Question

Q.3)
A6 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.
5.8
0.58
58
60​

Answer 1

Explanation:

volume of water =1×100 cc

mass of water =100×1 gm

volume of oil =1×50 cc

mass of oil = 0.8×50 =40gm

total mass = 100+40=140 gm =0.14 kg

so pressure intencity at the bottom per sqcm = 0.14×9.81 kgm/sec^2 =1.3734 Newton

so pressure intencity=1.3734N/sqcm

Answer 2

Answer:

The pressure at the bottom is $0.58kN/m^{2}$.

Explanation:

Given:

$h_{w} =4m$

$h_{o} =2m$

$\rho_{w}=10kN/m^{3}$

Here,

The relative density of the oil is denoted by $\rho$.

The height of the water is denoted by $h_{w}$.

The height of the oil is denoted by $h_{o}$.

The density of the oil is denoted by $\rho_{o}$.

The density of the water is denoted by $\rho_{w}$.

The pressure at the bottom is denoted by P.

The pressure at the bottom by the oil is denoted by $P_{o}$.

The pressure at the bottom by the water is denoted by $P_{w}$.

Now,

By the equation,

$\rho=\frac{\rho_{o} }{\rho_{w} } \\0.9=\frac{\rho_{o}}{10} \\\rho_{o}=9kN/m^{3}$

Then,

By the equation,

$P=P_{w} +P_{o} \\P = \rho_{w} gh_{w} +\rho_{o} gh_{o} \\P=10\times10\times4+9\times10\times2\\P=580N/m^{2} \\P=0.58kN/m^{2}$

So, the pressure at the bottom is $0.58kN/m^{2}$.