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Two spherical 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 $m_2$ makes with the vertical is approximately
$\frac{m_1 R}{\left(m_1+m_2\right) L}$
$\frac{2 m_1 R}{\left(m_1+m_2\right) L}$
$\frac{2 m_2 h}{\left(m_1+m_2\right) L}$
$\frac{m_2 R}{\left(m_1+m_2\right) L}$
Solution
(b)
Given arrangement of spheres is as shown below.
Free body diagram of spheres is
As, there is no rotation about point of contact $P$.
Equating moments of weights about centre line, we get
$\quad m_1 g \times r_1=m_2 g \times r_2$
$\text { where, } r_1+r_2=2 R$
$\Rightarrow \quad \frac{m_2 r_2+r_2=2 R}{m_1}$
$\Rightarrow \quad r_2\left(\frac{m_2+m_1}{m_1}\right)=2 R \Rightarrow r_2=\frac{2 m_1 R}{m_1+m_2}$
Now, if angle made by string of $m_2$ with vertical line is $\theta$, then
$\sin \theta=\frac{r_2}{L} \Rightarrow \sin \theta=\left(\frac{2 m_1}{m_1+m_2}\right)\left(\frac{R}{L}\right)$
As $R \ll L$, angle $\theta$ is small, therefore $\sin \theta \approx \theta$.
$\therefore \quad \theta=\frac{2 m_1 R}{\left(m_1+m_2\right) L}$
