Two wires of same diameter of the same material having the length l and 2l. If the force F is applied on each, the ratio of the work done in the two wires will be

The correct option is 1:2.

To find the ratio of the work done in stretching the two wires, let us recall the relationship between force, elongation, and work done for an elastic wire.

The work done $W$ in stretching a wire under an applied force $F$ by an extension $\Delta l$ is given by:
$W=\frac{1}{2}\times F\times \Delta l$

According to Hooke's Law and the definition of Young's modulus ($Y$):
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta l/l}$
where $A$ is the cross-sectional area of the wire, and $l$ is its original length.

Rearranging this formula to express the extension $\Delta l$ gives:
$\Delta l=\frac{F\times l}{A\times Y}$

Substituting this expression for $\Delta l$ into the work done formula, we get:
$W=\frac{1}{2}\times F\times \left(\frac{F\times l}{A\times Y}\right)=\frac{F^2\times l}{2\times A\times Y}$

We are given that:
1. Both wires have the same diameter, which means they have the same cross-sectional area $A$.
2. They are made of the same material, which means they have the same Young's modulus $Y$.
3. The same force $F$ is applied to both wires.

Therefore, the work done $W$ is directly proportional to the length $l$ of the wire:
$W\propto l$

Let $W_1$ and $W_2$ be the work done in the first and second wires, with lengths $l_1=l$ and $l_2=2l$ respectively. Taking their ratio gives:
$\frac{W_1}{W_2}=\frac{l_1}{l_2}=\frac{l}{2l}=\frac{1}{2}$

Thus, the ratio of the work done in the two wires is 1:2.