The acceleration of end $B$ when it has been displaced by distance $x$ , is:

\frac{x}{L} g

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\frac{2 x}{L} g

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\frac{x}{2} g

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g

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Solution

The correct option is A $\frac{x}{L} g$
Let $m$ be the mass per unit length
Considering the free body diagram on both sides we get
$T = m (L - x) g = m (L - x) a$
and
$m (L + x) g - T = m (L + x) a$
Adding the two we get
$m (L + x) g - m (L - x) g = m (L + x) a + m (L - x) a$
Solving we get
$a = \frac{g x}{L}$

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