A wire elongates by 1.0 mm when a load W is hanged from it. If this wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be

The correct answer is 1.0 mm.

To understand why the elongation remains the same, let us analyze the tension in the wire in both scenarios.

Scenario 1: Hanging a load W from a fixed support
When the wire is suspended from a rigid ceiling and a weight $W$ is hung from the free end, the wire is in equilibrium. The ceiling exerts an upward reacting force of magnitude $W$ on the top end of the wire. Consequently, the tension $T$ throughout the wire is equal to the load applied:
$T=W$

The formula for the elongation ($\Delta L$) of a wire of length $L$, cross-sectional area $A$, and Young's modulus $Y$ is given by Hooke's law:

$\Delta L=\frac{TL}{AY}=\frac{WL}{AY}=1.0\text{mm}$

Scenario 2: Passing the wire over a pulley with weights W on both ends
When the wire is placed over a frictionless, massless pulley and a weight $W$ is attached to each end, the system is balanced and remains in static equilibrium.
For either weight to remain stationary, the upward tension force exerted by the wire on each weight must balance the downward gravitational pull of the weight. Therefore, the tension $T\text{'}$ in the wire is:
$T\text{'}=W$

Just like in the first scenario, the wire is subjected to a tension force of $W$ at both ends (with the pulley acting as the support that holds the wire). Since the tension in the wire is unchanged ($T\text{'}=T=W$), the elongation will also remain exactly the same.

Thus, the elongation of the wire in the second case is 1.0 mm.