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Step-by-Step Solution
Step 1: Understand the Problem
When a load of weight $W$ is hung from a wire, it elongates by $l$ mm. The question is about what happens to the elongation if the wire is passed over a pulley and the same load $W$ is attached to both ends.
Step 2: Analyze the Tension in the Wire
In the first setup (a single weight $W$ at one end of the wire), the tension in the wire is $W$. In the second setup (two weights of $W$ each at both ends of the wire), the tension in the wire still remains $W$ because each side supports one weight $W$ and the system is in equilibrium. The tension does not add up to become $2W$ because each segment of the wire experiences the same tension, equal to the weight hanging from it.
Step 3: Relate Tension to Elongation
Elongation of a wire, $ \Delta l $, for small extensions in the elastic range, is given by:
$ \Delta l = \dfrac{FL}{AE} $
where:
$F$ is the tension in the wire
$L$ is the original length of the wire
$A$ is the cross-sectional area of the wire
$E$ is Young's modulus of the material
For the same wire and same tension $F = W$, the elongation $ \Delta l $ is the same in both setups.
Step 4: Conclude the Elongation
Because the tension in the wire remains $W$ in both cases, the elongation remains $l$ mm. Therefore, when two weights each of magnitude $W$ are hung on the two ends of the wire over a pulley, the elongation of the wire is still $l$ mm, and not doubled or halved.
The correct answer is therefore $l$.
