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A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire? Given that the Young's modulus for steel is Young's modulus of the material is 2.00×1011 N/m2)
- a)1.4 mm
- b)10 mm
- c)1.5 mm
- d)12.4 mm
Correct answer is option 'A'. Can you explain this answer?
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Understanding the Problem
To determine the minimum diameter required for a circular steel wire, we must consider the relationship between tensile stress, strain, and Young's modulus.
Key Given Data:
- Length of wire (L) = 2.00 m
- Maximum stretch (ΔL) = 0.25 cm = 0.0025 m
- Tensile force (F) = 400 N
- Young's modulus for steel (E) = 2.00 × 10^11 N/m²
Formulas to Use:
1. Strain (ε): ε = ΔL / L
2. Stress (σ): σ = F / A
3. Young's Modulus (E): E = σ / ε
Calculating Strain:
- Strain (ε) = ΔL / L
- ε = 0.0025 m / 2.00 m = 0.00125
Calculating Stress:
Using Young's modulus, we can rearrange the formula to find stress:
- σ = E × ε
- σ = (2.00 × 10^11 N/m²) × (0.00125) = 2.5 × 10^8 N/m²
Finding the Required Area:
From the stress formula, we can find the cross-sectional area (A):
- A = F / σ
- A = 400 N / (2.5 × 10^8 N/m²) = 1.6 × 10^-6 m²
Calculating Diameter:
The area of a circle is given by A = (π/4) × d². Rearranging gives:
- d² = (4A) / π
- d = √((4 × 1.6 × 10^-6 m²) / π)
- d ≈ 0.0014 m = 1.4 mm
Conclusion:
The minimum diameter required for the wire is 1.4 mm, confirming option 'A' as the correct answer.
To determine the minimum diameter required for a circular steel wire, we must consider the relationship between tensile stress, strain, and Young's modulus.
Key Given Data:
- Length of wire (L) = 2.00 m
- Maximum stretch (ΔL) = 0.25 cm = 0.0025 m
- Tensile force (F) = 400 N
- Young's modulus for steel (E) = 2.00 × 10^11 N/m²
Formulas to Use:
1. Strain (ε): ε = ΔL / L
2. Stress (σ): σ = F / A
3. Young's Modulus (E): E = σ / ε
Calculating Strain:
- Strain (ε) = ΔL / L
- ε = 0.0025 m / 2.00 m = 0.00125
Calculating Stress:
Using Young's modulus, we can rearrange the formula to find stress:
- σ = E × ε
- σ = (2.00 × 10^11 N/m²) × (0.00125) = 2.5 × 10^8 N/m²
Finding the Required Area:
From the stress formula, we can find the cross-sectional area (A):
- A = F / σ
- A = 400 N / (2.5 × 10^8 N/m²) = 1.6 × 10^-6 m²
Calculating Diameter:
The area of a circle is given by A = (π/4) × d². Rearranging gives:
- d² = (4A) / π
- d = √((4 × 1.6 × 10^-6 m²) / π)
- d ≈ 0.0014 m = 1.4 mm
Conclusion:
The minimum diameter required for the wire is 1.4 mm, confirming option 'A' as the correct answer.
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Question Description
A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?Given that the Youngs modulus for steel isYoungs modulus of the material is 2.00×1011N/m2)a)1.4 mmb)10 mmc)1.5 mmd)12.4 mmCorrect answer is option 'A'. Can you explain this answer? for NEET 2026 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?Given that the Youngs modulus for steel isYoungs modulus of the material is 2.00×1011N/m2)a)1.4 mmb)10 mmc)1.5 mmd)12.4 mmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?Given that the Youngs modulus for steel isYoungs modulus of the material is 2.00×1011N/m2)a)1.4 mmb)10 mmc)1.5 mmd)12.4 mmCorrect answer is option 'A'. Can you explain this answer?.
A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?Given that the Youngs modulus for steel isYoungs modulus of the material is 2.00×1011N/m2)a)1.4 mmb)10 mmc)1.5 mmd)12.4 mmCorrect answer is option 'A'. Can you explain this answer? for NEET 2026 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?Given that the Youngs modulus for steel isYoungs modulus of the material is 2.00×1011N/m2)a)1.4 mmb)10 mmc)1.5 mmd)12.4 mmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?Given that the Youngs modulus for steel isYoungs modulus of the material is 2.00×1011N/m2)a)1.4 mmb)10 mmc)1.5 mmd)12.4 mmCorrect answer is option 'A'. Can you explain this answer?.
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