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9-1.Fluid Mechanics
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A metallic body of material with density of $8000\ kg/m^3$ has a cavity inside. A spring balance shows its mass to be $10.0\ kg$ in air and $7.5\ kg$ when immersed in water. The ratio of the volume of the cavity to the volume of the material of the body must be
A
$\frac{2}{5}$
B
$\frac{1}{2}$
C
$1$
D
$\frac{3}{4}$
Solution
$\rho v_{2}=10(\text { in air })$
$\rho v_{2}-\rho_{\omega}\left(v_{1}+v_{2}\right)=7.5$ (in water)
$\rho_{\omega}\left(v_{1}+v_{2}\right)=2.5$
$\frac{\rho_{\omega}\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)}{\rho\left(\mathrm{v}_{2}\right)}=\frac{2.5}{10}=\frac{1}{4}$
$\frac{1000}{8000}\left(\frac{v_{1}}{v_{2}}+1\right)=\frac{1}{4}$
$\frac{v_{1}}{v_{2}}=1$
Standard 11
Physics
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