MCQ Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 9

Find the gravitational force of attraction between a particle of mass m and a uniform slender rod of mass M and length L for the two orientations shown in the figure.

0%
0%
0%
none of these
0%
both A and B.

A football rolls through the ground. The path followed by center of mass of football is:

0%
linear
0%
circular
0%
rotational
0%
all the above

Centre of mass is a point

0%
Which is geometric centre of a body
0%
From which distance of particles are same
0%
Where the whole mass of the body is supposed the
0%
none of these

A cylinder of mass m and radius R is rolling without slipping on a horizontal surface with angular velocity

${ \omega }_{ 0 }$ . The velocity of center of mass cylinder is

${ \omega }_{ 0 }R$ . The cylinder comes across a step of height

$\dfrac{R}{4}$ . Then the angular velocity of cylinder just after the collision is (Assume cylinder remains in contact and no slipping occurs on the edge of the step)

0%
$\dfrac{5\omega_o}{6}$
0%
${ \omega }_{ 0 }$
0%
$2\omega_o$
0%
$6{ \omega }_{ 0}$

Explanation

Given,

Mass of cylinder

$= m$

Radius of cylinder

$= R$

Angular velocity

$=\omega_0$

Velocity of center of mass

$=\omega_0 R$

Height

$h=\dfrac R4$

Using conservation of angular momentum

Initial angular momentum

$=$ Final angular momentum

The angular momentum is the product of the moment of inertia and angular velocity.

$L=r_{cm}\,m\,v_{cm}+I_{cm}\,\omega$

$\dfrac 34mR^2 \omega =(R-\dfrac R4)mv_0+\dfrac 12 mR^2(\dfrac{v_0}{R})$

$\dfrac 34 R \omega=\dfrac 34 v_0+\dfrac 12v_0$

$\omega =\dfrac 56 \times \dfrac{v_0}{R}$

$\omega=\dfrac{5\omega_0}{6}$

The object in figure is constructed of uniform wire bent into the shape from for the figure shown below the center of mass w.r.t. given frame is

0%
$\left( \frac { 1 } { 2 } \cos \frac { \theta } { 2 } , \frac { 1 } { 2 } \sin \frac { \theta } { 2 } \right)$
0%
$\left( \frac { 1 } { 4 } \cos ^ { 2 } \left( \frac { \theta } { 2 } \right) \cdot \frac { 1 } { 8 } \sin \theta \right)$
0%
$\left( \frac { 1 } { 2 } \sin ^ { 2 } \left( \frac { \theta } { 2 } \right) , \frac { 1 } { 2 } \right)$
0%
$\left( \frac { 1 } { 2 } , \frac { 1 } { 2 } \cos ^ { 2 } \theta \right)$

Explanation

$\begin{array}{l} Ycom=\dfrac { { { m_{ 1 } }{ y_{ 1 } }+{ m_{ 2 } }{ y_{ 2 } } } }{ { { m_{ 1 } }+{ m_{ 2 } } } } \\ =\dfrac { { { m_{ 1 } }\left( 0 \right) +m\left( { \dfrac { l }{ u } \sin \theta } \right) } }{ { 2m } } \\ =\dfrac { l }{ 8 } \sin \theta \\ Hence,\, the\, option\, B\, is\, the\, correct\, answer. \end{array}$

A thin wire of length l and uniform linear mass density p is bent in the form of a square. Its moment of inertia about an axis along one of its edge is

0%
$\frac{2p}{3} (\frac{l}{4})^3$
0%
$\frac{3p}{2} (\frac{l}{4})^3$
0%
$\frac{5p}{3} (\frac{l}{4})^3$
0%
$\frac{1p}{3} (\frac{l}{4})^3$

Centre of mass of two uniform rods of the same length but made up of different materials and kept as shown, if the meeting point is the origin of co-ordinates

0%
$(L/4),\, (L/4)$
0%
$(2L/3),\, (L/2)$
0%
$(L/3),\, (L/3)$
0%
$(L/3),\, (L/6)$

A thin uniform rod has moment of inertia l about an axis parallel to its length at a certain distance. If on heating its length increases by 20%, then moment of inertia of rod will change by

0%
0 %
0%
1 %
0%
$\frac 1 2$ %
0%
4%

Moment of inertia of a thin rod of length

$L$ and mass

$M$ about an axis passing through one of its end is

0%
$\dfrac {ML^{2}}{3}$
0%
$\dfrac {ML^{2}}{12}$
0%
$\dfrac {ML^{2}}{2}$
0%
None of these

Two particles of masses 2 kg and 3 kg start to move towards each other due to mutual forces of attraction. The speed of first particle is

$V_1$ and that of the other is

$V_2$ ata certain instant. The speed of centre of mass is

0%
$\frac{v_1 + v_2}{2}$
0%
$\frac{2v_1 + 3v_2}{5}$
0%
$\frac{3v_1 + 2 V_2}{5}$
0%
Zero

The moment of inertia of a uniform cylinder of length

$\ell$ and radius R about its perpendicular bisector is I. What is the ratio

$\ell /R$ such that the moment of inertia is minimum ?

