Let initial length and diameter be l₁ and d₁, whereas final length and diameter be l₂ and d₂.

Given, l₂ = 2l₁, d₂ = 2d₁, $$\Delta$$l₁ = 0.04 m

By using formula of Young's modulus of elasticity,

$$Y = {{F\,.\,l} \over {A\Delta l}}$$

$$\therefore$$ $${Y_1} = {Y_2}$$

$$ \Rightarrow {{F{l_1}} \over {{A_1} \times \Delta {l_1}}} = {{F{l_2}} \over {{A_2} \times \Delta {l_2}}}$$

$$ \Rightarrow {{F{l_1}} \over {\pi {{({d_1}/2)}^2} \times 0.04}} = {{F2{l_1}} \over {\pi {{(2{d_1}/2)}^2} \times \Delta {l_2}}}$$

$$ \Rightarrow {1 \over {1/4 \times 0.04}} = {2 \over {\Delta {l_2}}}$$

$$ \Rightarrow \Delta {l_2} = 0.02$$ m = 2 cm