Fluid Mechanics Viscosity Question 295
Question: There are two identical small holes P and Q of area of cross-section a on the opposite sides of a tank containing a liquid of density p. 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:
[c] Force $ F=\rho Qv=\rho av^{2}=\rho a\times 2gh $ $ =F_{2}-F_{1} $ $ =\rho av_{2}^{2}-\rho av_{1}^{2} $ $ =\rho a(v_{2}^{2}-v_{1}^{2}) $ $ =\rho a[2g(y+h)-2gy] $ $ =\rho a\times 2gh $
BETA
