Two identical objects each of radii $R$ and masses $m_{1}$ and $m_{2}$ are suspended using two strings of equal length $L$ as shown in the figure $(R< < L)$. The angle $\theta$ which mass $\mathrm{m}_{2}$ makes with the vertical is approximately

• A

  $\frac{m_{1} R}{\left(m_{1}+m_{2}\right) L}$
• Correct answer
  B

  $\frac{2 m_{1} R}{\left(m_{1}+m_{2}\right) L}$
• C

  $[\mathrm] \frac{2 m_{2} R}{\left(m_{1}+m_{2}\right) L}$
• D

  $\frac{m_{2} R}{\left(m_{1}+m_{2}\right) L}$

Solution

Distance of center of mass from $\mathrm{m}_{2}$ ? $$ \begin{array}{c} \mathrm{d}=\frac{m_{1}}{\left(m_{1}+m_{2}\right)} \times 2 R \\ \theta=\frac{\text { distance }(\mathrm{d})}{\mathrm{L}} \\ \theta=\frac{m_{1}}{\left(m_{1}+m_{2}\right)} \times \frac{2 R}{\mathrm{~L}} \end{array} $$

Hence, the correct answer is option: 'B'.