An object suspended by a wire stretches it by $10 \,mm$. When object is immersed in a liquid the elongation in wire reduces by $\frac{10}{3} \,mm$. The ratio of relative densities of the object and liquid is $............$
Medium
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$\Delta L=\frac{F L}{A Y}$ $\left\{\begin{array}{l}\text { Let density of liquid }=\rho \\ \text { Let density of object }=\sigma \\ \text { Mass of object }=M\end{array}\right.$
$\Rightarrow$ 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 $3: 1$
$\Rightarrow$ 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 $3: 1$
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