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A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.?
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Introduction:
In this problem, we are given a copper wire and a steel wire connected end to end. Both wires have the same diameter and are subjected to a load, resulting in elongation. We need to determine the load applied to the wires.

Given:
- Length of copper wire (Lc) = 2.2 m
- Length of steel wire (Ls) = 1.6 m
- Diameter of both wires (d) = 3.0 mm
- Net elongation (ΔL) = 0.70 mm

Formula:
The strain in a wire is given by the formula:
Strain (ε) = ΔL / L

The stress in a wire is given by the formula:
Stress (σ) = Young's Modulus (Y) * Strain (ε)

Young's Modulus for a material is given by:
Y = Stress / Strain

Calculations:
1. Calculate the cross-sectional area of the wires:
The cross-sectional area (A) of a wire is given by the formula:
A = π * (d/2)^2

For both copper and steel wires, the diameter (d) is 3.0 mm, so the radius (r) is 1.5 mm or 0.0015 m.

A = π * (0.0015 m)^2

2. Calculate the strain in the wires:
The strain (ε) is given by:
ε = ΔL / L

For the copper wire:
εcopper = 0.70 mm / 2.2 m

For the steel wire:
εsteel = 0.70 mm / 1.6 m

3. Calculate Young's Modulus for each wire:
Young's Modulus (Y) is given by:
Y = Stress / Strain

For the copper wire:
Ycopper = Stresscopper / εcopper

For the steel wire:
Ysteel = Stresssteel / εsteel

4. Calculate the stress in each wire:
The stress (σ) is given by:
σ = Y * ε

For the copper wire:
σcopper = Ycopper * εcopper

For the steel wire:
σsteel = Ysteel * εsteel

5. Calculate the load applied:
The load applied is equal to the stress multiplied by the cross-sectional area of the wires.

For the copper wire:
Loadcopper = σcopper * Acopper

For the steel wire:
Loadsteel = σsteel * Asteel

6. Calculate the total load applied:
The total load applied is the sum of the loads on the copper and steel wires.

Total Load = Loadcopper + Loadsteel

Conclusion:
By following the above calculations, we can determine the load applied on the copper and steel wires when they are stretched by a certain amount. The key parameters used in the calculations are the length, diameter, and net elongation of the wires, as well as the Young's Modulus of the materials. These calculations help in understanding the behavior of materials under load and can be applied to various engineering and structural analysis scenarios.
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A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.?
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A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.?.
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