Two wires, 1 and 2, are made of the identical material. Wire 1 has a length L and radius r , whereas wire 2 has a length 2L and radius 3r . They are subjected to longitudinal forces such that both wires experience the exact same elongation. The ratio of the applied force on wire 1 to that on wire 2 ( F₁ : F₂ ) is:

Step-by-step solution

The elongation L of a wire under a longitudinal force F is given by: L = FL AY = FL r^2 Y Rearranging for force F , we get: F = r^2 Y L L Since both wires are made of the identical material, they have the same Young's modulus Y . It is also given that they experience the same elongation L . For wire 1: F₁ = r^2 Y L L For wire 2: F₂ = (3r)^2 Y L 2L = 9 r^2 Y L 2L Taking the ratio of the forces: F₁ F₂ = r^2 Y L L 9 r^2 Y L 2L F₁ F₂ = 1 9 2 = 2 9 The ratio F₁ : F₂ is 2:9 . Answer: 2:9