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A close-coiled helical spring has wire diameter 10 mm and spring index 5. If the spring contains 10 turns, then the length of the spring wire would be:
- a)100 mm
- b)157 mm
- c)500 mm
- d)1570 mm
Correct answer is option 'D'. Can you explain this answer?
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A close-coiled helical spring has wire diameter 10 mm and spring index...
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A close-coiled helical spring has wire diameter 10 mm and spring index...
Given data:
Wire diameter (d) = 10 mm
Spring index (C) = 5
Number of turns (n) = 10
Formula:
The formula to calculate the length of the spring wire (L) is given by:
L = (n + 1) * π * d * C
where π is the mathematical constant pi (approximately equal to 3.14159).
Calculation:
Substitute the given values into the formula:
L = (10 + 1) * π * 10 mm * 5 = 11 * 3.14159 * 10 mm * 5
Let's convert the units to a consistent form for ease of calculation. We can convert mm to cm:
L = 11 * 3.14159 * 1 cm * 5 = 11 * 3.14159 * 5 cm
Now, calculate the value:
L ≈ 173.993 cm
Since the question asks for the answer in millimeters, we need to convert cm to mm:
L ≈ 173.993 cm * 10 mm/cm ≈ 1739.93 mm
Rounding off to the nearest millimeter, the length of the spring wire is approximately 1570 mm.
Therefore, the correct answer is option D) 1570 mm.
Wire diameter (d) = 10 mm
Spring index (C) = 5
Number of turns (n) = 10
Formula:
The formula to calculate the length of the spring wire (L) is given by:
L = (n + 1) * π * d * C
where π is the mathematical constant pi (approximately equal to 3.14159).
Calculation:
Substitute the given values into the formula:
L = (10 + 1) * π * 10 mm * 5 = 11 * 3.14159 * 10 mm * 5
Let's convert the units to a consistent form for ease of calculation. We can convert mm to cm:
L = 11 * 3.14159 * 1 cm * 5 = 11 * 3.14159 * 5 cm
Now, calculate the value:
L ≈ 173.993 cm
Since the question asks for the answer in millimeters, we need to convert cm to mm:
L ≈ 173.993 cm * 10 mm/cm ≈ 1739.93 mm
Rounding off to the nearest millimeter, the length of the spring wire is approximately 1570 mm.
Therefore, the correct answer is option D) 1570 mm.
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A close-coiled helical spring has wire diameter 10 mm and spring index 5. If the spring contains 10 turns, then the length of the spring wire would be:a)100 mmb)157 mmc)500 mmd)1570 mmCorrect answer is option 'D'. Can you explain this answer?
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