Two bodies are in equilibrium when suspended in water from the arms at a balance. The mass of one body is 30 g and its density is 6 g/cc. If the mass of the other body is 40g. Find its density ‘d’.

Given

Mass of one body $m_1=30\text{g}$

Density of body ${\rho }_1=6\text{g/cc}$

Mass of another body $m_2=40\text{g}$

Density${\rho }_2=?$

Apparent weight of 1st body $m_{1}g-F_b$ $=\left(30-\frac{30}{6}\right)\text{g}$  $\left(\because F_b=\text{buoyant}\text{force}\right)$  $\left(\because F_b=mg\right)$ $\left(\begin{array}{c}\because m=\rho v\\ \because v=\frac{m}{\rho }\end{array}\right)$

Apparent weight of 2nd body$m_{2}g-F_b\left(40-\frac{40}{{\rho }_2}\times 1\right)\text{g}$

Bodies are in equilibrium So, Apparent weights are equal

$\left(30-\frac{30}{6}\right)g=\left(40-\frac{40}{d}\right)g$

$25=40-\frac{40}{{\rho }_2}$

$\Rightarrow \frac{40}{{\rho }_2}=40-25$

$\Rightarrow {\rho }_2=\frac{40}{15}=2.66\text{g/cc}$