1. Given

• Hanging mass $= m$
• Pulley mass $= m$
• Pulley radius $= R$
• Pulley is a uniform disc, so its moment of inertia is $I=\frac{1}{2}mR^2$
• String does not slip, so $a=\alpha R$

We need the acceleration of the hanging mass.

2. Equation for the hanging mass

For the block of mass $m$, forces are:

• Weight downward: $mg$
• Tension upward: $T$

Taking downward direction as positive, $mg-T=ma \quad ...(1)$

3. Rotational equation for the pulley

The tension in the string produces torque on the pulley: $\tau = TR$

Using rotational dynamics, $TR=I\alpha$

Since $a=\alpha R$, we have $\alpha=\frac{a}{R}$

So, $TR=I\frac{a}{R}$

Substitute $I=\frac{1}{2}mR^2$: $TR=\frac{1}{2}mR^2\cdot \frac{a}{R}$ $TR=\frac{1}{2}mRa$

Cancelling $R$, $T=\frac{1}{2}ma \quad ...(2)$

4. Substitute into linear equation

From (1): $mg-T=ma$

Substitute $T=\frac{1}{2}ma$: $mg-\frac{1}{2}ma=ma$

$mg=ma+\frac{1}{2}ma$ $mg=\frac{3}{2}ma$

Therefore, $a=\frac{2}{3}g$

5. Check options

• A: $g$ ❌
• B: $\frac{2}{3}g$ ✅
• C: $\frac{g}{3}$ ❌
• D: $\frac{3}{2}g$ ❌

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.