Explanation

(d) Total moment of inertia will be equal to the sum of moment of inertia due to individual masses.
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$I = \sum_{i = 1}^{n}I_{i}$
where, $I_{1} = m_{1}r_{1}^{2},\ I_{2} = m_{2}r_{2}^{2}$.
Given, $r = r_{1} + r_{2}$
$m_{1}r_{1} = m_{2}r_{2}$
$\therefore$ $m_{1}r_{1} = m_{2}(r - r_{1})$
$\Rightarrow \ $ $m_{1}r_{1} + m_{2}r_{1} = m_{2}r$
$\Rightarrow$ $r_{1}\left( m_{1} + m_{2} \right) = m_{2}r$
$\Rightarrow$ $r_{1} = \frac{m_{2}r}{{(m}_{1} + m_{2})}$
Also, $r_{2} = r - r_{1}$
$r_{2} = r - \frac{m_{2}r}{\left( m_{1} + m_{2} \right)^{2}} = \frac{m_{2}r}{m_{1} + m_{2}}$
$\therefore$ $I = I_{1} + I_{2} = m_{1}r_{1}^{2} + m_{2}r_{2}^{2}$
$I = m_{1}\frac{m_{2}^{2}r^{2}}{\left( m_{1} + m_{2} \right)^{2}} + m_{2}\frac{m_{1}^{2}r^{2}}{\left( m_{1} + m_{2} \right)^{2}}$
$I = \frac{m_{1}m_{2}r^{2}}{(m_{1} + m_{2})} = \frac{m_{1}m_{2}}{(m_{1} + m_{2})}.r^{2}$