Question

A copper wire has a diameter of 0.5mm and resistivity 1.6*10^-8ohm m.

1) what will be the length of this wire to make the resistance of 12 ohm?

2) how much will be the resistance of another copper wire of the same length but half the diameter?
Pls reply soon

Answer 1

Given : A copper wire has a diameter of 0.5 mm and resistivity 1.6 × 10^-8 Ωm.

To find : (i) what will be the length of this wire to make the resistance of 12 Ω.

(ii) how much will be the resistance of another copper wire of the same length but half the diameter.

solution : using formula, R = ρL/A

here, ρ = 1.6 × 10^-8 Ω m

d = 0.5 mm = 5 × 10¯⁴ m

r = d/2 = 2.5 × 10¯⁴ m

so, cross sectional area of wire, A = πr²

= 6.25π × 10^-8 m²

(i) R = 12 Ω

so, length of wire, L = RA/ρ

= 12 × 6.25π × 10^-8/(1.6 × 10^-8)

= 12 × 6.25π/1.6

= 147.1875 m

Therefore length of copper wire is 147.1875 m

(ii) resistance , R ∝ 1/A = 1/πd²/4 so, R ∝ 1/d²

if diameter of the copper wire is halved, resistance of wire will be four times.

i.e., resistance of new wire , R' = 4R

= 48 Ω

Therefore the resistance of another wire is 48 Ω