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 of equal volumes of the liquids?

Answer 1

Hint: When a body is submerged in a liquid, it experiences an upward force. A body can float in a liquid only when the force due to weight of the solid is equal to the upthrust force on it which is the weight of the liquid displaced by the body.

Formula used:
The weight of a liquid is given as
$W = V\rho g$
where V denotes the volume of the liquid whose density is given as $\rho$ and g is the acceleration due to gravity.

Complete step-by-step answer:
We are given a solid. Let $\rho$ be the density of solid and V the volume of the solid.
It floats in liquid 1 with two-thirds of its volume immersed in it. Let ${\rho _1}$ be the density of liquid 1.
As we know that a body can float in a liquid only when the force due to weight of the solid is equal to the weight of the liquid displaced by the body. This means that solid is displacing two-thirds of the liquid. Therefore we can write the following expression for the weight of solid and the displaced liquid.
$\therefore V\rho g = \dfrac{2}{3} \times V{\rho _1}g \\\ \Rightarrow {\rho _1} = \dfrac{3}{2}\rho {\text{ }}...{\text{(i)}} \\\$
Similarly, there is a liquid 2. Let its density be ${\rho _2}$ and the solid displaces three-fourth of the volume of liquid 2. We can write the expression for this as follows:
$V\rho g = \dfrac{3}{4}V{\rho _2}g \\\ \Rightarrow {\rho _2} = \dfrac{4}{3}\rho {\text{ }}...{\text{(ii)}} \\\$
Now when equal volumes liquid 1 and liquid 2 are mixed homogeneously with each other, let the resultant density be $\rho '$. Since the two liquids are mixed in equal volumes, the new density is the average of the individual densities of liquid 1 and liquid 2 given as:
$\rho ' = \dfrac{{{\rho _1} + {\rho _2}}}{2}{\text{ }}...{\text{(iii)}}$
It is given that the solid floats in this mixture also. Let x be the fraction of volume of solid immersed in the mixture of two liquids. Now we can write the equation for this case as follows
$V\rho g = x \times V\rho 'g \\\ \Rightarrow x = \dfrac{\rho }{{\rho '}}{\text{ }}...{\text{(iv)}} \\\$
Substituting equation (iii) in (iv), we get
$x = \dfrac{{2\rho }}{{{\rho _1} + {\rho _2}}}{\text{ }}...{\text{(v)}}$
Using equations (i) and (ii) in equation (v), we get
$x = \dfrac{{2\rho }}{{\dfrac{3}{2}\rho + \dfrac{4}{3}\rho }} \\\ = \dfrac{{2\rho }}{{\dfrac{{9 + 8}}{6}\rho }} \\\ = \dfrac{{2 \times 6}}{{17}} \\\ \Rightarrow x = \dfrac{{12}}{{17}} \\\$

This is the required answer.

Note: 1. The floating bodies follow the Archimedes principle that the amount of upward force on a body immersed in a liquid is equal to the weight of the liquid displaced by the body.
2. Weight of a body is equal to mg. Since density is equal to mass per unit volume. Using this relation we get weight of a body $= V\rho g$