Question

Two pipelines, one carrying oil (mass density 900 kg/m${}^3$) and the other water, are connected to a manometer as shown in figure. By what amount the pressure in the water pipe should be increased so that the mercury levels in both the limbs of the manometer become equal? (Mass density of mercury =13550 kg/m${}^3$ and g = 9.81 m/s${}^2$)

• A. 24.7kPa
• B. 26.5 kPa
• C. 26.7 kPa
• D. 28.9 kPa

Answer 1

The correct option is A 24.7kPa

Initially, the pressure difference between water and oil can be given as

$P_{wi}+(1000\times 9.81\times 1.5)+(13550\times 9.81\times 20\times {10}^{-2})-(900\times 9.81\times 3)=P_0$

$\Rightarrow P_{wi}+14715+26585.1-26487=P_0$

$\Rightarrow P_0-P_{wi}=14813.1$

$\Rightarrow P_0-P_{wi}=14.8kPa...\left(i\right)$

To make the same level of mercury in both the limbs, the level of mercury should be decreased by 10 cm in right limb and increased by 10 cm in left limb

$P_{wi}+(1000\times 9.81\times 1.6)-(900\times 9.81\times 2.9)=P_0$

$\Rightarrow P_0-P_{wf}=-9.9kPa...\left(ii\right)$

Subtracting (ii)from (i) we get

$p_{wi}-p_{wf}=14.8+9.9$

$p_{wi}-p_{wf}=24.7kPa\left(increase\right)$