Question

Two solids $A$ and $B$ float in water. It is observed that $A$ floats in water with $\dfrac{1}{2}$ of its volume immersed in water while $B$ floats in water with $\dfrac{1}{4}$ of its volume above the water level. The ratio of density of $A$ to $B$ is
A. $4:3$
B. $2:3$
C. $3:4$
D. $1:2$

Answer 1

Archimedes principle states that any body partially or fully submerged in a liquid or gas at the state of rest is acted upon by an upward buoyant force the magnitude of which is equal to the fluid displaced by the body. In this question we shall equate the weight of the two bodies with the weight of the displaced liquid for both the objects. For the weight of the object, we shall calculate the downward force acting on it while for the weight of liquid displaced, we shall calculate the upthrust. Then finally we will take the ratio of the relative densities of both A and B to get the final answer.

Complete step by step answer:
We know that weight is given by $W = mg$ where m is the mass and g is the acceleration due to gravity.Also, we know that the density of a material is given as $\rho = \dfrac{m}{V}$ where $V$ is the volume of the object. And so, the mass can be given as $m = \rho V$.

Considering the body A first we have, we know that the weight of the body must be equal to the weight of the liquid displaced from the water.
Hence, ${W_A} = {W_{water}}$
This can be rewritten as ${\rho _A}Vg = {\rho _w}{V_{water}}g$ (Here ${V_{water}}$ is the volume of the object immersed)
The above expression simplifies to $\dfrac{{{V_{water}}}}{V} = \dfrac{{{\rho _A}}}{{{\rho _w}}}$

Given that A floats in water with $\dfrac{1}{2}$ of its volume immersed in water, $\dfrac{{{V_{water}}}}{V} = \dfrac{1}{2}$
Hence, $\dfrac{{{\rho _A}}}{{{\rho _w}}} = \dfrac{1}{2}\,\,\,\,\,\,\,............(1)$
Considering the body B we have,
We know that the weight of the body must be equal to the weight of the liquid displaced from the water.
Hence, ${W_B} = {W_{water}}$
This can be rewritten as ${\rho _B}Vg = {\rho _w}{V_{water}}g$ (Here ${V_{water}}$ is the volume of the object immersed)

The above expression simplifies to $\dfrac{{{V_{water}}}}{V} = \dfrac{{{\rho _B}}}{{{\rho _w}}}$
Given that B floats in water with $\dfrac{1}{4}$ of its volume above the water level, the volume immersed in the water will be $\dfrac{3}{4}$
Hence, $\dfrac{{{V_{water}}}}{V} = \dfrac{3}{4}$
Hence, $\dfrac{{{\rho _B}}}{{{\rho _w}}} = \dfrac{3}{4}\,\,\,\,\,\,\,\,\,\,.........(2)$
Taking ratio of (1) and (2)
$\dfrac{{{\rho _A}}}{{{\rho _B}}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{3}{4}}}$
$\therefore \dfrac{{{\rho _A}}}{{{\rho _B}}} = \dfrac{2}{3}$

Hence, option B is the correct answer.

Note: Very often weight is misunderstood as mass of the object. These are two different terms and must be noted carefully. Mass is the inherent property of an object which tells how much matter it contains while weight is the normal reaction acting on the body. Mass is constant while the weight may vary.