Question

Q.17)
A 6 m deep tank contains 4 m of water (
at bottom) and 2 m of oil of relative
density 0.90 above water. The pressure
at the bottom of the tank in kN / m^2 is -
------------(Assume Unit weight of water
as 10 kN/m^3) (1 marks )
Ans.
58
60
5.8
0.58​

Answer 1

Explanation:

60 is the correct answer

Answer 2

Answer:

The pressure at the bottom of the tank is 58kN/m².

Explanation:

The pressure at the bottom of the tank is calculated as,

$P=\rho_wgh_w+\rho_ogh_o$      (1)

Where,

P=pressure at the base of the tank

ρw=density of water

ρo=density of oil

hw=height up to which water is present in the tank

ho=height up to which oil is present  in the tank

g=acceleration due to gravity=10m/s²

From the question we have,

Total height of the tank(H)=6m

hw=4m

ho=2m

The relative density of oil is =0.9

The relative density(R.D) of oil is given as,

$R.D=\frac{\rho_o}{\rho_w}$

So,

$\rho_o=R.D\times \rho_w$

$\rho_o=0.9\times 1000=900kg/m^3$               (∵ρw=1000kg/m³)

By substituting the required values in equation (1) we get;

$P=1000\times 10\times 4+900\times 10\times 2$

$P=58000=58kN/m^2$

Hence, the pressure at the bottom of the tank is 58kN/m².