question_answer

There are two identical small holes of area of cross section a on the either sides of a tank containing a liquid of density \[\rho \] (shown in figure). The difference in height between the holes is ft. 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)  \[\frac{2gh}{\rho a}\]

B)  \[\frac{\rho gh}{a}\]

C)  \[gh\rho a\]

D)  \[2\rho agh\]

Correct Answer: D

Solution :

                Net force (reaction)                 \[F={{F}_{B}}-{{F}_{A}}\]                 \[=\frac{d{{p}_{B}}}{dt}-\frac{d{{p}_{A}}}{dt}\]                 \[=a{{v}_{B}}\rho \times {{v}_{B}}-a{{v}_{A}}\rho \times {{v}_{A}}\] \[\therefore \]  \[F=a\rho (v_{A}^{2}-v_{A}^{2})\] ... (i) According to Bernaullis theorem                 \[{{p}_{A}}+\frac{1}{2}\rho v_{A}^{2}+\rho gh={{p}_{B}}+\frac{1}{2}\rho v_{B}^{2}+0\] \[\Rightarrow \] \[\frac{1}{2}\rho (v_{B}^{2}-v_{A}^{2})=\rho gh\] \[\Rightarrow \] \[v_{B}^{2}-v_{A}^{2}=2gh\] ... (ii) From Eqs (i) and (ii), we get \[F=a\rho \,(2gh)=2a\rho gh\]