0%
$0.19\ kg\ {m}^{2}$

Explanation

Moment of inertia=

$\dfrac{ml^2}{12}+mr^2$

$\dfrac{2\times 1^2}{12}+2\times (\dfrac{1}{4})^2$ =0.29kgm²

Three identical thin rods each of length

$L$ and mass

$m$ are welded perpendicular to one another as shown in the figure. The assembly is rotated about an axis that passes through the end of one rod and is parallel to another. Find moment of inertia of this structure.

0%
$\dfrac { 11 }{ 12 } m{ L }^{ 2 }$
0%
$\dfrac { 9 }{ 16 } m{ L }^{ 2 }$
0%
$\dfrac { 8 }{ 11 } m{ L }^{ 2 }$
0%
$2m{ L }^{ 2 }$

Radius of gyration of a body about an axis at a distance 6 cm from its center of mass is 10 cm. Its radius of gyration about a parallel axis through its center of mass is :

0%
6 cm
0%
8 cm
0%
10 cm
0%
12 cm

A uniform rod is kept vertically on a horizontal smooth surface at a point

$O$ . If it is disturbed slightly and released, it falls down on the horizontal surface. The lower end will be

0%
At $O$
0%
At a distance less than $l/2$ from $O$
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At a distance $l/2$ from $O$
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At a distance larger than $l/2$ from $O$

Velocity of centre of mass of system (A+B) when block B just looses contact with the wall is :-

0%
$3g\sqrt{\dfrac{m}{k}}$
0%
$\dfrac{3g}{2}\sqrt{m}{k}$
0%
$2g\sqrt{\dfrac{m}{k}}$
0%
none of these

Explanation

Given that,

Mass of A =m

Mass of B=m

Spring constant of spring =k

Initial compression of spring

$=3mg/$ (x)

Assumption : Neglect friction anywhere.

Figure:- Rough sketch

Refer above image,

Assumed variables:

1) Natural length of spring = l

2) Final velocity of block A =

$V_A$

3) Velocity of centre of mass =

$V_{com}$

As we know that,

The compression of spring will increase the elastic potential energy of spring and when spring again regains its original length l the potential energy gets converted to kinetic energy of block A. by work energy theorem.

Block B remains stationary until the spring attains naturals length

$(V_{Block}\,B=0)$

Potential energy of spring

$=\dfrac{1}{2}kx^2$

k.E of block

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

$\boxed{\text{conservation of work-energy theorem} \dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2;}$

$mv^2=k(3mg/k)^2=\dfrac{k.9}{k^2}m^2g^2$

$\boxed{V_A=3g\sqrt{m/k}}$

$\boxed{v_{com}=\dfrac{mv_A+mv_B}{m+m}}$

$=\dfrac{mv_A}{2m}=\dfrac{v_A}{2}(v_B=0)$

$\boxed{v_{com}=\dfrac{3g}{2}\sqrt{m/k}}$

hence, velocity of centre of mass of A and B is

$\dfrac{3g}{2}\sqrt{\dfrac{m}{k}}$

$\therefore$ Option D is correct.

2079741_1058394_ans_0c114c9b3b2d45c2b3a79e8507f6bee0.png

Seven homogenous bricks, each of length L, are arranged as shown in figure. Each brick is displaced with respect to the one in contact by L/Find the x-ordinate of the centre of mass relative to the origin O shown.

0%
$\dfrac{22L}{5}$
0%
$\dfrac{2L}{35}$
0%
$\dfrac{22L}{35}$
0%
$\dfrac{12L}{35}$

AB and CD are two identical rods each of length l and mass m joined to form a cross is fixed inside a ring of mass m and radius

$\dfrac{l}{2}.$ Moment of inertia of the system about a bisector of the angle between the rods (xy) is:

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

A thin rod of length

$4l$ , mass

$4m$ is bent at the points as shown in figure. What is the moment of inertia of the rod about the axis passing point O and perpendicular to the plane of the paper?

