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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(c)Apparent weight $ = V(\rho - \sigma )g = \frac{m}{\rho }(\rho - \sigma )g$
where $m = $ mass of the body,
$\rho = $ density of the body
$\sigma = $ density of water
If two bodies are in equilibrium then their apparent weight must be equal.
$\therefore $ $\frac{{{m_1}}}{{{\rho _1}}}({\rho _1} - \sigma ) = \frac{{{m_2}}}{{{\rho _2}}}({\rho _2} - \sigma )$
==> $\frac{{36}}{9}(9 - 1) = \frac{{48}}{{{\rho _2}}}({\rho _2} - 1)$
By solving we get ${\rho _2} = 3$.
where $m = $ mass of the body,
$\rho = $ density of the body
$\sigma = $ density of water
If two bodies are in equilibrium then their apparent weight must be equal.
$\therefore $ $\frac{{{m_1}}}{{{\rho _1}}}({\rho _1} - \sigma ) = \frac{{{m_2}}}{{{\rho _2}}}({\rho _2} - \sigma )$
==> $\frac{{36}}{9}(9 - 1) = \frac{{48}}{{{\rho _2}}}({\rho _2} - 1)$
By solving we get ${\rho _2} = 3$.
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