Question

Let ρAl represent the density of aluminum and ρFe that of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius rFe on an equal-arm balance. (Use r_F for r Fe, rho_A for ρAl, and rho_F for ρ as necessary.)

Answer 1

r = [3 * (pAl/MAl)]/(4 * pi)]^1/3

r = [3 * (pFe / MFe)]/(4 * pi)]^1/3

Step-by-step explanation:

In the equilibrium state, aluminum and iron have the same mass. From the density equation and solving for the mass we have:

Mass = density/volume

MFe = pFe/V

MAl = pAl/V

In equilibrium, we have that MFe = MAl

Solving for the volume:

MFe = pFe/V

V = pFe/MFe

MAl = pAl/V

V = pAl/MAl

The equation for the volume of a sphere is equal to:

V = (4 * pi * r^3)/3

Replacing the volume of both iron and aluminum, we have:

V = (4 * pi * r^3)/3

r = [(3 * V)/(4 * pi)]^1/3

r = [3 * (pAl/MAl)]/(4 * pi)]^1/3

r = [3 * (pFe/MFe)]/(4 * pi)]^1/3

Answer 2

Viron = 4/3 πr Fr^3

Valu = 4/3 πr Al^3

pFr * 4/3πrfe^3 = pAl*4/3πrAl^3

or

rAl =
`\sqrt[3]{pFe/pAl}`

Hope this helps