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You hang a flood lamp from the end of a vertical steel wire. The flood lamp stretches the wire 0.18 mm and the stress is proportional to the strain. How much would it have stretched for a copper wire of the original length and diameter?
  • a)
    0.39 mm
  • b)
    0.37 mm
  • c)
    0.33 mm
  • d)
    0.18 mm
Correct answer is option 'C'. Can you explain this answer?
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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.
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Attempt All Questions sub parts from each question.Elasticity vs. plasticity: Objects get deformed when pushed, pulled, and twisted. Elasticity is the measure of the amount that the object can return to its original shape after these external forces and pressure are removed. The opposite of elasticity is plasticity. When something is stretched, and it stays stretched, the material is said to be plastic. Such deformation is said to be plastic deformation. In elastic deformation, atoms of the material are displaced temporarily from their original lattice site. They return back to their original position after the removal of external force. In plastic deformation, atoms of the solid are displaced permanently from their original lattice site. They don’t return back to the original position even after the removal of external load. So, elastic deformation is temporary, whereas plastic deformation is permanent. Amount of elastic deformation is very small. But the amount of plastic deformation is quite large. External force required for elastic deformation of solid is quite small. Force required for plastic deformation is much higher. Total energy absorbed by the material during elastic and plastic deformation region is called modulus of toughness. Energy absorbed by the material during elastic deformation is called module of resilience. Most materials have an amount of force or pressure for which they deform elastically. If more force or pressure is applied, then they undergo plastic deformation. Materials those have a fair amount of plastic deformation before breaking are said to be ductile. Materials those can't stretch or bend much without breaking are said to be brittle. Copper, aluminium etc. are ductile materials. For this reason those are used for making wires. Glass and ceramics are often brittle; they will not bend; they will break.Q. Aluminium is a ............... materials.

Attempt All Questions sub parts from each question.Elasticity vs. plasticity: Objects get deformed when pushed, pulled, and twisted. Elasticity is the measure of the amount that the object can return to its original shape after these external forces and pressure are removed. The opposite of elasticity is plasticity. When something is stretched, and it stays stretched, the material is said to be plastic. Such deformation is said to be plastic deformation. In elastic deformation, atoms of the material are displaced temporarily from their original lattice site. They return back to their original position after the removal of external force. In plastic deformation, atoms of the solid are displaced permanently from their original lattice site. They don’t return back to the original position even after the removal of external load. So, elastic deformation is temporary, whereas plastic deformation is permanent. Amount of elastic deformation is very small. But the amount of plastic deformation is quite large. External force required for elastic deformation of solid is quite small. Force required for plastic deformation is much higher. Total energy absorbed by the material during elastic and plastic deformation region is called modulus of toughness. Energy absorbed by the material during elastic deformation is called module of resilience. Most materials have an amount of force or pressure for which they deform elastically. If more force or pressure is applied, then they undergo plastic deformation. Materials those have a fair amount of plastic deformation before breaking are said to be ductile. Materials those can't stretch or bend much without breaking are said to be brittle. Copper, aluminium etc. are ductile materials. For this reason those are used for making wires. Glass and ceramics are often brittle; they will not bend; they will break.Q. Ceramic is a ............... material

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You hang a flood lamp from the end of a vertical steel wire. The flood lamp stretches the wire 0.18 mm and the stress is proportional to the strain. How much would it have stretched for a copper wire of the original length and diameter?a)0.39 mmb)0.37 mmc)0.33 mmd)0.18 mmCorrect answer is option 'C'. Can you explain this answer?
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You hang a flood lamp from the end of a vertical steel wire. The flood lamp stretches the wire 0.18 mm and the stress is proportional to the strain. How much would it have stretched for a copper wire of the original length and diameter?a)0.39 mmb)0.37 mmc)0.33 mmd)0.18 mmCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about You hang a flood lamp from the end of a vertical steel wire. The flood lamp stretches the wire 0.18 mm and the stress is proportional to the strain. How much would it have stretched for a copper wire of the original length and diameter?a)0.39 mmb)0.37 mmc)0.33 mmd)0.18 mmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for You hang a flood lamp from the end of a vertical steel wire. The flood lamp stretches the wire 0.18 mm and the stress is proportional to the strain. How much would it have stretched for a copper wire of the original length and diameter?a)0.39 mmb)0.37 mmc)0.33 mmd)0.18 mmCorrect answer is option 'C'. Can you explain this answer?.
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