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(A) Let the density of the object be $\sigma$ and the density of the liquid be $\rho$. Let the volume of the object be $V$.The elongation $\Delta L$ i

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$3: 1$

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(A) Let the density of the object be $\sigma$ and the density of the liquid be $\rho$. Let the volume of the object be $V$.The elongation $\Delta L$ in the wire is proportional to the tension $T$ in the wire,i.e.,$\Delta L \propto T$.Case $1$: When the object is in air,the tension $T_1 = Mg = V\sigma g$. The elongation is $\Delta L_1 = 10 \,mm$.Case $2$: When the object is submerged in the liquid,the tension $T_2 = Mg - F_B = V\sigma g - V\rho g$,where $F_B$ is the buoyant force. The elongation is $\Delta L_2 = 10 - \frac{10}{3} = \frac{20}{3} \,mm$.Taking the ratio of the elongations:$\frac{\Delta L_1}{\Delta L_2} = \frac{T_1}{T_2} = \frac{V\sigma g}{V\sigma g - V\rho g} = \frac{\sigma}{\sigma - \rho}$.Substituting the values:$\frac{10}{20/3} = \frac{\sigma}{\sigma - \rho} \Rightarrow \frac{3}{2} = \frac{\sigma}{\sigma - \rho}$.Cross-multiplying:$3(\sigma - \rho) = 2\sigma \Rightarrow 3\sigma - 3\rho = 2\sigma \Rightarrow \sigma = 3\rho$.Therefore,the ratio of the density of the object to the density of the liquid is $\frac{\sigma}{\rho} = 3: 1$.

An object suspended by a wire stretches it by $10 \,mm$. When the object is immersed in a liquid,the elongation in the wire reduces by $\frac{10}{3} \,mm$. The ratio of the relative densities of the object and the liquid is ............

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