A circular steel wire 2.00 m long must stretch no more than 0.25 cm wh...
Given:
- Length of circular steel wire, l = 2.00 m
- Maximum stretch allowed, Δl = 0.25 cm = 0.0025 m
- Tensile force applied, F = 400 N
To find: Minimum diameter required for the wire
Formula used:
- Young's modulus of elasticity, Y = (F/A) / (Δl/l)
- Cross-sectional area of wire, A = πd²/4
Calculation:
- Substitute the given values in the formula for Young's modulus:
Y = (400/A) / (0.0025/2.00)
- Substitute the formula for cross-sectional area in the above equation:
Y = (400/πd²/4) / (0.0025/2.00)
- Simplify the above equation:
Y = 32,000,000/d²
- Solve for diameter, d:
d² = 32,000,000/Y
d = √(32,000,000/Y)
- Substitute the value of Young's modulus for steel, Y = 2.0 x 10¹¹ N/m²:
d = √(32,000,000/2.0 x 10¹¹)
d = 0.0014 m = 1.4 mm
Therefore, the minimum diameter required for the wire is 1.4 mm.
Answer: (a) 1.4 mm
A circular steel wire 2.00 m long must stretch no more than 0.25 cm wh...
First we should know Young's Modulus of steel then apply its formula
change in length is 0.25cm
l=2m. T=400 A=πr2. then diameter
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