0%
1
0%
$\frac{3}{\sqrt 2}$
0%
$\sqrt {\frac{3}{2}}$
0%
$\frac{\sqrt 3}{2}$

If the linear momentum is increased by

$5$ % the kinetic energy will be increased by:

0%
$50$ %
0%
$100$ %
0%
$125$ %
0%
$10$ %

A mass of 3 Kg has a velocity of

$\hat{i}$ + 6

$\hat{j}$

$ms^{-1}$ . The momentum will be:

0%
$\left( 3\hat { i } +18\hat { j } \right) ms^{-1}$
0%
$\left( 2.3\hat { i } -4.1\hat { j } \right) ms^{-1}$
0%
$\left( 3.2\hat { i } -4.1\hat { j } \right) ms^{-1}$
0%
$\left( 2.3\hat { i } +4.1\hat { j } \right) ms^{-1}$

An object comprises of a uniform ring of radius R and a uniform chord AB (not necessarily made of the same material) as shown. Which of the following cannot be the centre of mass of the object?

0%
$( \mathrm { R } / 3 , \mathrm { R } / 3 )$
0%
$( R / 2 , R / 2 )$
0%
$( \mathrm { R } / 4 , \mathrm { R } / 4 )$
0%
none of the above

Explanation

$According\, to\, question\\ the\, chord\, is\, a\, uniform\, \, chord\, and\, the\, center\, of\, mass\, coordinate\, \, of\, \, system\, will\, be\, \\ displaced\, in\, x\, and\, y\, coordinate\, \, by\, same\, value.\, \\ Hence,\, \, X\, and\, Y\, coordinate\, will\, be\, same. \\ so\, that\, wrong\, opton\, is\, D.\, \, (None\, of\, the\, above)$

For a particle showing motion under the force

$F=-5{ \left( x-2 \right) }^{ 2 },$ the motion is

0%
Translatory
0%
Oscillatory
0%
SHM
0%
All of these

A homogeneous solid cylindrical roller of radius

$R$ and mass

$M$ is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is___?

0%
$\dfrac {3F}{2mR}$
0%
$\dfrac {F}{3mR}$
0%
$\dfrac {2F}{3mR}$
0%
$\dfrac {F}{2mR}$

Explanation

$FR = \dfrac {3}{2} MR^{2} \alpha$

$\alpha = \dfrac {2F}{3MR}$ .

1142003_1330933_ans_e5814cd88ea541eab4d9bfd6d95e1c48.png

Find ratio of radius of gyration of a disc and ring of same radii at their tangential axis in plane.

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

Explanation

For disc

$MK^2_1 = \dfrac{5}{4} MR^2 \Longrightarrow K_1 = \dfrac{\sqrt{5}}{2}R$ (Ref. Image [i])
For Ring

$MK^2_2 = \dfrac{3}{2}MR^2 \Longrightarrow K_2 = \sqrt{\dfrac{3}{2}}R$ (Ref. Image [ii])

So

$\dfrac{k_1}{k_2} = \dfrac{\sqrt{5} \times \sqrt{2}}{2 \times \sqrt{3}} = \sqrt{\dfrac{5}{6}}$
1161336_1349632_ans_5e89154961cd444284c31d42efa5e07d.png

The moment of inertia of NaCl molecules with bond length r about an axis perpendicular to the bond and passing through the centre of mass is

0%
$({ m }_{ Na }+{ m }_{ Cl }){ r }^{ 2 }$
0%
$\frac { { m }_{ Na }\times { m }_{ Cl } }{ { m }_{ Na }+{ m }_{ Cl } } { r }^{ 2 }$
0%
$\frac { { m }_{ Na }+{ m }_{ Cl } }{ { m }_{ Na }\times { m }_{ Cl } } { r }^{ 2 }$
0%
$\frac { { m }_{ Na }+{ m }_{ Cl } }{ { m }_{ Na }-{ m }_{ Cl } } { r }^{ 2 }$

A mass of

$200gm$ has initial velocity

$V_i=2\hat{i}+3\hat{j}$ and final velocity

$-2\hat{i}-3\hat{j}$ . Find magnitude of change in momentum.

