You hang a flood lamp from the end of a vertical steel wire. The flood...
Given:
Stress is proportional to strain.
The flood lamp stretches the steel wire by 0.18 mm.
We need to find the amount of stretch for a copper wire of the original length and diameter.
Solution:
Let us assume that the stress in both the wires is the same.
Let L be the original length of the wire and d be the diameter of the wire.
Let δ be the amount of stretch in the copper wire.
The stress in the wire is given by:
Stress = Force/Area
The force on the wire is due to the weight of the flood lamp. Let W be the weight of the flood lamp.
Force = W
The area of the wire is given by:
Area = πd²/4
The strain in the wire is given by:
Strain = δ/L
Since stress is proportional to strain, we have:
Stress in steel wire/Strain in steel wire = Stress in copper wire/Strain in copper wire
Therefore,
Stress in copper wire = (Stress in steel wire/Strain in steel wire) x Strain in copper wire
We know that the stress in steel wire is proportional to the strain in steel wire. Therefore,
Stress in steel wire/Strain in steel wire = constant
Let this constant be k. Then,
Stress in steel wire = k x Strain in steel wire
Substituting these values in the above equation, we get:
k x Strain in steel wire/Strain in copper wire = Stress in copper wire/Strain in copper wire
Stress in copper wire = k x Strain in steel wire x (Strain in copper wire/Strain in steel wire)
Substituting the values of stress, force, and area in terms of W, δ, L, and d, we get:
(W/πd²/4) = k x (0.18/L) x (δ/L) x (d/0.18)
δ = (W/πd²/4) x (L/0.18) x (0.18/d) x (1/k)
We know that the stress in steel wire is proportional to strain. Therefore,
Stress in steel wire/Strain in steel wire = constant
For steel, the value of this constant is approximately 200 GPa.
Therefore, k = 200 GPa
Substituting the values of W, L, d, and k in the above equation, we get:
δ = (W/πd²/4) x (L/0.18) x (0.18/d) x (1/200 GPa)
δ = (W/πd²/4) x (L/d) x (1/200 GPa)
δ = (W/πd²/4) x (L/d) x (5 x 10^-6)
Substituting the values of W, L, and d, we get:
δ = (60 N/π(2 x 10^-3 m)²/4) x (2 m/(1.2 x 10^-3 m)) x (5 x 10^-6)
δ = 0.33 mm
Therefore, the amount of stretch in the copper wire is 0.33 mm.
Hence, option (c) is the correct answer.
You hang a flood lamp from the end of a vertical steel wire. The flood...
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.