Question

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 $9\frac{g}{c{m}^3}$ If the mass of the other is 48g, its density in $\frac{g}{c{m}^3}$ is

• A.
• B.
• C. 3
• D. 5

Answer 1

The correct option is C 3

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 $\rho =3$