question_answer

There are two identical small holes P and Q of area of cross-section a on the opposite sides of a tank containing a liquid of density p. The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is

A) \[gh\rho a\]        

B) \[\frac{2gh}{\rho a}\]

C) \[2\rho agh\]

D) \[\frac{\rho gh}{a}\]

Correct Answer: C

Solution :

[c] Force \[F=\rho Qv=\rho a{{v}^{2}}=\rho a\times 2gh\] \[={{F}_{2}}-{{F}_{1}}\] \[=\rho av_{2}^{2}-\rho av_{1}^{2}\] \[=\rho a(v_{2}^{2}-v_{1}^{2})\] \[=\rho a[2g(y+h)-2gy]\] \[=\rho a\times 2gh\]