Question

Two objects suspended from the arms of a balance are in equilibrium when completely submerged in water. The mass of one body is $36g$ and its density is $9gc{m}^{-3}$ ​. If the mass of the other object is $48g$ , then its density in $gc{m}^{-3}$ is

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

Answer 1

Answer: (C)

Apparent weight $=\text{V}\left(\rho -\sigma \right)\text{g}=\frac{\text{m}}{\rho }\left(\rho -\sigma \right)\text{g}$
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.
$\begin{array}{l}\therefore \frac{m_1}{{\rho }_1}\left({\rho }_1-\sigma \right)g=\frac{m_2}{{\rho }_2}\left({\rho }_2-\sigma \right)g\\ \Rightarrow \frac{36}{9}(9-1)=\frac{48}{{\rho }_2}\left({\rho }_2-1\right)\end{array}$
By solving we get ${\rho }_2=3$.