0%
$|\Delta\overset{\rightarrow}{p}|=3.84$
0%
$|\Delta\overset{\rightarrow}{p}|=3.04$
0%
$|\Delta\overset{\rightarrow}{p}|=1.44$
0%
$|\Delta\overset{\rightarrow}{p}|=2.04$

Explanation

Given,

$\ |\Delta \vec P|=1.44$

Given,

$m= 200gm=0.2kg$

Initial velocity, $$

$\vec v_i=2\hat i+3\hat j$

Final velocity,

$\vec v_f=-2\hat i-3\hat j$

Change in magnitude,

$\Delta P= m(v_f-v_i)$

$\Delta P=0.2(-4\hat i-6\hat j)$

$\Delta P=-0.8\hat i-1.2\hat j$ in vector form.

The magnitude of change in momentum is

$|\Delta P|=\sqrt{(-0.8)^2+(-1.2)^2}$

$|\Delta P|=1.44$

A bullet of mass 0.01

$\mathrm { kg }$ and traveling at a speed of 500

$\mathrm { m } / \mathrm { sec }$ strikes a block of which suspended by a string of length 5

$\mathrm { m }$ . The centre of gravity of the block is found to vertical distance of 0.1

$\mathrm { m }$ . What is the speed of the bullet after it emerges from the block?

0%
359 $\mathrm { m } / \mathrm { s }$
0%
220 $\mathrm { m } / \mathrm { s }$
0%
204 $\mathrm { m } / \mathrm { s }$
0%
284 $\mathrm { m } / \mathrm { s }$

Explanation

By energy conservation of the suspended block we can say

initial

$KE=$ final potential energy

$\dfrac{1}{2}m{v^2} = mgH$

$\dfrac{1}{2}m{v^2} = gH$

$v = \sqrt {2gH}$

$v = \sqrt {2 \times 10 \times 0.1} = 1.41\,m/s$

now

$,$ by momentum conservation

${P_{bullet}} = {P_{block}} + {P_{bullet}}$

$0.01 \times 500 = 1 \times 1.41 + 0.01 \times {v_f}$

$5 - 1.41 = 0.01 \times {v_f}$

$0.01{v_f} = 3.59$

${v_f} = 359\,m/s$

Hence,

option

$(A)$ is correct answer.

A body having its centre of mass at the origin has three of its particles at

$(a,0,0),(0,a,0),(0,0,a)$ the moment of inertia of the body about

$X$ and

$Y$ axis are

$0.2kg\ {m^2}$ the moment of inertia about its

$Z$ axis is

0%
is $0.20\ kg-{m^2}$
0%
is $0.40\ kg-{m^2}$
0%
$0.20\sqrt 2 kg-{m^2}$
0%
cannot be deducted with this information

Explanation

Given,

the moment of inertia of the body about X and Y axis are

$0.2kg m^2$

$I_X=m_2a^2+m_3a^2=0.2$

and,

$I_Y=m_1a^2+m_3a^2=0.2$

and C.O.M is at origin.

And we have yo find ;

$I_Z=m_1a^2+m_2a^2=?$

here ,

we have 2 equation and 3-variable

$(m_1, m_2,m_3)$

hence,

$I_Z$ cannot be deducted with this information.

The moment of inertia of a thin scale of length L and mass M about an axis passing through the center of mass and perpendlcular to its lenqth would be

0%
$\frac { 7M L ^ { 2 } } { 48 }$
0%
$\frac { M L ^ { 2 } } { 4 }$
0%
$\frac { M L ^ { 2 } } { 3 }$
0%
$\mathrm { ML } ^ { 2 }$

Explanation

$I = {I_{cm}} + m{\left( {\dfrac{1}{4}} \right)^2}$

$= \dfrac{{M{l^2}}}{{12}} + \dfrac{{M{l^2}}}{{16}}$

$= \dfrac{{4M{l^2} + 3M{l^2}}}{{48}}$

$= \dfrac{{7M{l^2}}}{{48}}$

Hence,

option

$(A)$ is correct answer.

