Question

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 $9 \,g / cm ^{3}$. 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 $= V (\rho-\sigma) g =\frac{ m }{\rho}(\rho-\sigma) g$
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 $\rho_{2}=3$.