Question

A wire of length L and radius R has a resistance of 2 ohm,if the wire is stretched, so that its new diameter is half of the initial diameter,then the resistance of the stretched wire is...?​

Answer 1

Hello Dear,

◆ Answer -

Resistance of stretched wire = 32 ohm

◆ Explaination -

# Given -

R1 = 2 ohm

R' = R/2

# Solution -

Initially resistance of the wire will be -

R1 = ρ.L/A

R1 = ρ.L/πR^2

Volume of the wire will remain same after stretching.

L.A = L'.A'

L × 2πR^2 = L' × 2πR'^2

L' = L × (R/R')^2

L' = L × 2^2

L' = 4 L

New resistance of the wire will be -

R2 = ρ.L'/A'

R2 = ρ.L'/πR'^2

R2 = ρ × 4L / π(R/2)^2

R2 = 16 × (ρ.L/πR^2)

R2 = 16 R1

R2 = 16 × 2

R2 = 32 ohm

Resistance of stretched wire is 32 ohm.

Thanks dear...