1226425_1330932_ans_da651d6970ef4e40a44f413af66e74da.JPG

A particle with linear momentum of magnitude P is subjected to a force F= Kt (k > 0) which is directed along the directed of initial momentum. The time which its liner momentum changes to 3P is :

0%
$\sqrt { \dfrac { 2P }{ K } }$
0%
$2\sqrt { \dfrac { P }{ K } }$
0%
$\sqrt { \dfrac { 2K }{ P } }$
0%
$2\sqrt { \dfrac { K }{ P } }$

Explanation

Impulse is

$dP=Fdt$

integrating both sides we get

$\int _{P}^{3P}dP=\int _{t=0}^{t}Ktdt$

$2P=\dfrac{Kt^2}{2}$ so

$t=\sqrt{\dfrac{4P}{K}}$

Where will be the centre of mass on combining two masses

$m$ and

$M(M>m)$ ?

0%
$Towards \ m$
0%
$Towards \ M$
0%
$exactly \ between \ m \ and \ M$
0%
$None \ of \ the \ above$

Explanation

As we can see that the center of mass will be at

$r_c=\dfrac{mr+MR}{m+M}$ ,

where

$r$ and

$R$ are the position of the masses

$m$ and

$M$ respectively.

From the formula we notice that it will be between

$m$ and

$M$ and

$nearer$ to the higher mass

$M$ .

The centre of mass of a uniform thin hemispherical shell of radius R is located at a distance ?

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

Explanation

The centre of mass of a uniform thin hemispherical shell of radius R is located at a distance

$\dfrac{R}{2}$ from the center.

Two particle of mass

$2$ kg and

$4$ kg move a linear path in opposite direction with velocity

$2ms^-1$ and

$3ms^-1$ , then velocity of centre of mass of system is....

0%
$\frac{5}{3}m{s^{ - 1}}$ n the direction of second
0%
$\frac{4}{3}m{s^{ - 1}}$ n the direction of second
0%
$\frac{7}{3}m{s^{ - 1}}$ n the direction of second
0%
$\frac{8}{3}m{s^{ - 1}}$ in the direction of second object

In carbon monoxide molecules, the carbon and the oxygen atoms are separated by distance

$1.2 \mathring { A }$ . The distance of the centre of mass, from the carbon atom is

0%
$0.48\mathring { A }$
0%
$0.51\mathring { A }$
0%
$0.56\mathring { A }$
0%
$0.69\mathring { A }$

Two particles approach each other with different velocities. After collision , one of the particles momentum

$\overrightarrow { p }$ in their center of mass frame. in the same frame, the momentum of the other particles.

0%
0
0%
$-\overrightarrow { p }$
0%
$-\overrightarrow { p } /2$
0%
$-2\overrightarrow { p }$

Due to slipping, points

$A$ and

$B$ on the rim of the disc have the velocities shown. Distance between centre of disc and point on disc which is having zero velocity.

0%
0.04 $\mathrm { m }$
0%
0.05 $\mathrm { m }$
0%
0.06 $\mathrm { m }$
0%
None

Explanation

Angular velocity with respect to

$\frac{{{v_1}}}{r}$

$= \frac{{4.5}}{{0.4}}$

$A$ be at rest let a point at distance

$x$ from

$A$ has zero velocity . Then

$\begin{array}{l} { V_{ A } }+rq=0 \\ 1.5=-k\left( { \frac { { 4.5 } }{ { 0.4 } } } \right) \\ -\frac { { 0.4 } }{ 3 } =k \\ k=-0.133 \end{array}$

Distance of this point from center

$=0.2+0.133$ m

$=0.333$ m

1233798_1363328_ans_0c4896feecac4b88ae1ad3ae7c7813d8.JPG

which of these represent the centre of mass for a semicircular ring ?

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

Radius of gyration of the system of two rods each of mass '

$m ^ { \prime }$ and length

$l ^ { \prime }$ about an axis shown is:

0%
$l \sqrt { \frac { 1 } { 72 } }$
0%
$l \sqrt { 7 }$
0%
$l \sqrt { \frac { 7 } { 72 } }$
0%
None

Explanation

$\begin{array}{l} 2m{ K^{ 2 } }=\frac { { m{ l^{ 2 } } } }{ { 12 } } +\frac { { m{ l^{ 2 } } } }{ { 12 } } +\frac { { m{ l^{ 2 } } } }{ { 36 } } \\ K=l\sqrt { \frac { 7 }{ { 72 } } } \\ Ans.\, \, (C) \end{array}$
1215705_1368892_ans_a4b031b1d1e043539b0ec207d192d261.png

Two small kids weighing 10kg and 15kg are trying to balance a seesaw of total length 5m with the fulcrum at the centre. If one of the kids is sitting at an end, the other should sit at

0%
1.67 m from the centre
0%
2.50 m from the centre
0%
2 m from the centre
0%
1 m from the centre

