Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is $36 gm$ and its density is $9 gm / {cm}^{3}$ . If the mass of the other body is $48 gm$ . Its density in $gm / {cm}^{3}$ is _______.

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Detailed Solution

Here apparent weight of both the bodies should be equal. Let density of water is $σ$

$∴ T_{1} = T_{2}$

$m_{1} g (1 - \frac{σ}{ρ}) = m_{2} g (1 - \frac{σ}{ρ_{2}})$

$36 (1 - \frac{σ}{9}) = 48 (1 - \frac{σ}{ρ_{2}})$

$3 - \frac{σ}{3} = 4 - \frac{4 σ}{ρ_{2}} \Rightarrow \frac{4 σ}{ρ_{2}} - \frac{σ}{3} = 1$

$But σ = 1 g / {cm}^{3}$

$\Rightarrow \frac{4}{ρ_{2}} = 1 + \frac{1}{3} = \frac{4}{3}$

$\Rightarrow ρ_{2} = 3 g / {cm}^{3}$

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