1. Given system

• A block of mass $m$ hangs from a light string.
• The string is wound around a uniform hollow cylinder of mass $m$ and radius $R$.
• The string does not slip, so linear and angular accelerations are related.

We need the downward acceleration of the hanging mass.

2. For the falling mass

Let the mass accelerate downward with acceleration $a$.

Forces on the mass:

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

Applying Newton's second law:

$mg - T = ma ag{1}$

3. For the hollow cylinder

The string exerts tension $T$ at the rim, producing torque:

$\tau = TR$

For a uniform hollow cylinder (thin cylindrical shell), the moment of inertia about its axis is:

$I = mR^2$

Using rotational dynamics:

$TR = I\alpha$

Since there is no slipping,

$a = \alpha R \quad \Rightarrow \quad \alpha = \frac{a}{R}$

So,

$TR = mR^2\left(\frac{a}{R}\right)$

$TR = mRa$

$T = ma ag{2}$

4. Substitute into the linear equation

From (1):

$mg - T = ma$

Using $T = ma$:

$mg - ma = ma$

$mg = 2ma$

$a = \frac{g}{2}$

5. Check options

• A: $\dfrac{2g}{3}$ ❌
• B: $\dfrac{g}{2}$ ✅
• C: $\dfrac{5g}{6}$ ❌
• D: $g$ ❌

Thus the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, the derived answer agrees with the stored answer.