Two masses are attached to a rod end to end. If torque is applied...

Suppose the masses $m_1$ and $m_2$ are attached to the ends of a massless rod. $C_1$ is the axis of rotation which is at a distance $R$ from each of the masses. $\therefore$ the moment of inertia of the system, $\mathrm{I}=$m_1 R²+m_2 R²=\left(m_1+m_2\right) R² Now if the distance between the masses is doubled, then the moment of inertia \Gamma=m_1(2 R)^2+m_2(2 R)^2=\left(m_1+m_2\right) 4 R² Since the same torque is applied in both the cases,$\begin{aligned}& \tau=I \alpha=I^{\prime} \alpha^{\prime}\\ & \Rightarrow \alpha^{\prime}=\frac{I \alpha}{I^{\prime}}=\frac{\left(m_1+m_2\right) R^2 \alpha}{\left(m_1+m_2\right) 4 R^2}=\frac{\alpha}{4} \end{aligned}$$