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9-1.Fluid Mechanics
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There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density $\rho$. The difference in height between the holes is $h$. Tank is resting on a smooth horizontal surface. Horizontal force which will have to be applied on the tank to keep it in equilibrium is

A
$g\, h\, \rho a$
B
$\frac{2gh}{\rho a}$
C
$2g\, h\, \rho a$
D
$\frac{\rho gh}{a}$
Solution
$\mathrm{F}=\mathrm{u} \frac{\mathrm{dm}}{\mathrm{dt}}=\mathrm{v} \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{a} \rho \ell=\mathrm{a} \rho \mathrm{v} \frac{\mathrm{d} \ell}{\mathrm{dt}}=\mathrm{apv}^{2}$
$\mathrm{F}=2 \mathrm{a}$ $\rho gh$
Standard 11
Physics
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