Fluid Mechanics Viscosity Question 81
Question: 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
Options:
A) $ gh\rho a $
B) $ \frac{2gh}{\rho ,a} $
C) $ 2\rho agh $
D) $ \frac{\rho gh}{a} $
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Answer:
Correct Answer: C
Solution:
Net force (reaction) = $ F=F_{B}-F_{A} $ $ =\frac{dp_{B}}{dt}-\frac{dp_{A}}{dt} $ $ =av_{B}\rho \times v_{B}-av_{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. $
BETA
