1. Let the acceleration of the hanging mass be $a$ downward and the tension in the string be $T$.

2. Apply Newton's second law to the mass $m$: $mg - T = ma ag{1}$

3. For the disc, use rotational dynamics.

   The torque due to tension is: $\tau = TR$

   For a uniform disc, moment of inertia about its axis is: $I = \frac{1}{2}MR^2$

   Angular acceleration is related to linear acceleration by: $a = \alpha R \quad \Rightarrow \quad \alpha = \frac{a}{R}$

   So, $TR = I\alpha = \frac{1}{2}MR^2 \cdot \frac{a}{R}$

   Simplifying, $T = \frac{1}{2}Ma ag{2}$

4. Substitute equation (2) into equation (1): $mg - \frac{1}{2}Ma = ma$

   $mg = a\left(m + \frac{M}{2}\right)$

   $a = \frac{mg}{m + \frac{M}{2}}$

5. Simplify: $a = \frac{2mg}{2m + M}$

6. Match with the options: This is Option A.

Comparison with stored answer:

• Derived answer: A
• Stored correct answer: A
• Hence, they agree.