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A circular steel wire 2.00 m long must stretch no more than 0.25 m when a tensile force of 400N is applied to each of the wire what minimum diameter is required of the wire?
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A circular steel wire 2.00 m long must stretch no more than 0.25 m whe...
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A circular steel wire 2.00 m long must stretch no more than 0.25 m whe...
Given data:
- Length of the wire (L) = 2.00 m
- Maximum stretch (ΔL) = 0.25 m
- Tensile force (F) = 400 N
Formula:
The formula to calculate the stretch in a wire under tensile force is:
\[ \Delta L = \frac{FL}{A \cdot Y} \]
where,
ΔL = stretch in the wire,
F = tensile force,
L = original length of the wire,
A = cross-sectional area of the wire, and
Y = Young's modulus of the material
Finding the minimum diameter:
- Since the wire is circular, the cross-sectional area can be calculated as:
\[ A = \frac{\pi d^2}{4} \]
where,
d = diameter of the wire
- Substituting the values in the formula, we get:
\[ 0.25 = \frac{400 \cdot 2}{\frac{\pi d^2}{4} \cdot Y} \]
\[ 0.25 = \frac{800}{\pi d^2 \cdot Y} \]
\[ d^2 = \frac{800}{0.25 \cdot \pi \cdot Y} \]
\[ d^2 = \frac{800}{0.25 \cdot 3.14 \cdot Y} \]
\[ d^2 = \frac{800}{0.785 \cdot Y} \]
\[ d^2 = \frac{800}{Y} \]
\[ d = \sqrt{\frac{800}{Y}} \]
Conclusion:
To ensure that the wire does not stretch more than 0.25 m when a tensile force of 400 N is applied, the minimum diameter required for the wire can be calculated using the formula above.
- Length of the wire (L) = 2.00 m
- Maximum stretch (ΔL) = 0.25 m
- Tensile force (F) = 400 N
Formula:
The formula to calculate the stretch in a wire under tensile force is:
\[ \Delta L = \frac{FL}{A \cdot Y} \]
where,
ΔL = stretch in the wire,
F = tensile force,
L = original length of the wire,
A = cross-sectional area of the wire, and
Y = Young's modulus of the material
Finding the minimum diameter:
- Since the wire is circular, the cross-sectional area can be calculated as:
\[ A = \frac{\pi d^2}{4} \]
where,
d = diameter of the wire
- Substituting the values in the formula, we get:
\[ 0.25 = \frac{400 \cdot 2}{\frac{\pi d^2}{4} \cdot Y} \]
\[ 0.25 = \frac{800}{\pi d^2 \cdot Y} \]
\[ d^2 = \frac{800}{0.25 \cdot \pi \cdot Y} \]
\[ d^2 = \frac{800}{0.25 \cdot 3.14 \cdot Y} \]
\[ d^2 = \frac{800}{0.785 \cdot Y} \]
\[ d^2 = \frac{800}{Y} \]
\[ d = \sqrt{\frac{800}{Y}} \]
Conclusion:
To ensure that the wire does not stretch more than 0.25 m when a tensile force of 400 N is applied, the minimum diameter required for the wire can be calculated using the formula above.
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A circular steel wire 2.00 m long must stretch no more than 0.25 m when a tensile force of 400N is applied to each of the wire what minimum diameter is required of the wire?
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A circular steel wire 2.00 m long must stretch no more than 0.25 m when a tensile force of 400N is applied to each of the wire what minimum diameter is required of the wire? for NEET 2024 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 m when a tensile force of 400N is applied to each of the wire what minimum diameter is required of the wire? covers all topics & solutions for NEET 2024 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 m when a tensile force of 400N is applied to each of the wire what minimum diameter is required of the wire?.
A circular steel wire 2.00 m long must stretch no more than 0.25 m when a tensile force of 400N is applied to each of the wire what minimum diameter is required of the wire? for NEET 2024 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 m when a tensile force of 400N is applied to each of the wire what minimum diameter is required of the wire? covers all topics & solutions for NEET 2024 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 m when a tensile force of 400N is applied to each of the wire what minimum diameter is required of the wire?.
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