An object suspended by a wire stretches it by 10 ,mm . When object is immersed in a liquid elongatio

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9-1.Fluid Mechanics

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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 ............

$3: 1$

$1: 3$

$1: 2$

$2: 1$

Solution

(a)

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

$\Rightarrow$ 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 $3: 1$

Standard 11

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Solution

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