Question

The Specific gravity of a liquid column is 0.75 What height of that liquid is
needed to provide a pressure difference of 200 kPa? (1 marks)​

Answer 1

Answer:

Liquid column in a manometer rises to a height to balance the external pressure applied to it.

P

ext

=ρgh

Thus when height h of liquid decreases for the same external pressure, the density ρ of the liquid must have increased.

Answer 2

Answer:

The height, h, of the liquid calculated is $27.18m$.

Explanation:

Given,

The specific gravity of the liquid column = $0.75$

The pressure, P = $200kPa$ = $2\times 10^{5} Pa$

The height of the liquid, h =?

As we know,

• Pressure due to a column of liquid is given by:
• P = $\rho hg$      

Therefore,

• h = $\frac{P}{\rho g}$ -------equation (1)

Here,

• $\rho$ = density of liquid
• h = height of liquid
• g = acceleration due to gravity = $9.81m/s^{2}$

Now, we have to calculate the density of the liquid ( $\rho_{l}$ ) by using the specific gravity of the liquid.

As we know,

• The density of water, $\rho_{w}$ = $1000kg/m^{3}$

Now,

• Specific gravity of liquid = $\frac{\rho_{l}}{\rho_{w}}$
• $0.75 = \frac{\rho_{l}}{1000}$
• Density of liquid, $\rho_{l}$ = $750kg/m^{3}$

After putting the values in equation (1), we get:

• h = $\frac{2\times 10^{5} }{750\times 9.81}$
• h = $\frac{2\times 10^{5} }{7357.5}$
• h = $27.18m$

Hence, the height of the liquid, h = $27.18m$.