A uniform steel rod of length 1m and area of cross section 20 cm² is hanging from a fixed support. Find the increase in the length of the rod. 

$\left({\mathrm{Y}}_{\mathrm{steel}}=2.0\times {10}^{11}{\mathrm{Nm}}^{-2}, {\mathrm{\rho }}_{\mathrm{steel}}=7.85\times {10}^3{\mathrm{kgm}}^{-3}\right)$

Given,

$\mathrm{length}=1\mathrm{m}, \mathrm{area} \mathrm{of} \mathrm{cross} \mathrm{section}=20{\mathrm{cm}}^2, {\mathrm{Y}}_{\mathrm{steel}}=2.0\times {10}^{11}{\mathrm{Nm}}^{-2}, {\mathrm{P}}_{\mathrm{steel}}=7.85\times {10}^3{\mathrm{kgm}}^{-3}$

$\mathrm{Change} \mathrm{in} \mathrm{length} ∆\mathrm{L}=\frac{\mathrm{mgL}}{\mathrm{Ae}}$

$\mathrm{strain} \mathrm{e} =\frac{\mathrm{stress}}{\mathrm{young}\text{'}\mathrm{s} \mathrm{modulus}}=\frac{{\displaystyle \frac{\mathrm{force}}{\mathrm{area}}}}{\mathrm{young}\text{'}\mathrm{s} \mathrm{modulus}}$

$\mathrm{strain}=\frac{∆\mathrm{L}}{\mathrm{L}} \& \mathrm{force}=\mathrm{mg}$

$\frac{∆\mathrm{L}}{\mathrm{L}}=\frac{\mathrm{mg}}{\mathrm{AE}}\Rightarrow ∆\mathrm{L}=\frac{\mathrm{mgL}}{2\mathrm{AE}}$

now,

$$\begin{gathered} \mathrm{m}=\mathrm{density}\times \mathrm{volume}=\mathrm{S}\times \mathrm{A}\times \mathrm{L} \\ \Rightarrow ∆\mathrm{L}=\frac{\mathrm{SALgL}}{2\mathrm{AE}} \\ \Rightarrow \frac{{\mathrm{SL}}^2\mathrm{g}}{2\mathrm{E}}=\frac{7850\times 1^2\times 9.81}{2\times 2\times {10}^{11}} \\ \Rightarrow ∆\mathrm{L}=1.923\times {10}^{-5}\mathrm{cm} \end{gathered}$$

Hence the correct answer is $1.923\times {10}^{-5}\mathrm{cm}.$