Question

A child playing in a swimming pool realizes that it is easy to push a small inflated ball under the surface of the water whereas a large ball requires a lot of force. The child happens to have a styrofoam ball (the shape of the ball will not distort when it is forced under the surface), which he forces under the surface of the water. If the child needs to supply 5.00×102 N5.00×102 N to totally submerge the ball, calculate the diameter dd of the ball. The density of water is ρw=1.000×103 kg/m3ρw=1.000×103 kg/m3, the density of styrofoam is ρfoam=95.0 kg/m3ρfoam=95.0 kg/m3, and the acceleration due to gravity is g=9.81 m/s2g=9.81 m/s2.

Answer 1

Force required to totally submerge the ball, F = 5.00 x 10^2 N = 500 N

This 500 N (F, the buoyancy) is the weight of the displaced water minus the weight of the ball itself:

Suppose, volume of the ball = V

And, diameter of the ball = D

So,

F = (ρw - ρfoam) * V * g

V = 4/3 pi R^3 = 4/3 pi (D/2)^3 = 1/6 pi D^3

So

F = (ρw - ρfoam) * pi D^3 / 6 * g

D^3 = 6 F / ( pi*g*(ρw - ρfoam))

D = ( 6 F / ( pi*g*(ρw - ρfoam)) )^(1/3)

Now, plugging - in the values of the variables,

D = (6*500 / (3.141*9.81*(1000 - 95)))^(1/3)

= (3000 / 27886)^1/3

= 0.476 m (Answer)