Question

A solid floats with $\dfrac{2}{3}$ of its volume immersed in a liquid and with $\dfrac{3}{4}$ of its volume immersed in another liquid. What fraction of its volume will be immersed if it floats in a homogeneous mixture formed of equal volumes of the liquids?
(A) $\dfrac{6}{7}$
(B) $\dfrac{8}{{11}}$
(C) $\dfrac{{11}}{{16}}$
(D) $\dfrac{{12}}{{17}}$

Answer 1

The fraction of its volume will be immersed if it floats in a homogeneous mixture formed of equal volumes of the liquids can be determined by using the relation which shows the relation between the work done, volume, density and acceleration due to gravity. By using the two volume ratios which are given in the question, the fraction is determined.
Useful formula:
The relation between the work done, volume, density and acceleration due to gravity is given by,
$W = V\rho g$
Where, $V$ is the volume, $\rho$ is the density and $g$ is the acceleration due to gravity.

Complete answer:
Given that,
The solid floats with volume of, ${V_1} = \dfrac{2}{3}V$
The remaining immersed in the volume of, ${V_2} = \dfrac{3}{4}V$
Now,
The relation between the work done, volume, density and acceleration due to gravity is given by,
$W = V\rho g\,.................\left( 1 \right)$
By substituting the first condition of the volume in the above equation (1), then the above equation is written as,
$W = \dfrac{2}{3}V{\rho _1}g$
By rearranging the terms in the above equation, then the above equation is written as,
${\rho _1} = \dfrac{{3W}}{{2Vg}}\,...................\left( 2 \right)$
By substituting the second condition of the volume in the equation (1), then the equation (1) is written as,
$W = \dfrac{3}{4}V{\rho _2}g$
By rearranging the terms in the above equation, then the above equation is written as,
${\rho _2} = \dfrac{{4W}}{{3Vg}}\,...................\left( 3 \right)$
Now,
${\rho _{eff}} = \dfrac{{{\rho _1} + {\rho _2}}}{2}$
By substituting the equation (2) and equation (3) in the above equation, then
${\rho _{eff}} = \left( {\dfrac{{\left( {\dfrac{3}{2} + \dfrac{4}{3}} \right)}}{2}} \right)\left( {\dfrac{W}{{Vg}}} \right)$
By cross multiplying the terms in the above equation, then
${\rho _{eff}} = \left( {\dfrac{{\left( {\dfrac{{9 + 8}}{6}} \right)}}{2}} \right)\left( {\dfrac{W}{{Vg}}} \right)$
On further calculation in the above equation, then
${\rho _{eff}} = \dfrac{{17W}}{{12Vg}}$
Now,
$W = V{\rho _{eff}}g$
By substituting the terms in the above equation, then
$W = \left( {xV} \right)\left( {\dfrac{{17W}}{{12Vg}}} \right)g$
By cancelling the terms in the above equation, then
$W = x\dfrac{{17W}}{{12}}$
By rearranging the terms in the above equation, then
$x = \dfrac{{12W}}{{17W}}$
By cancelling the terms in the above equation, then
$x = \dfrac{{12}}{{17}}$

Hence, the option (D) is the correct answer.

Note:
The work done is directly proportional to the volume of the liquid, density of the liquid and the acceleration due to gravity. As the volume of the liquid or density of the liquid increases, then the work done is also increasing. As the volume of the liquid or density of the liquid decreases, then the work done also decreases.