Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is $36g$ and its density is $9g/c{m}^3$. If the mass of the other is $48g$, its density in $g/c{m}^3$ is

Given,
Mass of the body, $m=36g$

Density of the body, $p=9g/c{m}^3$

Density of water, $\sigma =1g/c{m}^3$

As we know, if two bodies are in equilibrium, it means both bodies have equal apparent weights. Therefore,

Apparent weight
$=V(p-\sigma )g=\frac{m}{p}(p-\sigma )g$

$\frac{m_1}{p_1}(p_1-\sigma )=\frac{m_2}{p_2}(p_2-\sigma )$

$\frac{36}{9}(9-1)=\frac{48}{p_2}(p_2-1)$

$\Rightarrow p_2=3$