Explanation

$\begin{array}{l}\text { Let, } B \text { of mass } 15\mathrm{~kg}\text { is sitting } \\\text { at distance } x \text { from centre. }\end{array}$

$\begin{array}{r}\text { By rotational equilibrium, }\\\qquad\begin{aligned}\left(\tau_{A}\right) &=\left(m_{A} g \times 2\cdot 5\right) \\\text { and }\left(\tau_{B}\right) &=\left(m_{B} g\times x\right)\end{aligned}\end{array}$

$\begin{array}{l}\text { Note that, } \tau_{A} \text { and }\tau_{B}\text { are in opposite directions. } \\\text { Thus, } m_{A} g \times 2 \cdot 5=m_{B} g \times x . \\\therefore x=\frac{10}{15}\times 2.5 \mathrm{~m}=1.67 \mathrm{~m}\end{array}$

2007263_1393907_ans_ef3e2f5eae384233acefd1fac085ee34.JPG

A ballot dancer is rotating about his own vertical axis on smooth horizontal floor with a time period

$0.5 sec.$ The dancer folds himself close to his axis of rotation due to which his radius of gyration decreases by

$20\%,$ then his new time period is

0%
$0.1$ sec
0%
$0.25$ sec
0%
$0.32$ sec
0%
$0.4$ sec

Explanation

Let initial radius of gyration is

$R.$

Final radius of gyration

$=(R-20)$ of

$R=\frac{{80}}{{100}}R$

${t_i}$ is initial time period

${t_f}$ is final time period

According to the law of conservation of angular momentum initial angular momentum

$=$ final angular momentum

$or,\,m \times {R^2} \times {W_i} = m \times {\left( {\frac{{80}}{{100}}R} \right)^2} \times {W_f}$

$0r,\,{\left( {\frac{{100}}{{80}}} \right)^2} \times \frac{{2\pi }}{{{t_i}}} = \frac{{2\pi }}{{{t_f}}}$

$or,\,\,{t_f} = 0.5 \times {\left( {\frac{4}{5}} \right)^2} = 0.32\,\sec$

Hence,

option

$(C)$ is correct answer.

Two thin rods each of mass

$m$ and length 1 are joined to form L shape as shown. The moment of inertia of rods about an axis passing through free end (O) of a rod end and perpendicular to both the ends is

0%
$\dfrac{{5m{l^2}}}{3}$
0%
$\dfrac { m l ^ { 2 } } { 6 }$
0%
$\dfrac { 5 } { 7 } M R ^ { 2 }$
0%
$\dfrac { 7 } { 12 } M R ^ { 2 }$

Explanation

$\begin{array}{l} I=\dfrac { { m{ L^{ 2 } } } }{ 3 } +\dfrac { { m{ L^{ 2 } } } }{ { 12 } } M\left( { \dfrac { { 5{ l^{ 2 } } } }{ 4 } } \right) \\ =\dfrac { { m{ L^{ 2 } } } }{ 3 } +\dfrac { { m{ L^{ 2 } } } }{ { 12 } } +\dfrac { { 5m{ l^{ 2 } } } }{ 4 } \\ =\dfrac { { 20m{ l^{ 2 } } } }{ { 12 } } \\ =\dfrac { { 5m{ l^{ 2 } } } }{ 3 } \end{array}$

$\therefore$ Option

$A$ is correct .

The moment of inertia of a thin uniform rod about a transverse axis passing through centre is given to be I. The moment of inertia of the same rod about a transverse axis passing through a point midway between the centre and the end of the rod is:

0%
$4I$
0%
$9I/4$
0%
$7I/4$
0%
$5I/2$

Explanation

$\begin{array}{l}\text { Let } L \text { be Length of rod, } m \text { be mas of rod } \\\text { we know that } I=\dfrac{m L^{2}}{12} \\\text { By using parallel axis theorem } \\\qquad I^{\prime}=I+\dfrac{m L^{2}}{16}\end{array}$

$\begin{aligned} &=\dfrac{m L^{2}}{12}+\dfrac{m L^{2}}{16} \\ &=\dfrac{7 m L^{2}}{48}=\dfrac{7 \times m L^{2}}{4 \times 12}=\dfrac{7I}{4} \\\text { Hence option } C \text { is correct }\end{aligned}$

2001221_1389266_ans_90bfc424ae5e409fa196c19cdc9de9f5.PNG

Two particles are shown in the figure. at

$t = 0$ a constant force

$F = 6N$ starts acting on the

$3kg$ man. Find the velocity of the center of mass of these particles at

$t = 5s$ .

