Fluid Mechanics Viscosity Question 287
Question: In the figure shown, a light container is kept on a horizontal rough surface of coefficient of friction $ \mu =\frac{Sh}{V} $ . A very small hole of area S is made at depth h. Water of volume V is filled in the container. The friction is not sufficient to keep the container at rest. The acceleration of the container initially is
Options:
A) $ \frac{V}{Sh}g $
B) $ g $
C) Zero
D) $ \frac{Sh}{V}g $
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Answer:
Correct Answer: D
Solution:
[d] Let the density of water be p, then the force by escaping liquid on container$ =\rho S{{\left( \sqrt{2gh} \right)}^{2}} $ $ \therefore $ Acceleration of container $ a=\frac{2\rho Sgh-\mu \rho Vg}{\rho V}=\left( \frac{2Sh}{V}-\mu\right)g $ Now, $ \mu =\frac{Sh}{V} $ $ \therefore ,,a=\frac{Sh}{V}g $
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