Question

A solid sphere of volume $V$ and density $\ell$ floats at the interface of two immiscible liquids of densities ${\ell _1}$ and ${\ell _2}$ respectively . If ${\ell _1} < \ell < {\ell _2}$ then the ratio of volume of the parts of the sphere in upper and lower liquid is:
A) $\dfrac{{\ell - {\ell _1}}}{{{\ell _2} - \ell }}$
B) $\dfrac{{{\ell _2} - \ell }}{{\ell - {\ell _1}}}$
C) $\dfrac{{\ell + {\ell _1}}}{{\ell + {\ell _2}}}$
D) $\dfrac{{\ell + {\ell _2}}}{{\ell + {\ell _1}}}$

Answer 1

We will first understand that solid sphere of volume $V$ and density $\ell$ floats at the interface of two types of immiscible liquids of densities ${\ell _1}$ and ${\ell _2}$ respectively. Then we have to calculate the ratio of volume of the parts of the sphere in upper and lower liquid then solve further by using the formula for law of floatation and then answers.

Complete step by step solution:
Given ,
Volume of the solid sphere = $$$$V
And, density of the solid sphere = $$$$\ell
We know
${U_1} =$ weight of liquid displaced
And ,${U_2} =$ weight of liquid displaced
Let volume of the part of the immersed in liquid = $$$${v_1} and
Density of the part of immersed sphere in liquid = $$$${\ell _1}
Also , let volume of the part of the sphere immersed in liquid = $$$${v_2}and
Density of the part of immersed sphere in liquid $= {\ell _2}$
Now ,${U_1} =$ weight of liquid displaced $= {v_1}{\ell _1}g$
And ,${U_2} =$ weight of liquid displaced $= {v_2}{\ell _2}g$
According to law of floatation
$V\ell g = {U_1} + {U_2}$ $\to \left( 1 \right)$
We know $V = {v_1} + {v_2}$and ${U_1} = {v_1}{\ell _1}g$, ${U_2} = {v_2}{\ell _2}g$
From equation $\left( 1 \right)$ , we get
$V\ell g = {U_1} + {U_2}$
Substituting the value of V,
\Rightarrow \left( {{v_1} + {v_2}} \right)\ell g$$$$ = {v_1}{\ell _1}g + {v_2}{\ell _2}g
\Rightarrow {v_1}\ell g + {v_2}\ell g$$$$ = {v_1}{\ell _1}g + {v_2}{\ell _2}g
Taking g common from both side
\Rightarrow g\left( {{v_1}\ell + {v_2}\ell } \right)$$$$ = g\left( {{v_1}{\ell _1} + {v_2}{\ell _2}} \right)
After cancelling g we get,
\Rightarrow $$$${v_1}\ell + {v_2}\ell $$$$ = {v_1}{\ell _1} + {v_2}{\ell _2}
$\Rightarrow {v_1}\ell - {v_1}{\ell _1} = {v_2}{\ell _2} - {v_2}\ell$
$\Rightarrow {v_1}\left( {\ell - {\ell _1}} \right) = {v_2}\left( {{\ell _2} - \ell } \right)$
So the ratio of $v_1$ and $v_2$
$\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{\ell _2} - \ell }}{{\ell - {\ell _1}}}$
Therefore , the ratio of volume of the parts of the sphere in upper and lower liquid is $\dfrac{{{\ell _2} - \ell }}{{\ell - {\ell _1}}}$.

Hence option B is correct.

Note: Alternate method of this question is $\dfrac{{{V^{th}}}}{n}$ part of the sphere is inside the liquid with density ${\ell _1}$ and ${\ell _2}$ and we can get the answer as $\dfrac{{{\ell _2} - \ell }}{{\ell - {\ell _1}}}$.