0%
$5$ $m/s$
0%
$4$ $m/s$
0%
$6$ $m/s$
0%
$3$ $m/s$

Explanation

Acceleration of

$3$ kg:

$a_2=\dfrac{6}{3}=2 m/s^2$
Velocity of

$3$ kg at

$t=5s$ :

$v_2=u_2+a_2t=0+2\times 5=10$ m/s

$v_{CM(t-5s)}=\dfrac{2\times 0+3\times 10}{2+3}=6$ m/s.
Method-

$2$ :

$a_{CM}=\dfrac{F}{m_1+m_2}=\dfrac{6}{2+3}=1.2m/s^2$

$v_{CM}=u_{CM}+a_{CM}t=0+1.2\times 5=6$ m/s

A circular platform is mounted on a vertical frictionless axle. Its radius is

$r = 2 { m }$ and its moment of inertia is

$I=200{ kg }-{ m }^{ { 2 } }.$ It is initially at rest. A

$70 { kg }$ man stands on the edge of the platform and begins to walk along the edge at speed

$v _ { 0 } = 1.0 { m } / { s }$ relative to the ground. The angular velocity of the platform is

0%
$1.2\quad { rad }/{ s }$
0%
$0.4\quad { rad }/{ s }$
0%
$2.0\quad { rad }/{ s }$
0%
$0.7\quad { rad }/{ s }$

Find I of the rod along the given axis. Suppose the mass of the rod is $m$ .

0%
$\dfrac{ mL^{2} }{ 12 }$
0%
$\dfrac{ mL^{2} }{ 16 }$
0%
$\dfrac{ mL^{2} }{ 20 }$
0%
$\dfrac{ mL^{2} }{ 24 }$

A thin wire of length L and uniform linear density

$\rho$ used to form a ring. The M.I. of this ring about tangential axis laying in its planes is:

0%
$\frac { \rho L ^ { 3 } } { 8 \pi ^ { 2 } }$
0%
$\frac { \rho L ^ { 3 } } { 16 \pi ^ { 2 } }$
0%
$\frac { 3 \rho L ^ { 3 } } { 12 \pi ^ { 2 } }$
0%
$\frac { 3 \rho L ^ { 3 } } { 8 \pi ^ { 2 } }$

Explanation

REF.Image

Mass of ring =

$\rho \times L$

Then,

MOI about

$OO'= (\rho L)(\frac{L}{2\pi })^{2}$

$=\frac{\rho L^{3}}{4\pi ^{2}}$

$2\pi r=L$ , density =

$\rho$

$r=\frac{L}{2\pi }$

Then,

MOI about

$NN'=\frac{\rho L^{3}}{4\pi ^{2}}\times \frac{1}{2}=\frac{\rho L^{3}}{8\pi ^{2}}$

(By perpendicular axis theorem)

Finally by parallel axis theorem

MOI about

$MM'=\frac{\rho L^{3}}{8\pi ^{2}}+(\rho L)(\frac{L}{2\pi })^{2}$

$=\frac{\rho L^{3}}{8\pi ^{2}}+\frac{\rho L^{3}}{4\pi ^{2}}=\frac{\rho L^{3}+2\rho L^{3}}{8\pi ^{2}}=\frac{3\rho L^{3}}{8\pi ^{2}}$

1217145_1402307_ans_1966aee807d14a14ab2cbbfdc55993a2.jpg

If the radius of solid sphere is

$\frac{1}{2}\sqrt{\frac{5}{7}}$ m, Thin its radius of gyration when the axis is along the tangent is -

0%
$\frac{5}{14}$ m
0%
$\frac{1}{2}$ m
0%
$\frac{5}{7}$ m
0%
1 m

Explanation

$\begin{array}{l} \text { Given, radius of solid sphere }=\frac{1}{2}\sqrt{\frac{5}{7}} \text { m. }(r) \text { .[ }\left.\operatorname{mass}=m\right] \\\text { we know, moment of inertia about diameter }=\frac{2}{5} m r^{2} \text { . }\end{array}$

$\begin{array}{l}\text { By parallel axis theorem: } \\\text { moment of inertia about tangent }(I)=\frac{2}{5} mr^{2}+m r^{2}=\frac{7}{5} mr^{2}\end{array}$

$\begin{array}{l}\text { Let, radius of gyration is } k \text { . }\\\text { Then, we can write, } I=m k^{2} \Rightarrow k=\sqrt{\frac{I}{m}} \\\text { So, } k=\sqrt{\frac{\frac{7}{5} m r^{2}}{m}}=\sqrt{\frac{7}{5}} r=\frac{1}{2} m\end{array}$

The momentum of body is numerically equal to its kinetic energy, The velocity of the body is :

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

A 5kg body collides with another stationary body. after the collision , the bodies moves in the same direction with one-third of the velocity of the first body . the mass of the second body will be-

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

A uniform rectangular plate is heated from

$0^o to 100^oC$ . Initial area of plate is

$10 cm^2$ the linear co-efficient of the material of the plate is

$18 \times 10^{-6}$ K what is the shift of the center of mass ?

