Fluid Mechanics Viscosity Question 263
Question: Figure shows a U-tube of uniform cross-sectional area A, accelerated with acceleration a as shown. If d is the separation between the limbs, then what is the difference in the levels of the liquid in the U-tube is
Options:
A) $ \frac{ad}{g} $
B) $ \frac{ag}{d} $
C) $ \frac{a}{d} $
D) $ \frac{dg}{a} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Mass of liquid in horizontal portion of U-tube$ =Ad\rho $ Pseudo force on this mass$ =Ad\rho a $ Force due to pressure difference in the two limbs $ =(h_{1}\rho g-h_{2}\rho g)A $ Equating both the forces $ (h_{1}-h_{2})\rho gA=Ad\rho a $ $ \Rightarrow ,,(h_{1}-h_{2})=\frac{Ad\rho a}{\rho gA}=\frac{ad}{g} $
BETA
