A wire of length L and radius r is rigidly fixed at one end. On stretching the other end of the wire with a force F, the increase in its length is l. If another wire of same material but of length 2L and radius 2r is stretched with a force of 2F, the increase in its length will be

The correct option is l.

Step-by-step Explanation:

Young's modulus ($Y$) of a material is defined as the ratio of tensile stress to tensile strain:

$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{l/L}=\frac{FL}{Al}$

where:
- $F$ is the stretching force applied,
- $L$ is the original length of the wire,
- $A$ is the cross-sectional area of the wire ($A=\pi r^2$, with $r$ being the radius),
- $l$ is the increase in length.

Substituting the area $A$ into the equation, we get:

$Y=\frac{FL}{\pi r^{2}l}$

Rearranging this equation to solve for the increase in length ($l$):

$l=\frac{FL}{\pi r^{2}Y}$

For the second wire, we have the following parameters:
- Same material, so Young's modulus remains $Y$
- New length, $L_2=2L$
- New radius, $r_2=2r$
- New force, $F_2=2F$

Let $l_2$ be the increase in length for the second wire. Substituting these values into our equation:

$l_2=\frac{F_2{L}_2}{\pi r_2^{2}Y}$

Substitute the relations into the equation:

$l_2=\frac{\left(2F\right)\left(2L\right)}{\pi {\left(2r\right)}^{2}Y}$

$l_2=\frac{4FL}{\pi \left(4r^2\right)Y}$

$l_2=\frac{4FL}{4\pi r^{2}Y}$

Simplifying the fraction by canceling the common factor of 4 in the numerator and denominator:

$l_2=\frac{FL}{\pi r^{2}Y}=l$

Thus, the increase in length of the second wire is also l.