Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl on applying a force F, how much force is needed to stretch the second wire by the same amount?

Let L be the original length and V be the volume of each wire. Since the wires are made of the same material and have the same volume, we have:
V₁ = V₂
A₁L₁ = A₂L₂
A₁ = A
A₂ = 3A
Therefore, L₂ = L₁/3 = L/3
From Hooke's law, the force F is proportional to the extension Δl and inversely proportional to the length L and directly proportional to the area A:
F = (Y * A * Δl) / L
where Y is Young's modulus.
For the first wire:
F = (Y * A * Δl) / L
For the second wire:
F₁ = (Y * 3A * Δl) / (L/3) = 9 * (Y * A * Δl) / L = 9F
Therefore, the force needed to stretch the second wire by the same amount is 9F.
Option (2).