Two cylindrical bars, each with diameter of 2.40 cm, are welded together end-to-end. One of the original bars is copper (resistivity 1.72e-8 ohm*m) and is 0.330 m long. The other bar is platinum (resistivity 10.60e-8 ohm*m) and is 0.125 m long. What is the resistance between the ends of the welded bar at 20 degrees Celsius?

The resistance of a constant-diameter length of conductive material is:

R = ρL/A

R is the resistance, ρ is the material's resistivity, L is the length, and A is the cross-sectional area.

We know that a cylindrical bar's cross sectional area A is given by:

A = πr²

where r is the radius.

The resistance is then given by:

R = ρL / (πr²)

Copper bar:

ρ = 1.72*10⁻⁸Ωm

r = 1.20*10⁻²m (half of its diameter 2.40cm)

L = 0.330m

R = (1.72*10⁻⁸) (0.330) / (π (1.20*10⁻²) ²)

R = 1.25*10⁻⁵Ω

Platinum bar:

ρ = 10.60*10⁻⁸Ωm

r = 1.20*10⁻²m (half of its diameter 2.40cm)

L = 0.125m

R = (10.60*10⁻⁸) (0.125) / (π (1.20*10⁻²) ²)

R = 2.93*10⁻⁵Ω

Add up the resistances to find the total resistance:

1.25*10⁻⁵Ω + 2.93*10⁻⁵Ω

= 4.18*10⁻⁵Ω

=