A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be $\_\_\_\_$ cm 
 

Let $l_1$ and $d_1$ represent the starting length and diameter, and $l_2$ and $d_2$ represent the end length and diameter.

$l_2=2l_1, {\mathrm{d}}_2=2{\mathrm{d}}_1, ∆l_1=0.04\mathrm{m}$

Using the Young's modulus of elasticity formula,

$\mathrm{Y}=\frac{\mathrm{F}l}{\mathrm{A}∆l}$

${\mathrm{Y}}_1={\mathrm{Y}}_2$

$\frac{\mathrm{F}{l}_1}{{\mathrm{A}}_1∆l_1}=\frac{\mathrm{F}{l}_2}{{\mathrm{A}}_2∆l_2}$

$\frac{\mathrm{F}{l}_1}{\mathrm{\pi }{\left({\displaystyle \frac{{\mathrm{d}}_1}{2}}\right)}^2\times 0.04}=\frac{\mathrm{F}2l_1}{\mathrm{\pi }{\left({\displaystyle \frac{2{\mathrm{d}}_1}{2}}\right)}^2\times ∆{\mathrm{l}}_2}\Rightarrow \frac{1}{{\displaystyle \frac{1}{4}\times 0.04}}=\frac{2}{∆l_2}\Rightarrow ∆l_2=0.02\mathrm{m}=2\mathrm{cm}$

Hence the correct answer is 2.