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(D) Since no external torque acts on the system about the axis of rotation,the angular momentum of the system is conserved.Let $I_0$ be the initial mo

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$\frac{m_0}{2\mu}$

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(D) Since no external torque acts on the system about the axis of rotation,the angular momentum of the system is conserved.Let $I_0$ be the initial moment of inertia of the disc,$I_0 = \frac{1}{2} m_0 r^2$.Let $\omega_0$ be the initial angular velocity.The initial angular momentum is $L_i = I_0 \omega_0$.After time $t$,the mass of water absorbed is $m_w = \mu t$. This water is at the rim of the disc,so its moment of inertia is $I_w = m_w r^2 = (\mu t) r^2$.The new moment of inertia of the system is $I_f = I_0 + I_w = \frac{1}{2} m_0 r^2 + \mu t r^2$.The final angular velocity is $\omega_f = \frac{\omega_0}{2}$.By conservation of angular momentum,$L_i = L_f \Rightarrow I_0 \omega_0 = I_f \omega_f$.Substituting the values: $(\frac{1}{2} m_0 r^2) \omega_0 = (\frac{1}{2} m_0 r^2 + \mu t r^2) \frac{\omega_0}{2}$.Dividing both sides by $\frac{\omega_0 r^2}{2}$: $m_0 = \frac{1}{2} m_0 + \mu t$.Solving for $t$: $\mu t = \frac{1}{2} m_0 \Rightarrow t = \frac{m_0}{2\mu}$.

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