0%
$zero$
0%
$1 cm$
0%
$2 cm$
0%
$3 cm$

A system of binary stars of masses

$m_{A}$ and

$m_{B}$ are moving in circular orbits of radii

$r_{A}$ and

$r_{B}$ respectively. If

$T_{A}$ and

$T_{B}$ are the time periods of masses

$m_{A}$ and

$m_{B}$ respectively, then

0%
$\frac{T_{A}}{T_{B}}=(\frac{r_{A}}{r_{B}})^{3/2}$
0%
$T_{A} > T_{B} (if r_{A} > r_{B})$
0%
$T_{A} > T_{B} (if m_{A} > m_{B})$
0%
$T_{A} = T_{B}$

if a force

$10\hat { i } +15\hat { j } +25\hat { K }$ acts on a system and gives an acceleration

$2\hat { i } +3\hat { j } -5\hat { K }$ to the centre of mass of the system, the mass of the system is :

0%
5 units
0%
$\sqrt { 38 } units$
0%
$5\sqrt { 38 } units$
0%
None

Explanation

Given,

${ F }_{ cm }=10\hat { i } +15\hat { j } +25\hat { k } \\ { a }_{ cm }=2\hat { i } +3\hat { j } -5\hat { k }$

We know that force is multiplication of mass and acceleration therefore,

${ F }_{ cm }=m\hat { a } \\ 10\hat { i } +15\hat { j } +25\hat { k } =m(2\hat { i } +3\hat { j } -5\hat { k } )\\ m=\frac { 10\hat { i } +15\hat { j } +25\hat { k } }{ 2\hat { i } +3\hat { j } -5\hat { k } } \\ m=\frac { \sqrt { { 10 }^{ 2 }+{ 15 }^{ 2 }+{ 25 }^{ 2 } } }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 } } } \\ m=\frac { \sqrt { 950 } }{ \sqrt { 38 } } \\ m=\frac { 5\sqrt { 38 } }{ \sqrt { 38 } } =5\quad units$

Hence, correct option is

$(A)$

Three man

$A, B \& C$ of mass

$40 \mathrm{kg}, 50 \mathrm{kg} \& 60 \mathrm{kg}$ are standing on a plank of mass

$90 \mathrm{kg}$ .which is kept on a smooth horizontal plane. If

$\mathrm{A} \& \mathrm{C}$ exchange their positions then mass

$\mathrm{B}$ will shift

0%
$1 / 3 \mathrm{m}$ towards left
0%
$1 / 3 m$ towards right
0%
will not move w.r.t. ground
0%
$7 / 3 \mathrm{m}$ towards left

From a disc of radius R, a concentric circular portion of radius r is cut out so as leave an annular disc of mass M. The moment of inertia of this annular disc about the axis perpendicular to its plane and passing through its center of gravity is

0%
$1/2 M(R^{2}+r^{2})$
0%
$1/2 M(R^{2}-r^{2})$
0%
$1/2 M(R^{4}+r^{4})$
0%
$1/2 M(R^{4}-r^{4})$

Explanation

$\begin{array}{l}\text { Given, mass of left annular disc }=M .\\\qquad \text { so, }\text { density }=\frac{M}{\pi\left(R^{2}-r^{2}\right)}\end{array}$

$\begin{array}{l}\text { Now, moment of inertia }\left(I_{1}\right)\text { of disc of radius } R \\\qquad \text { about central axis }=\frac{m_{1} R^{2}}{2} \quad\left(m_{1} \rightarrow \text { its mass }\right) \\\text { and, moment of inertia }\left(I_{2}\right) \text { of removed disc } \\\text { about the central axis }=\frac{m_{2} r^{2}}{2}\left(m_{2} \rightarrow\right.\text { its mass) }\end{array}$

$\begin{aligned}\therefore \text { moment of inertia of remaluing portion } &=I_{1}-I_{2} \\\text { (I) } &=\frac{1}{2}\left[M_{1} R^{2}-M_{2} r^{2}\right]\end{aligned}$

