# NEET Physics Mechanical Properties of Fluids Questions Solved

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 / cm3. If the mass of the other is 48 g, its density in g / cm3 is
(a)$\frac{4}{3}$                                         (b) $\frac{3}{2}$
(c) 3                                           (d) 5

(c) Apparent weight = $\mathrm{V}\left(\mathrm{\rho }-\mathrm{\sigma }\right)\mathrm{g}=\frac{\mathrm{m}}{\mathrm{\rho }}\left(\mathrm{\rho }-\mathrm{\sigma }\right)\mathrm{g}$
where m = mass of the body,
$\mathrm{\rho }$ = density of the body
$\mathrm{\sigma }$ = density of water
If two bodies are in equilibrium then their apparent weight must be equal.

$\therefore \frac{{\mathrm{m}}_{1}}{{\mathrm{\rho }}_{1}}\left({\mathrm{\rho }}_{1}-\mathrm{\sigma }\right)=\frac{{\mathrm{m}}_{2}}{{\mathrm{\rho }}_{2}}\left({\mathrm{\rho }}_{2}-\mathrm{\sigma }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{36}{9}\left(9-1\right)=\frac{48}{{\mathrm{\rho }}_{2}}\left({\mathrm{\rho }}_{2}-1\right)$

By solving we get ${\mathrm{\rho }}_{2}=3.$

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