Physics - Fluid Question with Solution | TestHub

Apparent weight$=\text{V}\left(\mathrm{\rho }-\mathrm{\sigma }\right)\text{g}=\frac{\text{m}}{\mathrm{\rho }}\left(\mathrm{\rho }-\mathrm{\sigma }\right)\text{g}$

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.

$\therefore \frac{{\text{m}}_1}{{\rho }_1}\left({\mathrm{\rho }}_1-\mathrm{\sigma }\right)\text{g}=\frac{{\text{m}}_2}{{\mathrm{\rho }}_2}\left({\mathrm{\rho }}_2-\mathrm{\sigma }\right)\text{g}$

$\Rightarrow \frac{36}{\text{9}}\left(9-1\right)=\frac{48}{{\mathrm{\rho }}_2}\left({\mathrm{\rho }}_2-1\right)\text{g}\text{.}$

By solving we get${\rho }_2=3$.