0%
$\cfrac{m{l}^{2}}{3}$
0%
$\cfrac{10m{l}^{2}}{3}$
0%
$\cfrac{m{l}^{2}}{12}$
0%
$\cfrac{m{l}^{2}}{24}$

Explanation

Since the mass of rod is 54m and length is 4l.

so massC = m

and length of AB = BO=OC= CD = l

WE, know moment of Inertia of a rod about to end =

$\dfrac{ml^2}{3}$

So, moment of Inertia of AB,BO,OC,CD about B,O,O,C respectively =

$\dfrac{ml^2}{3}$

FRom parallel axis theorem

Moment of Inertia of AB about O.

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

Similerly od CD about O

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

SO moment of Inertia Rod about O

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

$= \dfrac{10ml^2}{3}$

Hence option(B) is correct.

Two particles of equal masses have velocities

$\vec{V_1} = 2\hat{i}$ m/s. The first particle has an acceleration

$\vec{a_1} = (3\hat{i} + 3\hat{j}) \frac{m}{s^2}$ . while the accelration of the other particle is zero. The centre of mass of the particle moves in a :

0%
circle
0%
parabola
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straight line
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ellipse

A man of mass

$80kg$ is riding on a small cart of mass

$40kg$ which is rolling along a level floor at a speed of

$2m/s$ . He is running on the cart so that his velocity relative to the cart is

$3m/s$ in the direction opposite to the motion of cart .What is the speed of the centre of mass of the system :-

0%
$1.5m/s$
0%
$1 m/s$
0%
$3m/s$
0%
$0$

A uniform solid right circular cone of base radius R is joined to a uniform solid hemisphere of radius R and of the same density, as shown. The centre of mass of the composite solid lies at the centre of base the cone. The height of the cone is

0%
1.5 R
0%
$\sqrt 3$ R
0%
3 R
0%
$2\sqrt 3$ R

Explanation

Hint: Use the formula of center of mass as:

${X}=\dfrac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}$

Step 1: Finding the masses of the cone and hemisphere:

let,

Volume of cone

$\Rightarrow$

$V_1=\dfrac{1}{3} \pi r^{2} h$

Mass of cone,

$\Rightarrow$

$m_{1}=\rho \times \dfrac{1}{3} \pi r^{2} h$ (

$\rho$ is density)

Volume of hemisphere,

$\Rightarrow$

$V_2= \dfrac {1}{2} \times \dfrac{4}{3} \pi r^{3}$

$\Rightarrow$

$\dfrac{2}{3} \pi r^{3}$

Mass of hemisphere,

$\Rightarrow$

$m_{2}=\rho \times \dfrac{2}{3} \pi r^{3}$

Step 2: Showing the coordinate of the center of masses of hemisphere and cone:

Step 3: Calculation of height h

Now formula of center of mass in

$Y$ coordinate is given by:

$\Rightarrow$

${Y}=\dfrac{m_{1} y_{1}+m_{2} y_{2}}{m_{1}+m_{2}}$
(Taking

$G$ as origin)

Putting the values:

$\Rightarrow0=\dfrac{\rho\times\dfrac{1}{3}\pi r^2h\times (\dfrac{-h}{4})+\rho\times\dfrac{2}{3}\pi r^3\left(\dfrac{3r}{8}\right)}{\rho\times\dfrac{1}{3}\pi r^2h+\rho\times\dfrac{2}{3}\pi r^3}$

$\Rightarrow$

$\quad \rho \times \dfrac{1}{3} \pi r^{2}\left[-\dfrac{h^{2}}{4}+2 r \times \dfrac{3 r}{8}\right]=0$

$\Rightarrow$

$-\dfrac{h^2}{4}+\dfrac{3r^2}{4}=0$

$\Rightarrow$

$h=\sqrt{3}r$

Therefore, the correct answer is 'B' is

$\sqrt{3} r$ .

