Question

Two objects suspended from the arms of a balance are in equilibrium when completely submerged in water. The mass of one body is $36 \,g$ and its density is $9 \,g / cm^3$​. If the mass of the other is $48 \,g$, its density in $g / cm^3$ 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.
$\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 $\rho_2=3$.