Question

A solid weighs $75gm$ in air and $55.6gm$ in water. Find the relative density of the solid. Also state if the object will float or sink when immersed in water.

Answer 1

Hint: This problem can be solved by realizing the fact that the loss in weight of an object when put in water from air is due to the buoyant force applied by the water on the air. This buoyant force is equal to the weight of the water displaced by the object.

Formula used:
${{F}_{B}}={{W}_{water}}$
$\text{ Relative density = }\dfrac{\text{Density of an object}}{\text{Density of water at }{{\text{4}}^{0}}C}$

Complete step-by-step answer:
We will solve this problem by finding out the loss in weight of the object in water and equating it with the weight of the water displaced, thereby obtaining a relation for the density of the object from which we can find out the relative density.
Now, the buoyant force ${{F}_{B}}$ experienced by a body in water is equal in magnitude to the weight ${{W}_{water}}$ of the water displaced by it.
${{F}_{B}}={{W}_{water}}$ --(1)
Also, the relative density of an object is the ratio of the density of the object to the density of water at ${{4}^{0}}C$.
$\text{ Relative density = }\dfrac{\text{Density of an object}}{\text{Density of water at }{{\text{4}}^{0}}C}$ --(2)
Now, let us analyze the question.
The weight of the solid in air is $W=75g$
The weight of the object in water is $W'=55.6g$
Therefore, the loss in weight of the solid when placed in water form air is $\Delta W=W-W'=75g-55.6g=19.4g$ --(3)
Now, this loss in weight is due to the buoyant force exerted by the water on the body.
Therefore, the magnitude of the buoyant force on the body is equal to the loss in weight of the body in water.
Therefore, the magnitude of the buoyant force exerted is ${{F}_{B}}=\Delta W=19.4g$.
Now, using (1), we get the weight of the water displaced by the solid as
${{W}_{water}}={{F}_{B}}=19.4g$ --(4)
Now, this weight can also be expressed as
${{W}_{water}}={{M}_{water}}g={{V}_{water}}{{\rho }_{water}}g$ --(5) $\left( \because \text{Weight = Mass}\times \text{Acceleration due to gravity, Mass = Density}\times \text{Volume} \right)$
where ${{M}_{water}}$ is the mass of the liquid displaced, ${{V}_{water}}$ is the volume of the liquid displaced, ${{\rho }_{water}}$ is the density of water and $g$ is the acceleration due to gravity.
Using (4) and (5), we egt
$19.4g={{V}_{water}}{{\rho }_{water}}g$
$\therefore {{V}_{water}}{{\rho }_{water}}=19.4$ --(6)
Now, since the solid is submerged in water, the volume of the solid is actually the volume of the liquid displaced.
$\therefore {{V}_{water}}={{V}_{solid}}$ --(7)
where ${{V}_{solid}}$ is the volume of the solid.
Putting (7) in (6), we get
${{V}_{solid}}{{\rho }_{water}}=19.4$
$\therefore {{V}_{solid}}=\dfrac{19.4}{{{\rho }_{water}}}$ --(8)
Also, the weight of the solid in air is $W=75g$.
This can also be expressed as
$W=Mg={{V}_{solid}}{{\rho }_{solid}}g$ $\left( \because \text{Weight = Mass}\times \text{Acceleration due to gravity, Mass = Density}\times \text{Volume} \right)$
where $M$ is the mass of the solid and ${{\rho }_{solid}}$ is the density of the solid.
$\therefore 75g={{V}_{solid}}{{\rho }_{solid}}g$
$\therefore 75={{V}_{solid}}{{\rho }_{solid}}$ --(9)
Putting (8) in (9), we get
$75=\dfrac{19.4}{{{\rho }_{water}}}{{\rho }_{solid}}$
$\therefore \dfrac{{{\rho }_{solid}}}{{{\rho }_{water}}}=\dfrac{75}{19.4}\approx 3.86$ --(10)
Now, using (2), we can say that the relative density $RD$ of the solid is
$RD=\dfrac{{{\rho }_{solid}}}{{{\rho }_{water}}}$ --(11)
Putting (10) in (11), we get
$RD=3.86$
Therefore, the relative density of the solid is $3.86$.
For objects that have a relative density greater than one, the objects sink when immersed in water. Therefore, since the solid has a relative density greater than one, it will sink when immersed in water.

Note: Students must note that the air also being a fluid also exerts a buoyant force on the object. However, the density of air is negligible in comparison to the density of the solid. Since the buoyant force exerted by a fluid depends directly upon the density of the fluid, therefore the buoyant force exerted by air on the solid will be negligible and will cause so significant change in its weight. Therefore, we have neglected the buoyant force exerted by air in the problem. However, if the density of the body given is such that it is comparable to that of air, then the buoyant force exerted by air should also be taken into account.