$\begin{array}{l}\text { Now, } M_{1}=\pi R^{2} \times \frac{M}{\pi \left(R^{2}-r^{2}\right)}=M\left(\frac{R^{2}}{R^{2}-r^{2}}\right) \\\text { and, } m_{2}=\pi r^{2} \times \frac{M}{\pi \left(R^{2}-r^{2}\right)}=M\left(\frac{r^{2}}{R^{2}-r^{2}}\right)\end{array}$

$\begin{aligned}\text { Thus, } I=\frac{1}{2}\left[\frac{M R^{4}}{\left(R^{2}-r^{2}\right)}-\frac{M r^{4}}{\left(R^{2}r^{2}\right)}\right] &=\frac{M}{2\left(R^{2}-r^{2}\right)}\left(R^{4}-r^{4}\right) \\&=\frac{1}{2} M\left(R^{2}+r^{2}\right)\end{aligned}$

2008740_1435495_ans_b8a7520cf491417ca21feb0bf119522b.JPG

A uniform cube of side a and mass m rests on a rough horizontal surface. A horizontal force F is applied normal to face at a point that is directly above the centre of the face at a height

$\cfrac{\alpha}{4}$ above the centre. The minimum value of F for which the cube begins to topple above an edge without sliding is:

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

The velocity of centre of mass of the system as shown in the figure is

0%
$\left( \frac { 2 - 2 \sqrt { 3 } } { 3 } \right) \hat { i } - \frac { 1 } { 3 } \hat { j }$
0%
$\left( \frac { 2 + 2 \sqrt { 3 } } { 3 } \right) \hat { i } - \frac { 3 } { 3 }\hat j$
0%
both
0%
None of these

Explanation

Let

$x$ component of velocity for mass

$1kg$ is

${ v }_{ 1x }$

Similarly

$x$ component of velocity for mass

$2kg$ is

${ v }_{ 2x }$

$y$ component of velocity for mass

$1kg$ is

${ v }_{ 1y }$

Similarly

$y$ component of velocity for mass

$2kg$ is

${ v }_{ 2y }$

Now

$x$ component of velocity of mass

$2kg$ is,

$2cos{ 30 }^{ 0 }=2\frac { \sqrt { 3 } }{ 2 } =\sqrt { 3 }$

$y$ component of velocity is

$2sin{ 30 }^{ 0 }=2\frac { 1 } { 2 } =1$

$M=\frac { { m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \\ M=\frac { 1\left( 2\hat { i } \right) +2\left( \sqrt { 3 } \hat { i } -\hat { j } \right) }{ 1+2 } \\ M=\frac { 2\hat { i } +2\sqrt { 3 } \hat { i } -2\hat { j } }{ 3 } \\ M=\left( \frac { 2+2\sqrt { 3 } }{ 3 } \right) \hat { i } -\frac { 2 }{ 3 } \hat { j }$

Correct option is

$(D)$

1929844_1455257_ans_78657d279c5d4bc991600cbe741fb4b3.PNG

What is the moment of force about the apex of triangle, if 3 forces of 40 N each acting along the sides of equilateral triangle of side 2 m taken in order

0%
51.96 Nm
0%
69.3 Nm
0%
30.6 Nm
0%
6.67 Nm

Explanation

$\begin{array}{l}\text { Force moment due to single Foru is } \\\qquad\tau=40\mathrm{~h}\mathrm{Nm} \\\text { on calculating the value of } h \text { - } \\\qquad\tan 30^{\circ}=\frac{h}{1} \quad n=\frac{1}{\sqrt{3}} \mathrm{~m} \\\Rightarrow\quad t=\frac{40}{\sqrt{3}} \mathrm{~N} m\end{array}$

$\begin{array}{l}\text { Force moment of three such forces will be } \\\qquad \begin{array}{c} t_{\text {Tota }}=3 x \frac{40}{\sqrt{3}}\mathrm{Nm}=40 \sqrt{3} \mathrm{Nm} \\ T_{\text {Total }}=69.3\mathrm{Nm}\end{array}\end{array}$

2007080_1475103_ans_c31e0162bcb0445191ec2191b4e89ff8.PNG

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 9 - MCQExams.com

At initial position both A and B are at rest, hence, momentum is zero. It is given that no external force is acting on them. So, momentum remains zero. Since mass cannot be zero, therefore velocity is zero.

0:0:3

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 9 - MCQExams.com

At initial position both A and B are at rest, hence, momentum is zero. It is given that no external force is acting on them. So, momentum remains zero. Since mass cannot be zero, therefore velocity is zero.

0:0:3

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.