\(\Delta L=\frac{F L}{A Y}\)   \(\left\{\begin{array}{l}\text { Let density of liquid }=\rho \\ \text { Let density of object }=\sigma \\ \text { Mass of object }=M\end{array}\right.\)

\(\Rightarrow\) Elongation \(\propto\) force and force is due to weight

So elongation \(\propto\) weight

\(\Delta L_1 \propto\) weight   \(\quad \ldots (1)\)  {When not submerged in liquid }

\(\Delta L_2 \propto\) apparant weight \(\ldots .(2)\)  {When submerged in liquid }

Dividing \((1)\) by \((2)\)

\(\frac{10}{10-\frac{10}{3}}=\frac{M g}{M g-\frac{M g \rho}{\sigma}}\)

\(\Rightarrow \frac{1}{1-\frac{1}{3}}=\frac{1}{1-\frac{\rho}{\sigma}}\)

Solving this we get

\(\frac{\rho}{\sigma}=\frac{1}{3}\)

So relative densities of object \((\sigma)\) and liquid \((\rho)\) is \(3: 1\)