$\textbf{Hence option B correct}$

A rigid of mass

$M$ slides along semi circular track (in vertical plane) followed by a flat track. At the given instant velocity of end

$'B'$ is

$V$ along the horizontal plane. Then at the given instant

0%
Angular speed of the rod $\dfrac {v}{r}$
0%
$V_{CM} = \dfrac {V} { \sqrt{3}}$
0%
$\vec {L}$ at $'O'$ is $\dfrac {4}{3} mvr$
0%
$\dfrac {KE_r}{KE_t} = \dfrac {4}{3} mvr$

The angular velocity of earth about its axis of rotation is -

0%
$2 \pi / (60 \times 60 \times 24) rad/sec$
0%
$2 \pi / (60 \times 60) rad/sec$
0%
$2 \pi / 60 \times rad/sec$
0%
$2 \pi / (365 \times 24 \times 60 \times 60) rad/sec$

Explanation

The angular velocity is given as,

$\omega = \dfrac{{\Delta {\rm{\theta }}}}{{\Delta t}}$

$\omega = \dfrac{{2\pi }}{{1\;{\rm{day}}}}$

$\omega = \dfrac{{2\pi }}{{24 \times 60 \times 60}}\;{\rm{rad/sec}}$

Thus, the angular velocity of earth about its axis of rotation is $\dfrac{{2\pi }}{{24 \times 60 \times 60}}\;{\rm{rad/sec}}$ .

A wheel of grind stone has applied at its axle

$2\ cm$ in radius, a constant tangential force of

$600\ N$ . Calculate the torque acting on it, and angular momentum acquired by it after

$8\ sec$ starting from rest.

0%
$24 m^2/s$
0%
$108 m^2/s$
0%
$72 m^2/s$
0%
$96 m^2/s$

Two solid spheres of iron have their radii in the ratio

$2:1$ . Then ratio of the moments of inertia about their diameters is :

0%
$4:1$
0%
$8:1$
0%
$16:1$
0%
$32:1$

Explanation

The moment of inertia is given as,

$I = \dfrac{2}{5}M{R^2}$

$I = \dfrac{2}{5} \times \dfrac{4}{3}\pi {R^3} \times \rho \times {R^2}$

$I \propto {R^5}$

$\dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{{{R_1}}}{{{R_2}}}} \right)^5}$

The ratio of moments of inertia about their diameters is given as,

$\dfrac{{{I_1}}}{{{I_2}}} = {\left( {\dfrac{2}{1}} \right)^5}$

$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{32}}{1}$

The time period of a bar pendulum when suspended at distances

$30\ cm$ and

$50\ cm$ from its centre of gravity comes out to be the same. If the mass of the body is

$2kg$ . Find out its moment of inertia about an axis passing through first point.

0%
$0,24\ kg-m^{2}$
0%
$0.72\ kg-m^{2}$
0%
$0.48\ kg-m^{2}$
0%
Data insufficient

Explanation

For point

$-1$

$\tau =mg \left (\dfrac {30}{100}\right)\sin \Theta$

$\Theta$ is very small

$\tau =\dfrac {3\ mg}{10}\Theta$

$C_1=\dfrac {3\ mg}{10}$

where

$t$ point

$2$

$\tau =\dfrac {5}{10}mg \Theta$

$C_2=\dfrac {5}{10}mg$

As the period of

$SHM(T)=\Theta n\sqrt {\dfrac {I}{C}}$

Let at

$COM\ I$ is the moment of inertia

By parallel axis theorem

$I_1=I+(2)\left (\dfrac {3}{10}\right)^2$

$I_2=I+2\left (\dfrac {5}{10}\right)^2$

Because time period of blue points are equal

$2\pi \sqrt {\dfrac {I_1}{c_1}}=2\pi \sqrt {\dfrac {T_2}{c_2}}\ \Rightarrow \dfrac {I_1}{I_2}=\dfrac {c_1}{c_2}$

$\Rightarrow \ \dfrac {I+2\left (\dfrac {3}{10}^2 \right)}{I+2\left (\dfrac {5}{10}\right)^2 }=\dfrac {\dfrac {3\ mg}{10}}{\dfrac {5\ mg}{10}}=\dfrac {3}{5}$

$\Rightarrow \ I=0.30\ kg\ m^2$

At point

$-1, I_1=0.30+2\left (\dfrac {3}{10}\right)^2 =0.48\ kg\ m^2$

Option

$C$ is correct

Two small sphere of masses

$10kg$ and

$30kg$ are joined by a rod of length

$0.5m$ and of negligible mass . The M.I of the system about a normal axis through centre of mass of the system is

0%
$1.875kgm^2$
0%
$245kgm^2$
0%
$0.75khm^2$
0%
$1.75kgm^2$

$O$ is the centre of a an equilateral triangle

$ABC$ .

