1. Set up the system

Let the linear mass density of the uniform chain be $\lambda = \frac{m}{L}.$

At some instant, lengths hanging on the two sides are:

• left side: $l$
• right side: $L-l$

Assume $l > L-l$, so the side with length $l$ moves downward.

2. Find the net driving force

Since the pulley is massless and frictionless, the only unbalanced external force causing motion is the difference in weights of the two hanging parts.

Weight of left part: $W_1 = \lambda l g$

Weight of right part: $W_2 = \lambda (L-l) g$

So net force on the chain is $F = W_1 - W_2 = \lambda g\big(l-(L-l)\big)$ $F = \lambda g(2l-L).$

3. Use Newton's second law for the whole chain

Total mass of the chain is $m = \lambda L$. Hence acceleration is $a = \frac{F}{m} = \frac{\lambda g(2l-L)}{\lambda L}.$

Therefore, $a = g\left(\frac{2l-L}{L}\right) = g\left(\frac{2l}{L}-1\right).$

4. Given condition

At the instant when $l = \frac{L}{x},$ the acceleration is $a = \frac{g}{2}.$

Substitute into the formula: $g\left(\frac{2l}{L}-1\right) = \frac{g}{2}.$

Cancel $g$: $\frac{2l}{L}-1 = \frac{1}{2}.$

So, $\frac{2l}{L} = \frac{3}{2}$ $\frac{l}{L} = \frac{3}{4}.$

But $l = \frac{L}{x}$, hence $\frac{1}{x} = \frac{3}{4}$ $x = \frac{4}{3}.$

5. Check with options

The computed value is $x = \frac{4}{3} \approx 1.33,$ which is not present in the options.

If the intended meaning was the magnitude of acceleration, $\left|g\left(\frac{2l}{L}-1\right)\right| = \frac{g}{2},$ then $\left|\frac{2l}{L}-1\right| = \frac{1}{2}$ which gives $\frac{l}{L} = \frac{3}{4} \quad \text{or} \quad \frac{1}{4},$ so $x = \frac{4}{3} \quad \text{or} \quad 4.$

Among the given options, only $4$ appears. So the stored answer likely assumes the acceleration magnitude and the case $l < L/2$.

6. Conclusion

From the direct equation for signed acceleration, the answer is $x = \frac{4}{3}.$

From the given options, the only matching choice is $4$, if acceleration is interpreted as magnitude.