Question

Two bodies are in equilibrium when suspended in water from the arms (equal length) of a balance (in balanced condition). The mass of one body is $36\text{g}$ and its density is $9\frac{g}{c{m}^3}$. If the mass of the other body is $48\text{g}$, its density in $\frac{g}{c{m}^3}$ is

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

Answer 1

The correct option is C $3$

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 weights must be equal.
$\frac{m_1}{{\rho }_1}({\rho }_1-1)=\frac{m_2}{{\rho }_2}({\rho }_2-1)$
$\frac{36}{9}(9-1)=\frac{48}{{\rho }_2}({\rho }_2-1)$
$8{\rho }_2=12{\rho }_2-12$
$\Rightarrow {\rho }_2=3$
Hence, the correct answer is option (c).