One end of a thin uniform rod of length $L$ and mass $M_{1}$ is rivetted to the centre of a uniform circular disc of radius $r$ and mass $M_{2}$ , so that both are coplanar. The distance of centre of mass of the combination from the centre of the disc is

$\frac{L (M_{1} + M_{2})}{2 M_{1}}$

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$\frac{L M_{1}}{2 (M_{1} + M_{2})}$

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$\frac{2 (M_{1} + M_{2})}{L M_{1}}$

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$\frac{2 L M_{1}}{(M_{1} + M_{2})}$

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Solution

The correct option is B
$\frac{L M_{1}}{2 (M_{1} + M_{2})}$

Taking origin at the centre of the disk and x-axis along the length of the rod, the mass of the rod can be considered to be concentrated at its centre $(x = \frac{L}{2})$ .
Now, $R_{C M} = \frac{M_{1} \times \frac{L}{2} + M_{2} \times 0}{M_{1} + M_{2}} = \frac{L M_{1}}{2 (M_{1} + M_{2})}$
Hence, the correct choice is (b).

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One end of a thin uniform rod of length L and mass M1 is rivetted to the centre of a uniform circular disc of radius r and mass M2, so that both are coplanar. The distance of centre of mass of the combination from the centre of the disc is

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