Connections are led from A & B to a U-tube containing mercury. If the difference of pressure between A &B is 0.18 $kg/cm^{2}$. Find the reading shown by the differential manometer.

Given:

For the given configuration,

"specific gravity of the oil=$\frac{ρ _{oil}}{ρ _{water}} =0.8⟹{ρ}_{oil}=800 (since \ ρ_{water}≈1000 kg/m^{3})$

H=30 cm (potential difference between point A and B)

$∆P=P _A -P_B=0.18 kg/cm^2$(Static pressure difference between the two point A & B)

∴$∆P=0.18∙g 10^4 N/m^2$

To find: h (Reading shown by the mercury U-tube manometer)"

Sol: Consider the diagram of the mercury containing U-tube manometer connected to the vertical pipe,

This is a problem on fluid statics, since the tube which tap to the vertical pipe is of constant cross sectional area, we can equate the pressures rather than the forces.

Pressure at point 1= $P_1=P_B+ρ _{oil}∙g∙ X_B$

"Pressure at point 2= $P_2=P_1 +ρ _Hg∙g∙ h=P_B+ρ_oil∙g∙X_B+ρ _Hg∙g∙h$

Where,$ρ _Hg=13600kg/m ^3(Density \ of \ Mercury(Hg))$

"Pressure at point 4= $P_4=P _A +ρ_oil∙g∙ X_A$

"Pressure at point 3= $P _3 =P _4 +\rho _oil ∙g∙ h=P_A +ρ _oil∙g∙ X_A+ρ _oil ∙g∙ h$

Now, we know that at equal levels within a continuous liquid column (of the same fluid) the pressure is a constant for a static fluid.

"⟹ $P_2=P _3$

Equating the two equations,

$P _B+ρ _oil ∙g∙ X_B+ρ_Hg∙g∙ h=P _A +ρ _oil∙g∙ X_A+ρ _oil ∙g∙ h$

∴$P_A-P_B=ρ_oil∙g∙(X_B-X_A)+(ρ_Hg -ρ_oil)∙g∙h$

Now, It can be seen from the diagram,

$X_B-X_A=H$

Substituting in the above equation,

∴0.18∙g∙$10^4$=800∙g∙H+(13600 -800)∙g∙h

∴(0.18)$10^4=800∙(0.3)+(12800)∙h$

∴12800∙h=1800-240

∴h=0.121875 m=12.1875 cm

…Which is the reading shown by the Mercury U-Tube differential manometer.