Question

An air filled balloon is at a depth of 1 km below the water level in an ocean . The normal stress of the balloon (in Pa) is (Given, rho_("water") = 10^(3) kgm^(-3), g = 9.8 ms^(-2) and P_(atm) = 10^(5) Pa)

Answer 1

The normal stress of the balloon is 99X$10^{5}$ Pa

• As pressure is increased below the surface given by the formula P = ρgh

• Hence putting the values(ρ = 10³kg/m³, g = 9.8m/s², P(atm) = $10^{5}$ Pa) we get 10³X9.8X10³ = 98X$10^{5}$ Pa

• As the atmospheric pressure is $10^{5}$ Pa
• Therefore the pressure below 1 km is 99X$10^{5}$ Pa ($10^{5}$ Pa+99X$10^{5}$ Pa)

Answer 2

• Given : rho(water) = 10^ 3 kg/m^3

g = 9.8 m/ s ^2 ,h = 1000m

Atmospheric pressure ,P( atm) = 10^5 Pa

• the net pressure at a depth of 1 km in the ocean is p = P ( atm) + rho× g × h

p = 10^ 5 + 10^3×9.8× 10^3

p= 9.9× 10^ 6 Pa

• this pressure is acting uniformly on all the sides of the ballon, which make sthe ballon to be in equilibrium. Therefore , the normal stress will become same as the external pressure.

• therefore the normal stress of the balloon is 9.9× 10^ 6 Pa