Question

State the law of floatation of bodies in a liquid. Arrive at an expression for the fraction of volume of the floating body submerged in the liquid.

Answer 1

Hint: Archimedes has calculated the amount of buoyant force exerted by the liquid on the body. The weight of the liquid displaced will be equivalent to the weight of the object. Using this find the fraction of the volume of the body submerged. This will help you in answering this question.

Complete step by step answer:
Archimedes has calculated the amount of buoyant force exerted by the liquid on the body. According to the Archimedes, if a body is completely or partially immersed in a liquid at rest, it will experience an upthrust, identical to the weight of the liquid displaced by the body.
Let us assume that $x$ be the fraction of volume of the body floating above the surface of the liquid. We can write that,
$\text{weight of the liquid displaced}=\text{weight of the object}$
This can be written as,
$\therefore \left( {{V}_{0}}-x{{V}_{0}} \right)dg={{V}_{0}}{{d}_{0}}g$
Where ${{V}_{0}}$ be the volume of the liquid, $d$ be the density of the liquid, ${{d}_{0}}$ be the density of the object and $g$ be the acceleration due to gravity.
The equation can be simplified as,
$\left( 1-x \right)d={{d}_{0}}$
Therefore the fraction of volume of the floating body submerged in the liquid can be found by rearranging this equation,
$x=1-\dfrac{{{d}_{0}}}{d}=\dfrac{d-{{d}_{0}}}{d}$
The answer has been obtained.

Note: A body will be sunk in a liquid if its density is more than that of the liquid in which it is immersed. The body is floating being completely immersed in water when the density of the body will be identical to the density of the liquid. The body is floating being partially immersed when the density of the body will be less than the density of liquid.