Question

There are identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density r. 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. ghra
• B. $\frac{2gh}{\rho a}$
• C. 2pagh
• D. $\frac{\rho gh}{a}$

Answer 1

Answer: (C)

2pagh

Answer 2

Thrust force

F = F₁ – F₂ = ran₁² – ran₂²

= ra (2gh₁) – ra (2gh₂)

= 2rag (h₁ – h₂)

=2 ragh