Two bodies are in equilibrium when suspended | vedclass

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(C) The apparent weight of a body submerged in a fluid is given by $W_{app} = V(\rho - \sigma)g = \frac{m}{\rho}(\rho - \sigma)g$,where $m$ is the mas

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$3$

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(C) The apparent weight of a body submerged in a fluid is given by $W_{app} = V(\rho - \sigma)g = \frac{m}{\rho}(\rho - \sigma)g$,where $m$ is the mass of the body,$\rho$ is the density of the body,and $\sigma$ is the density of the fluid (water,$\sigma = 1 \ g/cm^3$).Since the two bodies are in equilibrium,their apparent weights must be equal:$\frac{m_1}{\rho_1}(\rho_1 - \sigma) = \frac{m_2}{\rho_2}(\rho_2 - \sigma)$Given $m_1 = 36 \ g$,$\rho_1 = 9 \ g/cm^3$,$m_2 = 48 \ g$,and $\sigma = 1 \ g/cm^3$:$\frac{36}{9}(9 - 1) = \frac{48}{\rho_2}(\rho_2 - 1)$$4(8) = \frac{48}{\rho_2}(\rho_2 - 1)$$32 = \frac{48(\rho_2 - 1)}{\rho_2}$$32\rho_2 = 48\rho_2 - 48$$16\rho_2 = 48$$\rho_2 = 3 \ g/cm^3$.

Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is $36 \ g$ and its density is $9 \ g/cm^3$. If the mass of the other is $48 \ g$,its density in $g/cm^3$ is:

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