Question

A manometer using oil (density 0.900 g/cm3) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases by 8.33 mmHg. Density of mercury is 13.6 g/cm3.A .By how much does the fluid level rise in the side of the manometer that is open to the atmosphere? (cm)B. By how much does the fluid level rise in the side of the manometer that is open to the atmosphere if the manometer used mercury instead? (cm)

Answer 1

(a) 6.3 cm

(b) 0.42 cm

Explanation:

Given:

Density of oil = 0.9 g/cm^3

Density of mercury = 13.6 g/cm^3

Rise in the height of the mercury = 8.33 mm of Hg = 0.833 cm of Hg

(a)

The change in pressure is:

`\Delta P=\rho\cdot g\cdot\Delta h`

The change in pressure when oil is used is:

`\Delta P=\rho_{\text{oil}}\cdot g\cdot d`

Equating the above two equation

`\begin{gathered} \rho\cdot g\cdot\Delta h=\rho_{\text{oil}}\cdot g\cdot d \\ \text{ solving for d:} \\ d=\frac{\rho\cdot\Delta h}{\rho_{\text{oil}}} \\ d=(13.6\cdot0.833)/(0.9) \\ d=12.59\text{ cm} \end{gathered}`

The amount of fluid rise is half the difference of level, so the height of rise in the oil

`\begin{gathered} \Delta h_(oil)=(d)/(2) \\ \text{ replacing} \\ \Delta h_(oil)=(12.59)/(2)=6.295\cong6.3 \\ \Delta h_(oil)=6.3\text{ cm} \end{gathered}`

(b) If the manometer uses mercury then

`\begin{gathered} \Delta h^(\prime)=\Delta h_(oil)\cdot\mleft((\rho_(oil))/(\rho)\mright) \\ \text{ replacing:} \\ \Delta h^(\prime)=6.3\cdot\mleft((0.9)/(13.6)\mright) \\ \Delta h^(\prime)=0.42\text{ cm} \end{gathered}`