Question

A thin spherical shell has a mass of 2 kg when it is empty, radius 0.2 meters, and a volume of 0.0335 m3. It is filled with helium (density=0.179 kg/m3) and released from rest at the bottom of a pool of water that is 4 meters deep. (density of water= 1000 kg/m3).

1. Show that the shell rises with constant acceleration and determine the acceleration.

2. How long does it take the top of the shell to reach the water surface?

Answer 1

radius of spherical shell= r = 0.2 m mass of the helium in shell=mhe =VP He PHe = density of helium=0.179 kg mº volume of thin spherical shell =V=0.0335m? we plug the known values in the relation gives mHe=VPHe=0.0059965 kg total mass of the shell=M=mshell + m He=2.0060 kg total buoyance force on shell=Fy F =(PwVg- Phe Vg) P = water density = 1000 kg/m g=9.8." we plug the known values in the relation gives Fb = 328.24 N the force due to Gravity on the shell = FG = Mg and Fy > FG sin ce the net force on the shell is in upward direction of buoyance force, the net upward force on the shell =Fnet Fnet = Fb - Mg = 308.58 N 1) net upward acceleration= a = 2) if the depth of the lake = d=4.0 m the time it will take for the sheel to reach top surface=t 2d /40; weplug the known values in the relation gives ya => [t=0.22805 sec