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9-1.Fluid Mechanics
hard
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
A
$\frac{4}{3}$
B
$\frac{3}{2}$
C
$3$
D
$5$
Solution
(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 ${\rho _2} = 3$.
Standard 11
Physics
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