Fluid Mechanics Viscosity Question 294
Water is flowing through a horizontal tube having cross-sectional areas of its two ends being A and A’ such that the ratio A/A’ is 5. If the pressure difference of water between the two ends is $ 3\times 10^{5}N{{m}^{-2}} $, the velocity of water with which it enters the tube will be (neglect gravity effects)
Options:
A) $ 5m{{s}^{-1}} $
B) $ 10m{{s}^{-1}} $
C) $ 25,m{{s}^{-1}} $
D) $ 50\sqrt{10},m{{s}^{-1}} $
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Answer:
Correct Answer: A
Solution:
According to Bernoulli’s theorem $ P_{1}+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2} $ …(i) From question, $ p_{1}-p_{2}=3\times 10^{5},\frac{A_{1}}{A_{2}}=5 $ According to equation of continuity $ A_{1}v_{1}=A_{2}v_{2},,or,,\frac{A_{1}}{A_{2}}=\frac{v_{2}}{v_{1}}=5\Rightarrow v_{2}=5v_{1} $ From equation (i) $ P_{1}-P_{2}=\frac{1}{2}\rho \left( v_{2}^{2}-v_{1}^{2} \right) $ or $ 3\times 10^{5}=\frac{1}{2}\times 1000\left( 25v_{1}^{2}-v_{1}^{2} \right) $ $ \Rightarrow ,,,3\times 10^{5}=2000v_{1}^{2}\Rightarrow v_{1}^{2}=150 $ $ \therefore ,,,,v_{1}=12.25m/s $
BETA
