Question

# 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 has to be applied on the tank to keep it in equilbrium is :

A
ghρa
B
2ghρa
C
2ρagh
D
ρgha

Solution

## The correct option is D $$2\rho agh$$Momentum of the mass of water coming out of the hole per sec is thrust.Resultant thrust $$F= F_1-F_2=-\rho av_1^2 - \rho av_2^2=-\rho a \sqrt{2gh_1} - \rho a \sqrt{2gh_2} =-2a\rho g(h_1-h_2)$$$$\therefore$$ horizontal force that has to be applied to the tank to keep it in equilibrium is$$2a\rho g(h_1-h_2)$$or$$2a\rho gh$$Physics

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