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) $\frac{2\mathrm{~L}}{3} - \frac{\lambda_0 \mathrm{~L}^2}{6\mathrm{M}}$

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) $\left( \frac { \sqrt { 3 } } { 8 } \mathrm {~m} , \frac { 7 } { 12 } \mathrm {~m} \right)$

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