Question

A tall, open container is full of glycerine. At what depth h below the surface of the glycerine is the pressure 2970 Pa greater than atmospheric pressure? The density of glycerine is 1.26 × 10 3 kg/m 3 .

Answer 1

Answer:

At the depth of 24 cm below the surface of the glycerine the pressure is  2970 Pa.

Explanation:

Given that,

Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.

The density of glycerine, $\rho=1.26\times 10^3\ kg/m^3$

We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :

$P=\rho gh$

$h=\dfrac{P}{\rho g}$

$h=\dfrac{2970\ Pa}{1.26\times 10^3\ kg/m^3\times 9.8\ m/s^2}$

h = 0.24 meters

or

h = 24 cm

So, at the depth of 24 cm below the surface of the glycerine the pressure is  2970 Pa. Hence, this is the required solution.