Gujarati
Hindi
9-1.Fluid Mechanics
hard

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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