## hollow iron spherical shell submerged in water, find the inner diameter

1. The problem statement, all variables and given/known data
A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 66.0 cm, and the density of iron is 7.87 g/cm3. Find the inner diameter.

2. Relevant equations
Fb=MG
4/3pir^3=Vsphere

3. The attempt at a solution

Not sure where to begin, im guessing volume of sphere submerged=V displaced water, but if a certain amount of space in the sphere is massless, how do you calculate the inner diameter? I dont see the connection I guess
 Recognitions: Homework Help First find the mass of the sphere (use buoyancy). If the sphere was solid iron what would its mass be?
 mass of the sphere=dv d=7.87 g/cm^3 v= 4/3pir^3 mass=1184691.197g if solid sphere Fb=(1000kg/m^3)*(150532cm^3)(9.8m/s)

Recognitions:
Homework Help

## hollow iron spherical shell submerged in water, find the inner diameter

What is the actual mass of the sphere as determined via the buoyancy?

What spherical volume of iron do you have to subtract from the larger solid sphere in order to achieve the actual mass of the spherical shell?
 Fb=1475.2136N/9.8=actual mass of sphere mass of sphere=150.532 mass whole iron sphere-mass of actual sphere=mass of inner sphere=1184540.665 7.87=1184540.665/V rinner=32.998 cm but thats incorrect, where is my error?
 Recognitions: Homework Help Remember that they want the diameter of the inner hollow, not the radius.
 wow lol, thank you for all the help you are a life saver.
 it says that my answer is incorrect, I multiplied 32.998 by 2 and got 65.996, but its not correct

Recognitions:
Homework Help
 Quote by tigers4 it says that my answer is incorrect, I multiplied 32.998 by 2 and got 65.996, but its not correct
There's probably some difficulty with rounding and significant figures through the calculations. This can happen when results depend upon small differences between large numbers, and constants with too few significant figures, like taking g = 9.8 (two figures), or pi = 3.14 through the calculations.

In this case it might be best to carry out all the operations symbolically right up to the end. A lot of the constants and calculations will cancel out.

Suppose we let v1 be the volume of iron comprising the shell, vs be the overall volume of the spherical shell, and vh be the volume of the hollow. Then

$$v_s = \frac{4}{3} \pi (r_s)^3$$

$$v_h = \frac{4}{3} \pi (r_h)^3$$

$$v1 = v_s - v_h$$

Now, v1 is also determined by the density of iron and the mass of the iron shell as determined by the volume of displaced water.

$$m_s = v_s \rho_w$$

$$v1 = m_s/\rho_{Fe}$$ Volume of iron comprising the shell

$$v1 = v_s \frac{\rho_w}{\rho_{Fe}}$$

So now, putting the parts together,

$$v_s \frac{\rho_w}{\rho_{Fe}} = v_s - v_h$$

Divide through by vs to yield

$$\frac{\rho_w}{\rho_{Fe}} = 1 - v_h/v_s$$

Note that we still haven't used g, and it looks like the constants in the volume calculations for vs and vh are in a position to cancel out. Can you carry on from here to solve for the radius (and then diameter)?
 yes everything looks good, the diameter is 63.1 cm

 Tags buoyancy, density