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 7.28 mmHg. Density of mercury is 13.6 g/cm3. By how much does the fluid level rise in the side of the manometer that is open to the atmosphere?What would your answer be if the manometer used mercury instead?

Answer 1

Answer:

Rise in level of fluid is 0.11 m

Rise in level of fluid in case of mercury is 0.728 cm or 7.28 mm

Solution:

As per the question:

Density of oil, $\rho_{o} = 0.900\ g/cm^{3} = 900\ kg/m^{3}$

Change in Pressure in the tank, $\Delta P = 7.28\ mmHg$

Density of the mercury, $\rho_{m} = 13.6\ g/cm^{3} = 13600\ kg/m^{3}$

Now,

To calculate the rise in the level of fluid inside the manometer:

We know that:

1 mmHg = 133.332 Pa

Thus

$\Delta P = 7.28\ times 133.332 = 970.656\ Pa$

Also,

$\Delta P = \rho_{o} gh$

where

g = acceleration due to gravity

h = height of the fluid level

$970.656 = 900\times 9.8\times h$

h = 0.11 m

Now, if mercury is used:

$\Delta P = \rho_{m} gh$

$970.656 = 13600\times 9.8\times h$

h = 0.00728 m = 7.28 mm