JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Current Electricity

A wire of

1Ω

has a length of 1 m. It is stretched till its length increases by

25%

. The percentage change in resistance to the nearest integer is:

Answer: (A)

Solution

Resistance

R = ρ L / A

. When the wire is stretched, volume remains constant:

A L = A^{'} L^{'}

. New length

L^{'} = 1.25 L

, so

A^{'} = A /1.25

. New resistance

R^{'} = ρ L^{'} / A^{'} = ρ (1.25 L) / (A /1.25) = ρ L \times 1.5625/ A = 1.5625 R

. Percentage change

= (R^{'} - R) / R \times 100 = 56.25% \approx 56%

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option