Fluid Mechanics Viscosity Question 262
Question: What is the absolute pressure of the gas above the liquid surface in the tank shown in fig. Density of $ oil=820kg/m^{3} $ , density of mercury$ =13.6\times 10^{3}kg/m^{3} $ Given 1 atmospheric pressure$ =1.01\times 10^{5}N/m^{2} $
Options:
A) $ 3.81\times 10^{5}N/m^{2} $
B) $ 6\times 10^{6}N/m^{2} $
C) $ 5\times 10^{7}N/m^{2} $
D) $ 4.6\times 10^{2}N/m^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a]Suppose $ P_{gas} $ is the pressure of the gas on the oil. As the points A and B are at the same level in the mercury columns, so $ P_{A}=P_{B} $ or$ P_{gas}+{\rho_{oil}}gh_{oil}=P_{a}+{\rho_{Hg}}g{h_{^{Hg}}} $ or $ P_{gas}+820\times 9.8\times (1+1.50) $ $ =P_{a}+13.6\times 10^{3}\times 9.8\times (1.5+0.75) $ or$ P_{gas}+20.09\times 10^{3}=P_{a}+299.88\times 10^{3} $ $ \therefore ,P_{gas}-Pa=299.88\times 10^{3}-20.09\times 10^{3} $ or $ {{[P_{gas}]}{gauge}}=279.8\times 10^{3}N/m^{2}=2.8\times 10^{5}N/m^{2} $ Absolute pressure of gas $ {{[P{gas}]}{gauge}}={{[P{gas}]}{gauge}}+P{a} $ $ =2.8\times 10^{5}+1.01\times 10^{5}=3.81\times 10^{5}N/m^{2} $
BETA
