question_answer

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 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 :

                   Net force (reaction) = \[F={{F}_{B}}-{{F}_{A}}\]\[=\frac{d{{p}_{B}}}{dt}-\frac{d{{p}_{A}}}{dt}\]                                \[=a{{v}_{B}}\rho \times {{v}_{B}}-a{{v}_{A}}\rho \times {{v}_{A}}\]                    \                        \[F=a\rho \left( v_{B}^{2}-v_{A}^{2} \right)\]      ?(i)                    According to Bernoulli's theorem                    \[{{p}_{A}}+\frac{1}{2}\rho v_{A}^{2}+\rho gh={{p}_{B}}+\frac{1}{2}\rho v_{B}^{2}+0\]                    Þ \[\frac{1}{2}\rho \left( v_{B}^{2}-v_{A}^{2} \right)=\rho gh\]Þ \[v_{B}^{2}-v_{A}^{2}=2gh\]             From equation (i), \[F=2a\rho gh.\]