Two bodies placed on a balance are in equilib | vedclass

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(C) The apparent weight of a body immersed in a liquid is given by $W_{app} = V(\rho - \sigma)g$,where $V$ is the volume,$\rho$ is the density of the

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(C) The apparent weight of a body immersed in a liquid is given by $W_{app} = V(\rho - \sigma)g$,where $V$ is the volume,$\rho$ is the density of the body,and $\sigma$ is the density of the liquid.Since $V = m/\rho$,the apparent weight is $W_{app} = \frac{m}{\rho}(\rho - \sigma)g = m(1 - \frac{\sigma}{\rho})g$.Given that the two bodies are in equilibrium in water,their apparent weights must be equal:$m_1(1 - \frac{\sigma}{\rho_1}) = m_2(1 - \frac{\sigma}{\rho_2})$Given: $m_1 = 36 \, g$,$\rho_1 = 9 \, g/cm^{3}$,$m_2 = 48 \, g$,and $\sigma = 1 \, g/cm^{3}$ (density of water).Substituting the values:$36(1 - \frac{1}{9}) = 48(1 - \frac{1}{\rho_2})$$36(\frac{8}{9}) = 48(\frac{\rho_2 - 1}{\rho_2})$$32 = 48(\frac{\rho_2 - 1}{\rho_2})$$\frac{32}{48} = \frac{\rho_2 - 1}{\rho_2}$$\frac{2}{3} = 1 - \frac{1}{\rho_2}$$\frac{1}{\rho_2} = 1 - \frac{2}{3} = \frac{1}{3}$$\rho_2 = 3 \, g/cm^{3}$.

Two bodies placed on a balance are in equilibrium when immersed in water. If the mass of the first body is $36 \, g$ and its density is $9 \, g/cm^{3}$,and the mass of the second body is $48 \, g$,what is the density of the second body in $g/cm^{3}$?

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