The required pressure of 2 atmospheric pressure 2.02 x 105 N/m².
One atmosphere is also equal to the pressure created by the weight of a freshwater column that is approximately 10.3 metres tall (101.325 kPa or 14.7 psi) (33.8 ft). As a result, at a depth of 10.3 metres, a diver encounters pressure of around 2 atmospheres (1 atm of air plus 1 atm of water).
p=density*g*h
2*10^5=1000*10*h
h=20 meters
depth to feel the pressure of 2atm = ρgh
now solving for h
2*1.01 * 105 = 1000 * 10 * h
h = 20.2 m
here the answer is required pressure of 2 atm which is only due to the water on its top so
ρgh = 2atm
2 * 1.01 x 105 N/m2. [1atm = 1.01 x 105 N/m2. ]
= 2.02 x 105 N/m2.
s0 ρgh= 2.02 x 105 N/m2.
(1x10 3 kg/m³) (9.8 m/s²)h=2.02 x 105 N/m².
The required pressure of 2 atmospheric pressure 2.02 x 105 N/m².
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