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

The correct option is (C) 3

We know,

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

where m = mass of the body,

$\rho$ = density of the body

$\sigma$ = density of water

If two bodies are in equilibrium then their apparent weight must be equal.

$\therefore \frac{m_1}{{\rho }_1}({\rho }_1-\sigma )$ = $\frac{m_2}{{\rho }_2}({\rho }_2-\sigma )$

$\Rightarrow \frac{36}{9}(9-1)$ = $\frac{48}{{\rho }_2}({\rho }_2-1)$

By solving we get ${\rho }_2$ = 3