Question

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

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

Answer 1

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

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

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

Using Archimedes' principle, the weight of the body is equal to the weight of water displaced 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 $1/3^{\text{rd}}$ of the density of the water in the container.