$F_{1}, F_{2}$ and

$F_{3}$ are three force acting along the sides

$AB, BC$ and

$AC$ as shown in the figure. What should be the magnitude of

$F_{3}$ , so that the total torque about

$O$ is zero?

0%
$(F_{1} + F_{2})$
0%
$2(F_{1} + F_{2})$
0%
$(F_{1} + F_{2})/2$
0%
$(F_{1} - F_{2})$

A thin rod of length L and mass M is bend at the middle point O as shown in figure. Consider an axis passing through two middle point O and perpendicular to the plane of the bent rod.Then moment of inertia about this axis is :

0%
$\dfrac{2}{3} mL^2$
0%
$\dfrac{1}{3} mL^2$
0%
$\dfrac{1}{12} mL^2$
0%
$\dfrac{1}{24} mL^2$

Two bodies have their moments of inertia I and

$2$ I respectively about their axis of rotation.If their kinetic energies of rotation are equal, their angular momenta will be in the ratio-

0%
$\sqrt { \dfrac { 3 }{ 1 } }$
0%
$\sqrt { \dfrac { 6 }{ 1 } }$
0%
$\sqrt { \dfrac { 2 }{ 1 } }$
0%
$\sqrt { \dfrac { 5 }{ 1 } }$

Explanation

$\dfrac { 1 }{ 2 } \ast { i }_{ 1 }\ast { \omega }_{ 1 }^{ 2 }=\dfrac { 1 }{ 2 } \ast { i }_{ 2 }\ast { \omega }_{ 2 }^{ 2 }\\$

here

${ i }_{ 1 }=l,{ i }_{ 2 }=2l\\$

$\frac { { \omega }_{ 1 } }{ { \omega }_{ 2 } } =\sqrt { \dfrac { 2 }{ 1 } }$

The moment of inertia of a rod about an axis through its centre and perpendicular to it is

$\dfrac {ML^{2}}{12}$ (where

$M$ is the mass and

$L$ , the length of the rod). The is bent in the middle so that the each halves make an angle of

$30^{o}$ with that axis. The moment of inertia of the bent rod about the sum axis would be (The axis lies in the plane of structure)

0%
$\dfrac {ML^{2}}{96}$
0%
$\dfrac {ML^{2}}{48}$
0%
$\dfrac {ML^{2}}{12}$
0%
$\dfrac {ML^{2}}{8\sqrt {3}}$

A thin uniform wire is bent to form the two equal sides

$AB$ and

$AC$ of triangle

$ABC$ , where

$AB=AC=5\ cm$ . The third side

$BC$ , of length

$6\ cm$ , is made from uniform wire of twice the density of the first. Find the centre of mass of the framework.

0%
$\left(3,\dfrac{10}{11} \right)$
0%
$\left(\dfrac{21}{16},\dfrac{20}{16} \right)$
0%
$\left(\dfrac{10}{8},\dfrac{21}{8} \right)$
0%
$\left( 5,6\right)$

Two thin rods of mass

$m$ and length

$l$ , each are joined to form a

$L$ shape as shown. The moment of inertia of rods about an axis passing through free end (O) of a rod and perpendicular to both the rods is:

0%
$\dfrac{2}{7} ml^2$
0%
$\dfrac{ml^2}{6}$
0%
$ml^2$
0%
$\dfrac{5ml^2}{3}$

Explanation

Moment of inertia of rod 1 about O perpendicular to both the rods

${ I }_{ 1 }=\dfrac { m{ l }^{ 2 } }{ 12 } +m{ x }^{ 2 }$ where x is the distance between O and center of mass of rod.

