JEE Main 2023 · 12 April · morning · Single correct · +4 / −1
A wire of resistance 160 Ω is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be
- (a) 640 Ω
- (b) 40 Ω
- (c) 16 Ω
- (d) 10 Ω
Answer and solution
Answer (d) 10 Ω
Solution
Physics sir · 3 steps
Melting keeps the volume fixed, so a wire made four times shorter also becomes four times thicker.
- Volume stays fixed. Melting and redrawing does not change the amount of metal, so the volume V = AL stays the same. If the length becomes L/4, the area must become 4A.
- Two effects together. In R = (ρ L)/A, the shorter length cuts R by 4. The larger area cuts it by another 4. The resistivity ρ does not change, because the material is the same.
R' = (ρ (L/4))/4A = 1/16 (ρ L)/A = R/16
- So. R' = 160/16 = 10 Ω, which is option (d).
The idea to keep. When a wire is stretched or compressed at constant volume, R ∝ L². Change the length by a factor n and the resistance changes by n².
Answer from the NTA final answer key. Solution by PrepWiser.