A wire elongates by $$\ell\,mm$$ when a load $$W$$ is hanged from it. If the wire goes over a pulley and two weights $$W$$ each are hung at the two ends, the elongation of the wire will be (in $$mm$$)

Case 1: Wire fixed at one support and loaded with $$W$$

Let the original length of the wire be $$L$$, its cross-sectional area be $$A$$, and the Young's modulus of its material be $$Y$$

In this setup, the tension $$T$$ in the wire is uniform and equal to the applied load $$W$$.

According to Hooke's Law, the initial elongation $$\ell$$ is given by:

$$\ell = \frac{T L}{A Y} = \frac{W L}{A Y}$$

Case 2: Wire over a smooth pulley with weights $$W$$ at both ends

$$T' = W$$.

Even though the wire is bent over the pulley, the total length subjected to this tension is still $$L$$. Since the tension $$T'$$ is exactly the same as in Case 1 ($$T' = W$$), the total elongation $$\Delta L$$ depends entirely on this same tension acting over the entire length $$L$$:

$$\Delta L = \frac{T' L}{A Y} = \frac{W L}{A Y}$$

Comparing this with our equation from Case 1, we clearly see that:

$$\Delta L = \ell$$ (option B)