To solve this problem, we need to find the diameter of the copper wire that will cause the same elongation as the aluminium wire under the same weight.
Formula for Elongation:
The elongation (ΔL) of a wire under a load is given by the formula:
ΔL = (F * L) / (A * Y)
Where:
F is the force (weight) applied on the wire,
L is the original length of the wire,
A is the cross-sectional area of the wire,
Y is the Young's modulus of the material.
We are given:
Length of both wires, L = 5 m,
Force on the wire, F = 40 kg * 9.8 m/s² = 392 N,
Young's modulus for aluminium, Y_al = 7 * 10¹⁰ N/m²,
Young's modulus for copper, Y_cu = 12 * 10¹⁰ N/m²,
Diameter of the aluminium wire, d_al = 3 mm = 0.003 m.
Let the diameter of the copper wire be d_cu, which we need to find.
Step 1: Elongation for Aluminium Wire
For the aluminium wire, the cross-sectional area A_al is:
A_al = π * (d_al / 2)² = π * (0.003 / 2)² ≈ 7.07 * 10⁻⁶ m².
Now, using the formula for elongation:
ΔL_al = (F * L) / (A_al * Y_al)
ΔL_al = (392 * 5) / (7.07 * 10⁻⁶ * 7 * 10¹⁰)
ΔL_al ≈ 0.000394 m.
Step 2: Elongation for Copper Wire
For the copper wire, we want the same elongation (ΔL_cu = ΔL_al).
The cross-sectional area of the copper wire A_cu is:
A_cu = π * (d_cu / 2)².
Using the elongation formula for the copper wire:
ΔL_cu = (F * L) / (A_cu * Y_cu).
Substituting the values for ΔL_cu = ΔL_al, we get:
(392 * 5) / (A_cu * 12 * 10¹⁰) = 0.000394.
Now, solving for A_cu:
A_cu = (392 * 5) / (0.000394 * 12 * 10¹⁰) ≈ 6.61 * 10⁻⁶ m².
Step 3: Solving for Diameter of Copper Wire
A_cu = π * (d_cu / 2)².
Solving for d_cu:
d_cu = 2 * √(A_cu / π) = 2 * √(6.61 * 10⁻⁶ / π) ≈ 0.0025 m = 2.5 mm.
Final Answer:
The diameter of the copper wire should be 2.5 mm.
So, the correct answer is C) 2.5.