$l=\sqrt { { l }^{ 2 }+{ (\dfrac { l }{ 2 } ) }^{ 2 } } =\sqrt { \dfrac { { 5l }^{ 2 } }{ 4 } }$

Moment of inertia of rod 2 about axis through O perpendicular to both rods

${ I }_{ 2 }=\dfrac { m{ l }^{ 2 } }{ 3 }$

Total moment of inertia I

$={ I }_{ 1 }{ +I }_{ 2 }$ -------(1)

${ I }_{ 1 }=\dfrac { m{ l }^{ 2 } }{ 12 } +\dfrac { 5m{ l }^{ 2 } }{ 4 } =\dfrac { 4m{ l }^{ 2 } }{ 3 }$

Thus 1 becomes

$=\dfrac { 4m{ l }^{ 2 } }{ 3 } +\dfrac { m{ l }^{ 2 } }{ 3 } =\dfrac { 5m{ l }^{ 2 } }{ 3 }$

A sphere

$(C.O.M)$ is moving at

$v_{0}$ has angular velocity

$\dfrac {3v_{0}}{4\ R}$ as show. Now it placed on a rough horizontal surface. Its final velocity (of

$C.O.M$ ) is:

0%
$\dfrac {v_{0}}{7}$
0%
$\dfrac {v_{0}}{2}$
0%
$\dfrac {13v_{0}}{14}$
0%
$\dfrac {5v_{0}}{7}$

A body is dropped from a height h, if it acquires a momentum p, then the mass of the body is:

0%
$\dfrac{p}{\sqrt{2gh}}$
0%
$\dfrac{P^2}{2gh}$
0%
$\dfrac{2gh}{p}$
0%
$\sqrt{2gh/p}$

Explanation

momentum (p) = mv

So mass of the body(m) =

$\dfrac{p}{v}$

$v = \sqrt{2gh}$

$\therefore$ mass

$=\dfrac{p}{\sqrt{2gh}}$

hence (A) is correct answer

Three identical rods are hinged at point A as shown. The angle made by rod AB with vertical is

0%
$tan^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right)$
0%
$tan^{ -1 }\left( \frac { 3 }{ 4 } \right)$
0%
$tan^-1 (1)$
0%
$tan^{ -1 }\left( \frac { 4 }{ 3 } \right)$

A ladder

$PQ$ of length

$5 m$ inclined to a vertical wall is slipping over a horizontal surface with a velocity of

$2 ms^{-1}$ , when

$Q$ is at a distance

$3 cm$ from the ground. Calculate the velocity of centre of mass of the rod at this moment.

0%
$2.667 \ ms^{-1}$
0%
$1.25 \ ms^{-1}$
0%
$1.75 \ ms^{-1}$
0%
$2.75 \ ms^{-1}$

Explanation

In order to find the height of wall when calculated comes to

$3$ .

Through Pythagoras theorem, we come to know that

$x=5$ and

$y=3$ .

Now with respect to differentiation, the equation comes to

$(2x) \times (dx/dt) + (2 y) \times (dy/dt) = 0$ .

When you substitute

$x$ and

$y$ , you will know velocity will be

${\dfrac{8}{3}}$ .

The moment of inertia of a rod and length

$l$ about an axis passing through its centre of mass and perpendicular to rod is

$i$ . The moment of inertia of hexagonal shape formed by six such rods, about an axis passing through its centre of mass and perpendicular to its plane will be?

0%
$15$ $I$
0%
$40$ $I$
0%
$60$ $I$
0%
$80$ $I$

A thin rod of length

$4l$ , mass

$4$ m is bent at the points as shown in the figure. What is the moment of inertia of the rod about the axis passing through

$O$ and perpendicular to the plane of the paper?

0%
$\dfrac{ml^2}{3}$
0%
$\dfrac{10ml^2}{3}$
0%
$\dfrac{ml^2}{12}$
0%
$\dfrac{ml^2}{24}$

Explanation

Since, the mass of the rod is

$4 m$ and the length is

$4 l$

so, mass of

$AB = BO = OC = CD = m$

and, length of

$AB = BO = OC = CD = l$

We know, Moment of inertia of a rod about its end =

${\dfrac{(ml^2)}{3}}$

Moment of inertia of

$AB$ about

$B (I_1)$ =

${\dfrac{ml^2)}{3}}$

Moment of inertia of

$BO$ about

$O (I_2)$ =

${\dfrac{(ml^2)}{3}}$

Moment of inertia of

$OC$ about

$O (I_3)$ =

${\dfrac{(ml^2)}{3}}$

Moment of inertia of

$CD$ about

$C (I_4)$ =

${\dfrac{(ml^2)}{3}}$

Now, from parallel axis theorem,

Moment of inertia of

$AB$ about

$O (I_5)$ :

$= I_1 + ml^2$ ————(parallel axis theorem)

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

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

Similarly, Moment of inertia of

$CD$ about

$O (I_6)$

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

So, moment of inertia of the rod about

$O$

$I_2 + I_3 + I_5 + I_6$

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

${\dfrac{(10 ml^2)}{3}}$

A uniform rod AB of length l and mass m is free to rotate about A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is

$\dfrac{ml^2}{3}$ , the initial angular acceleration of the rod will be?

