The Young's modulus of a rubber string of length 12   cm and density 1 . 5   kg   m - 3…

Hooke's law for elasticity, $\sigma =\gamma \epsilon$

$\gamma =\frac{\sigma }{\epsilon }=\frac{F}{A}\frac{l}{∆l}$here, $l$is the length of string undergone through tension. 

Here, the string is suspended vertically and the string is elongated due to its own weight therefore, weight is acting at the centre of mass which in the given case is located at the half-length of string, therefore, $l=\frac{12 \mathrm{cm}}{2}=6 \mathrm{cm} \text{or }0.06 \mathrm{m}$

$$\begin{gathered} \gamma =\frac{\sigma }{\epsilon }=\frac{F}{A}\frac{l}{∆l} \\ \gamma =\frac{F}{Al}\frac{l^2}{∆l}=\frac{mg}{V}\frac{{\left(0.06\right)}^2}{∆l} \cdots \cdots \left(A \text{area}\times l \text{length}=V \text{volume}\right) \\ \gamma =\rho g\frac{{\left(0.06\right)}^2}{∆l} \cdots \cdots \left(\rho =\frac{m}{V}\right) \\ \Rightarrow ∆l=\rho g\frac{{\left(0.06\right)}^2}{\gamma }=\left(1.5 \mathrm{kg} {\mathrm{m}}^{-3}\right)\left(10 \mathrm{m} {\mathrm{s}}^{-2}\right)\frac{{\left(0.06 \mathrm{m}\right)}^2}{5\times {10}^8 \mathrm{N} {\mathrm{m}}^{-2}} \\ \Rightarrow ∆l=2.16\times {10}^{-10} \mathrm{m} \end{gathered}$$