Calculate Depth of Floating Instrument: 0.219 kg in Water

[DDS]

A hollow brass tube (diameter = 3.23 cm) is sealed at one end and loaded with lead shot to give a total mass of 0.219 kg. When the tube is floated in pure water, what is the depth, z, of its bottom end?

through manipulation of some laws i have broken it down to this stage:

z = m/(πr2ρ)

where:

m=0.219 kg
r=0.0323/2=0.01615m
p=1000 kg/m^3

thus i have :

z= 0.219/ n(0.01615)2(1000)

but what the hell is n... i know its not the incidencs for water because nothing is getting reflected and i just tried to plug it into see what would happen and it doesn't work any help?

 

[Chi Meson]

That's not"n", it's "pi" (3.1415926535...)

It came from the formula for the area of a circle.

 

[DDS]

thanx so mucn, silly mistake on my part