A wire of resistance R₁ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is :

Solution

Physics sir · 3 steps

Increased by twice means the new length is 3L, and stretching thins the wire, so R grows as length squared.

1. New length. The length is increased by 2L, so the new length is L + 2L = 3L.
2. Volume stays fixed. Drawing out a wire does not change the amount of metal, so A × L stays constant. With length 3L, the new area must be A/3.

   R₂ = ρ 3L/(A/3) = 9 ρ L/A = 9R₁

3. So. The resistance becomes 9 times larger, so R₂ : R₁ = 9 : 1, option (a).

The idea to keep. When a wire is stretched, its volume stays the same, so the area falls as the length rises. That makes R ∝ L²: stretch it n times and R becomes n² times.

Answer from the NTA final answer key. Solution by PrepWiser.