There are two identical small holes of area of cross section a on the opposite sides of a tank containing liquid of density `rho`. The differences in height between the holes is `h`. The tank is resting ona smooth horizontal surface. The horizontal force which will have to be applied on the tank to keep it inequilibrium is

Options

A. `ghrhoa`
B. `(2gh)/(rhoa)`
C. `2rhoagh`
D. `(rhogh)/a`

Answer

C

Explanation

Thrust force `F=F_(1)-F_(2)=rhoav_(1)^(2)-rhoav_(2)^(2)`
`=rhoa(2gh_(1))-rhoa(2gh_(2))`
`=2rhoag(h_(1)-h_(2))=2rhoagh`