Question

Two metal wires have the resistivity 1.62 10^-8 and 4.86 x 10^-8. Compare the resistance of the two wires if wire A is twice as long as wire B and half of its diameter.

Answer 1

Hello friend,

◆ Answer -
R1/R2 = 1/6

◆ Explaination-
# Given-
ρ1 = 1.62
×10^-8 ohm-m
ρ2 = 4.86
×10^-8 ohm-m
l1/l2 = 2
d1/d2 = 1/2

# Solution-
Resistance of wire A,
R1 = ρ1l1 / A1
Resistance of wire B,
R2 = ρ2l2 / A2

Ratio of resistances is -
R1/R2 = (ρ1l1/A1) / (ρ2l2/A2)
R1/R2 = (ρ1/ρ2) × (l1/l2) × (d2/d1)^2
R1/R2 = (1.62×10^-8 / 4.86×10^-8) × (2) × (1/2)^2
R1/R2 = (1/3) × (2) × (1/4)
R1/R2 = 1/6

Therefore, resistance of wire B is six times that of B.

Hope it helps..