Q. 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 $9\ g/cm^{3}$. 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
Explanation
(c) 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
$\therefore\frac{m_{1}}{\rho_{1}}\left( \rho_{1} - \sigma \right) = \frac{m_{2}}{\rho_{2}}\left( \rho_{2} - \sigma \right) \Rightarrow \frac{36}{9}(9 - 1) = \frac{48}{\rho_{2}}(\rho_{2} - 1)$
By solving we get $\rho_{2} = 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.