Question

A closed spherical shell that has a radius of 0.17m is lowered into a liquid with a density of 1260 kg / (m^3). The air pressure inside the shell is 1 atmosphere . When the shell is lowered 12.3m below the surface , what is the net force on the shell in Newtons ? The surface area of a spherical shell is 4Pi * R^2 the volume of a spherical shell is 4Pi * R^3/3

Answer 1

The pressure on the shell due to the liquid is

pgh

where rho = 1260 kg/m^3 is the density of the liquid

g = 9.8 m/s^2

h = 12.3 m

The pressure on the shell due to the atmosphere is 1 atm. Hence the net pressure on the shell is

pgh

The pressure inside the shell is 1 atm. This cancels the atmospheric pressure. therefore the total force on the shell is pressure times surface area of the shell

$F=\rho gh(4\pi R^2)$

$F=1260\times 9.8\times 12.3(4\pi \times 0.17^2)=55158.1\,\,N$

The answer to 4 significant digits is

$F=5.516\times 10^4 \,N$