A uniform metallic wire is elongated by 0.04m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be ______cm.
Question
A uniform metallic wire is elongated by 0.04m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be ______cm.
Solution
The elongation of a wire under a given force is given by the formula:
ΔL = (F/A) * (L/Y)
where:
- ΔL is the elongation,
- F is the force,
- A is the cross-sectional area of the wire,
- L is the original length of the wire,
- Y is the Young's modulus of the material of the wire.
In this case, the force F and the material of the wire (thus, the Young's modulus Y) remain constant. However, the length L is doubled and the diameter d (thus, the cross-sectional area A) is also doubled.
The cross-sectional area A of a wire with diameter d is given by the formula:
A = π * (d/2)^2
So, if the diameter is doubled, the new cross-sectional area A' will be:
A' = π * ((2d)/2)^2 = 4 * π * (d/2)^2 = 4A
So, the new elongation ΔL' under the same force F will be:
ΔL' = (F/A') * (2L/Y) = (F/(4A)) * (2L/Y) = (1/2) * (F/A) * (L/Y) = (1/2) * ΔL
Given that the original elongation ΔL is 0.04 m = 4 cm, the new elongation ΔL' will be:
ΔL' = (1/2) * 4 cm = 2 cm.
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