A body $A$ of mass $M$ while falling vertically downwards under gravity breaks into two parts; a body B of mass $1/3\,M$ and a body C of mass $2/3\,M$. The centre of mass of bodies B and C taken together shifts compared to that of body A towards
(1) depends on height of breaking
(2) does not shift
(3) body C
(4) body B

Consider a two particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance $d$, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?
(1) $d$
(2) $\frac{m_2}{m_1}$ d
(3) $\frac{m_1}{m_1 + m_2} d$
(4) $\frac{m_1}{m_2} d$

A thin rod of length ' $L$ ' is lying along the $x$-axis with its ends at $x = 0$ and $x = L$. Its linear density (mass/length) varies with $x$ as $k\left( \frac { x } { L } \right) ^ { n }$, where $n$ can be zero or any positive number. If the position $x _ { \mathrm { CM } }$ of the centre of mass of the rod is plotted against ' $n$ ', which of the following graphs best approximates the dependence of $x _ { \mathrm { CM } }$ on $n$?
(1), (2), (3), (4) [see graphs in original]

A thin bar of length L has a mass per unit length $\lambda$, that increases linearly with distance from one end. If its total mass is $M$ and its mass per unit length at the lighter end is $\lambda_\mathrm{O}$, then the distance of the centre of mass from the lighter end is:
(1) $\frac{\mathrm{L}}{2} - \frac{\lambda_0 \mathrm{~L}^2}{4\mathrm{M}}$
(2) $\frac{\mathrm{L}}{3} + \frac{\lambda_0 \mathrm{~L}^2}{8\mathrm{M}}$
(3) $\frac{\mathrm{L}}{3} + \frac{\lambda_0 \mathrm{~L}^2}{4\mathrm{M}}$
(4) $\frac{2\mathrm{~L}}{3} - \frac{\lambda_0 \mathrm{~L}^2}{6\mathrm{M}}$

A uniform thin rod AB of length $L$ has linear mass density $\mu ( x ) = a + \frac { b x } { L }$, where $x$ is measured from A. If the CM of the rod lies at a distance of $\left( \frac { 7 } { 12 } L \right)$ from A, then $a$ and $b$ are related as:
(1) $2 a = b$
(2) $a = 2 b$
(3) $a = b$
(4) $3 a = 2 b$

The position vector of the center of mass $\overrightarrow { \mathrm { r } } _ { \mathrm { cm } }$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:
(1) $\overrightarrow { \mathrm { r } } _ { \mathrm { cm } } = \frac { 13 } { 8 } \mathrm {~L} \hat { \mathrm { x } } + \frac { 5 } { 8 } \mathrm {~L} \hat { \mathrm { y } }$
(2) $\vec { r } _ { \mathrm { cm } } = \frac { 5 } { 8 } \mathrm {~L} \hat { \mathrm { x } } + \frac { 13 } { 8 } \mathrm {~L} \hat { \mathrm { y } }$
(3) $\overrightarrow { \mathrm { r } } _ { \mathrm { cm } } = \frac { 3 } { 8 } \mathrm {~L} \hat { \mathrm { x } } + \frac { 11 } { 8 } \mathrm {~L} \hat { \mathrm { y } }$
(4) $\overrightarrow { \mathrm { r } } _ { \mathrm { cm } } = \frac { 11 } { 8 } \mathrm {~L} \hat { \mathrm { x } } + \frac { 3 } { 8 } \mathrm {~L} \hat { \mathrm { y } }$

Three particles of masses $50 \mathrm {~g} , 100 \mathrm {~g}$ and 150 g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure). The $( \mathrm { x } , \mathrm { y } )$ coordinates of the centre of mass will be:
(1) $\left( \frac { 7 } { 12 } \mathrm {~m} , \frac { \sqrt { 3 } } { 4 } \mathrm {~m} \right)$
(2) $\left( \frac { 7 } { 12 } \mathrm {~m} , \frac { \sqrt { 3 } } { 8 } \mathrm {~m} \right)$
(3) $\left( \frac { \sqrt { } 3 } { 4 } \mathrm {~m} , \frac { 5 } { 12 } \mathrm {~m} \right)$
(4) $\left( \frac { \sqrt { 3 } } { 8 } \mathrm {~m} , \frac { 7 } { 12 } \mathrm {~m} \right)$

An $L$-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If $A B = B C$, and the angle made by $A B$ with downward vertical is $\theta$, then:
(1) $\tan \theta = \frac { 2 } { \sqrt { 3 } }$
(2) $\tan \theta = \frac { 1 } { 3 }$
(3) $\tan \theta = \frac { 1 } { 2 }$
(4) $\tan \theta = \frac { 1 } { 2 \sqrt { 3 } }$

The position of the centre of mass of a uniform semi-circular wire of radius $R$ placed in $x - y$ plane with its centre at the origin and the line joining its ends as $x$-axis is given by, $\left( 0 , \frac { x R } { \pi } \right)$. Then, the value of $| x |$ is $\_\_\_\_$ .

Three identical spheres each of mass $M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $\sqrt{x}$ m. The value of $x$ is

The identical spheres each of mass $2M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is $\frac { 4 \sqrt { 2 } } { x }$, where the value of $x$ is $\_\_\_\_$.

Q22. In a system two particles of masses $m _ { 1 } = 3 \mathrm {~kg}$ and $m _ { 2 } = 2 \mathrm {~kg}$ are placed at certain distance from each other. The particle of mass $m _ { 1 }$ is moved towards the center of mass of the system through a distance 2 cm . In order to keep the center of mass of the system at the original position, the particle of mass $m _ { 2 }$ should move towards the center of mass by the distance $\_\_\_\_$ cm.