Let us consider a copper wire having radius r and length l. Let its resistance be R. If the radius of another copper wire is 2r and the length is l/2 then the resistance of this wire will be
[amp_mcq option1=”R” option2=”2R” option3=”R/4″ option4=”R/8″ correct=”option4″]
This question was previously asked in
UPSC NDA-1 – 2019
For the second wire, the radius is r’ = 2r and the length is l’ = l/2.
The new cross-sectional area A’ = π(r’)² = π(2r)² = π(4r²) = 4πr².
The new resistance R’ = ρ * (l’ / A’) = ρ * ((l/2) / (4πr²)).
R’ = ρ * (l / (8πr²)) = (1/8) * [ρ * (l / (πr²))].
Since R = ρ * (l / (πr²)), we have R’ = R/8.