A wire of length 2L and radius r is stretched between A and B with out the application of any tension. If Y is the young’s modulus of the wire and it is stretched like, ACB, then the tension in the wire will be

Given,

The wire's initial length = 2L

The wire's final length is = $2\sqrt{{\mathrm{L}}^2+{\mathrm{d}}^2}$

Change in length of the wire, 

$∆\mathrm{L}=2\sqrt{{\mathrm{L}}^2+{\mathrm{d}}^2}-2\mathrm{L}$

By applying Hooke's law,

$\frac{\mathrm{T}}{{\mathrm{\pi r}}^2}=\mathrm{Y}\frac{∆\mathrm{L}}{2\mathrm{L}}$

$\mathrm{T}=\frac{{\mathrm{\pi r}}^2\mathrm{Y}\left(2\sqrt{{\mathrm{L}}^2+{\mathrm{d}}^2}-2\mathrm{L}\right)}{2\mathrm{L}}$

$\mathrm{T}=\frac{{\mathrm{\pi r}}^2\mathrm{Y}\left(2\mathrm{L}\left({\left(1+{\displaystyle \frac{{\mathrm{d}}^2}{{\mathrm{L}}^2}}\right)}^{{\displaystyle \frac{1}{2}}}-1\right)\right)}{2\mathrm{L}}$

$\mathrm{T}={\mathrm{\pi r}}^2\mathrm{Y}\left({\left(1+\frac{{\mathrm{d}}^2}{{\mathrm{L}}^2}\right)}^{\frac{1}{2}}-1\right)$

$\mathrm{T}={\mathrm{\pi r}}^2\mathrm{Y}\left(\left(1+\frac{1}{2}\frac{{\mathrm{d}}^2}{{\mathrm{L}}^2}\right)-1\right)=\frac{{\mathrm{\pi r}}^2{\mathrm{Yd}}^2}{2{\mathrm{L}}^2}$

Hence the correct answer is $\frac{{\mathrm{\pi r}}^2{\mathrm{Yd}}^2}{2{\mathrm{L}}^2}.$