A solid uniform ball having volume $ V $ and density $ \rho $ floats at the interface of two unmixible liquids as shown in the figure. The densities of the upper and the lower liquids are $ \rho_{1} $ and $ \rho_{2} $ respectively, such that $ \rho_{1}<\rho<\rho_{2} . $ What fraction of the volume of the ball will be in the lower liquid?
(a) $ \frac{\rho-\rho_{2}}{\rho_{1}-\rho_{2}} $
(b) $ \frac{\rho}{\rho_{1}-\rho_{2}} $
(c) $ \frac{\rho_{1}-\rho}{\rho_{1}-\rho_{2}} $
(d) $ \frac{\rho_{1}-\rho_{2}}{\rho_{2}} $

Academic Physics NCERT Class 8

As known for a stationary body in a fluid, buoyant force$=$weight of the body immersed.

Force acting on the body immersed in the fluid, $F=V\rho g$

Let the volume of the body$=V$

The density of the body$=\rho$ [given]

Density of upper liquid$=\rho_{1}$ [given]

Density of lower liquid$= \rho_{2}$ [given]

Let us assume that the volume of the body in upper liquid be $v_1$ and the volume in lower liquid be $v_2$.

Therefore, $V=v_1+v_2$

Since the ball is stationary at the interface as given in the question, therefore the net buoyant force on the body will be equal to the weight of the body.

Let the buoyant force due to upper liquid be $F_1$ and due to the lower liquid be $F_2$.

$F_1+F_2=mg$ …..$( i)$

$F_1=v_1\rho_{1}g$

And $F_2=v_2\rho_{2}g$

Putting these value in $( i)$

$v_1\rho_{1}g+v_2\rho_{2}g=(v_1+v_2)\rho g$

On eliminating $g$ on both sides, we havee

$v_1\rho_{1}+v_2\rho_{2}=v_1\rho+v_2\rho$

$v_1\rho_{1}+(V-v_1)\rho_{2}=v_1\rho+(V-v_1)\rho$

$v_1(\rho_{1}-\rho_{2})+V\rho_{2}=v_1(\rho-\rho)+V\rho$

$v_1(\rho_{1}-\rho_{2})=V(\rho-\rho_{2})$

$\frac{v_1}{V}=\frac{(\rho-\rho_{2})}{(\rho_{1}-\rho_{2})}$

And $\frac{v_2}{V}=1-\frac{v_1}{V}=1-\frac{(\rho-\rho_{2})}{(\rho_{1}-\rho_{2})}$

$\frac{v_2}{V}=\frac{\rho_{1}-\rho}{\rho_{1}-\rho_{2}}$

Thus, option c is correct.

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Updated on: 10-Oct-2022

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