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

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

Step-by-step 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* 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=(P)/(\rho g)`

`h=(2970\ Pa)/(1.26* 10^3\ kg/m^3* 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.