When a tension 'F' is applied, the elongation produced in uniform wire of length l, radius r is e, when tension 2F is applied, the elonga- tion produced in another uniform wire of length 2l and radius 2r made of same material is 

Let's assume that x represents the elongation that the application of 2F to a wire of radius 2r and length 2l causes.

Young's modulus is equal since the wire is made of the same material.

$\mathrm{Y}=\mathrm{Y} \text{'}$

$\frac{{\displaystyle \frac{\mathrm{F}}{\mathrm{A}}}}{{\displaystyle \frac{\mathrm{e}}{l}}}=\frac{{\displaystyle \frac{2\mathrm{F}}{\mathrm{A}}}}{{\displaystyle \frac{\mathrm{x}}{2l}}}$

$\frac{1}{\mathrm{Ae}}=\frac{4}{\mathrm{A}\text{'}\mathrm{x}}$

$\frac{1}{{\mathrm{\pi r}}^2\mathrm{e}}=\frac{4}{4{\mathrm{\pi r}}^2\mathrm{x}}$

$\frac{1}{\mathrm{e}}=\frac{1}{\mathrm{x}}$

$\mathrm{x}=1.0\mathrm{e}$

Hence the correct answer is 1.0e.