Question

A body is floating in the water. $1/3^{rd}$ part of the body is inside the water and $2/3^{rd}$ is in the air as shown in the diagram. Find out the density of the body.
${\rho }_w=\text{density of water}$
${\rho }_b=\text{density of body}$

• A. ${\rho }_b=\frac{{\rho }_w}{3}$
• B. ${\rho }_b=\frac{{\rho }_w}{2}$
• C. ${\rho }_b=\frac{{\rho }_w}{4}$
• D. ${\rho }_b=\frac{2{\rho }_w}{3}$

Answer 1

The correct option is A ${\rho }_b=\frac{{\rho }_w}{3}$

It is given that $1/3^{rd}$ of body is submersed in the water.

So, the volume of displaced water,
$V_w=\frac{V}{3}$
where:
$V:$ Volume of body

Using Archiemedes' principle, the weight of the body is equal to the weight of displaced water by the body.
$W_{\text{body}}=W_{\text{displaced water}}$
$⟹{\rho }_{b}Vg={\rho }_w{V}_{w}g$
${\rho }_{b}V={\rho }_w\frac{V}{3}$
${\rho }_b=\frac{{\rho }_w}{3}$

Hence, the density of the body is $1/3^{rd}$ of the density of water.