0%
$\dfrac{3g}{2l}$
0%
$\dfrac{2g}{3l}$
0%
$mg\dfrac{l}{2}$
0%
$\dfrac{3}{2}gl$

Explanation

Given that moment of inertia about A is

$I={ ml }^{ 3 }/3$

Now, torque about A is given as

$\tau=F×r$

$\tau=mg\frac { l }{ 2 }$

$I\alpha=mg\frac { l }{ 2 }$

$\dfrac { { ml }^{ 2 } }{ 3 } \alpha=mg\dfrac { l }{ 2 }$

$\alpha=\dfrac { 3g }{ 2l }$

The moment of inertia of a meter stick of mass

$400 g$ about an axis passing through

$20 cm$ mark and perpendicular to its length is

0%
$3.9 \times 10^5 cm^2$
0%
$2.9 \times 10^5 cm^2$
0%
$4.9 \times 10^5 cm^2$
0%
$6.9 \times 10^5 cm^2$

Explanation

Moment of inertia of metre stick about its centre is given by:

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

$I = \dfrac{{0.4 \times {1^2}}}{{12}}$

Now moment of inertia about axis perpendicular to length of stick at 20 cm mark

$I = \dfrac{{0.1}}{3} + 0.4 \times {\left( {\dfrac{3}{{10}}} \right)^2}$

$I = \dfrac{{0.1}}{3} + \dfrac{{3.6}}{{100}}$

$I = \dfrac{{10 + 10.8}}{{300}}$

$I = \dfrac{{20.8}}{{300}}$

$I = 0.069\;{\rm{kg}}{{\rm{m}}^{\rm{2}}}$

$I = 6.9 \times {10^5}\;{\rm{gc}}{{\rm{m}}^{\rm{2}}}$

The centre of mass of non uniform rod of length

$L$ whose mass per unit length

$\lambda$ varies as

$\lambda=k{x}^{2}$ where

$k$ is a constant and

$x$ is the distance of any point on rod from left end

$A$ is (from the same end)

0%
at the centre of the rod
0%
is at $x=\cfrac{3L}{4}$
0%
is at $x=\cfrac{4L}{5}$
0%
is at $x=\cfrac{5L}{6}$

Let

$\xrightarrow [ P ]{ }$ be the liner momentum of a particle whose whose position fector is

$\xrightarrow [ r ]{ }$ with respect to the origin and

$\xrightarrow [ L ]{ }$ be the angular momentum at this particle about the origin then.

0%
$\xrightarrow [ r ]{ p} \xrightarrow [ I ]{ } =\quad 0\quad and\quad \xrightarrow [ p ]{ } \xrightarrow [ L ]{ } =0$
0%
$\xrightarrow [ r ]{ } \xrightarrow [ I ]{ } =\quad 0\quad and\quad \xrightarrow [ p ]{ } \xrightarrow [ L ]{ } =0$
0%
$\xrightarrow [ r ]{ k} \xrightarrow [ I ]{ } =\quad 0\quad and\quad \xrightarrow [ p ]{ } \xrightarrow [ L ]{ p} =0$
0%
$\xrightarrow [ r ]{ f} \xrightarrow [ I ]{ } =\quad 0\quad and\quad \xrightarrow [ p ]{ } \xrightarrow [ L ]{ } =0$

A small block of mass

$m$ is moving on a horizontal table surface at initial speed

$v_0$ . It then moves smoothly onto a sloped wedge of mass

$M$ . The wedge can also move on the table surface. Assume that everything moves without friction. Which of the following statement(s) is (are) correct :-

0%
COM of the system always moves rightward
0%
COM of the system moves upwards for some time
0%
Speed of M always increases during contact with $m$ .
0%
Speed of COM of the system is always constant

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 12 - 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 12 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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