Question

two metal wires have the resistivity of 1.62 × 10^-18 ωm and 4.86 × 10^-18 ωm. compare the resistance of 2 wires if a is twice as long as b and half of its diameter

Answer 1

Hello friend,

◆ Answer -

R1/R2 = 1/6

◆ Explaination-

# Given-

ρ1 = 1.62

×10^-18 ohm-m

ρ2 = 4.86

×10^-18 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^-18 / 4.86×10^-18) × (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 A.

Hope it helps...