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 $9gcm^{- 3}$ ​. If the mass of the other object is $48g$ , then its density in $gcm^{- 3}$ 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 $\rho_{2}=3$.