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 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)  \[\frac{2gh}{\rho a}\]   

B)         \[\frac{\rho gh}{a}\]

C)  \[gh\rho a\]     

D)         \[2\rho agh\]

Correct Answer: D

Solution :

Net force (reaction) \[F={{F}_{B}}-{{F}_{A}}\] \[=\frac{d{{p}_{B}}}{dt}-\frac{d{{p}_{A}}}{dt}\] \[=a{{v}_{B}}\rho \times {{v}_{B}}-a{{v}_{A}}\rho \times {{v}_{A}}\] \[\therefore \]  \[F=a\rho (v_{B}^{2}-v_{A}^{2})\]                           ?(i) According to Bemaullis 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                 \[F=a\rho (2gh)\] \[=2a\rho gh\]