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

• A. $\frac{4}{3}$
• B. $\frac{3}{2}$
• C. 3
• D. 5

Answer 1

Answer: (C)

Apparent weight $=V(\rho -\sigma )g=\frac{m}{\rho }(\rho -\sigma )g$
where $m=$ mass of the body,
$\rho =$ density of the body and
$\sigma =$ density of water If two bodies are in equilibrium then their apparent weight must be equal.
$\therefore \frac{(m{)}_1}{(\rho {)}_1}\left((\rho {)}_1-\sigma \right)g=\frac{(m{)}_2}{(\rho {)}_2}\left((\rho {)}_2-\sigma \right)g$
$\Rightarrow \frac{36}{9}(9-1)=\frac{48}{(\rho {)}_2}\left((\rho {)}_2-1\right)g.$
By solving we get ${\rho }_2=3$.