The ratio of diameters of two wires of same material is n : 1. The length of wires are 4 m each. On applying the same load, the increase in length of thin wire will be

The correct option is n² times.

To find the increase in length of the thin wire, we can use the definition of Young's modulus of elasticity ($Y$), which is given by the formula:

$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}$

Rearranging the formula to solve for the extension or increase in length ($\Delta L$):

$\Delta L=\frac{F·L}{A·Y}$

Here,
- $F$ is the applied load (force)
- $L$ is the original length of the wire
- $A$ is the cross-sectional area of the wire
- $Y$ is the Young's modulus of the material of the wire

The cross-sectional area $A$ of a wire with diameter $d$ is:

$A=\frac{\pi d^2}{4}$

Substituting this area into our extension equation gives:

$\Delta L=\frac{4·F·L}{\pi ·d^2·Y}$

According to the problem description:
- Both wires are of the same material, which means their Young's modulus $Y$ is the same.
- Both wires have the same initial length ($L=4\text{m}$).
- The same load ($F$) is applied to both wires.

Since $F$, $L$, and $Y$ are constants for both wires, the extension $\Delta L$ is inversely proportional to the square of the diameter:

$\Delta L\propto \frac{1}{d^2}$

Let $d_1$ and $\Delta L_1$ be the diameter and extension of the thick wire, and $d_2$ and $\Delta L_2$ be the diameter and extension of the thin wire. The ratio of the diameters is given as $d_1:d_2=n:1$, so:

$\frac{d_1}{d_2}=n$

Using the inverse square relation, we can compare the extension of the thin wire to the thick wire:

$\frac{\Delta L_2}{\Delta L_1}={\left(\frac{d_1}{d_2}\right)}^2$

Substituting the ratio $\frac{d_1}{d_2}=n$ into the equation:

$\frac{\Delta L_2}{\Delta L_1}=n^2$

Rearranging the expression, we find:

$\Delta L_2=n^2·\Delta L_1$

Thus, the increase in length of the thin wire will be n² times that of the thicker wire.