JEE Main & Advanced
Physics
Fluid Mechanics, Surface Tension & Viscosity / द्रव यांत्रिकी, भूतल तनाव और चिपचिपापन
Question Bank
Critical Thinking
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), \[F=2a\rho gh.\]