Question

1. How deep can a submarine that is able to support a gauge pressure of 20atm dive in the ocean (r1g1=1.02)?

2.A balloon weighs 150 N and carries a gondola, gear, and load of another 1800 N. If the density of air is 1.2 kg/m^3. What should the radius of the spherical balloon be to carry the load.

Note: The air inside the balloon has a density of 1.0 kg/m^3.

Answer 1

1)

Given,

density of the water, = 1.02 g/cm³ = 1020 kg/m³

Now,

Gauge pressure, P = 20 atm = 20 * 101.325 * 10³ Pa

                              = 2026.5 * 10³ Pa

Now,

Let the depth be h

=> P = *h*g

=> 2026.5 * 10³ = 1020*h*9.8

=> h = 2026.5 * 10³/(1020*9.8)

         = 202.731 m

Hence, submarine can go upto the depth of 202.731 m

2)

Given,

weight of the balloon, W = 150 N

Other weights, W₁ = 1800

Total load, L = 1800+150 = 1950 N

Density of air inside the balloon, = 1.0 kg/m³

Density of the air outside, = 1.2 kg/m³

Let the volume of balloon be V

Now,

To lift the balloon, buoyant force must be equal to the total load + weight of air inside the balloon

=> *V*g + L = *V*g

=> 1*V*9.8 + 1950 = 1.2*V*9.8

=> (1.2*V*9.8) - (1*V*9.8) = 1950

=> 0.2*V*9.8 = 1950

=> V = 1950/(0.2*9.8)

          = 994.897

Now,

Let radius be R

=> V = 4/3**R³

=> 994.898 = 1.33*3.14*R³

=> R³ = 994.898/(1.33*3.14)

=> R³ = 237.635

Or R = = 6.194 m

or radius of the balloon is 6.194 m