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NEET PHYSICSMedium

The following four wires are made of the same material. Which of them will have the largest extension when the same tension is applied?

A

Length=50 cm, diameter=0.5 mm

B

Length=100 cm, diameter=1 mm

C

Length=200 cm, diameter=2 mm

D

Length=300 cm, diameter=3 mm

Step-by-Step Solution

  1. Formula: The extension (ΔL\Delta L) in a wire is given by the formula for Young's Modulus (YY): ΔL=FLAY\Delta L = \frac{F L}{A Y}, where FF is the tension, LL is the length, and AA is the cross-sectional area .
  2. Area: The cross-sectional area A=πr2=π(d/2)2=πd24A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}, where dd is the diameter.
  3. Proportionality: Since the material (YY) and tension (FF) are the same for all wires, the extension is directly proportional to length and inversely proportional to the square of the diameter: ΔL∝Ld2\Delta L \propto \frac{L}{d^2}
  4. Comparison: Calculating the ratio Ld2\frac{L}{d^2} for each option (keeping units consistent relative to each other):
  • Option A: 50(0.5)2=500.25=200\frac{50}{(0.5)^2} = \frac{50}{0.25} = 200
  • Option B: 100(1)2=100\frac{100}{(1)^2} = 100
  • Option C: 200(2)2=2004=50\frac{200}{(2)^2} = \frac{200}{4} = 50
  • Option D: 300(3)2=3009≈33.3\frac{300}{(3)^2} = \frac{300}{9} \approx 33.3
  1. Conclusion: The wire with length 50 cm and diameter 0.5 mm has the highest ratio, and thus the largest extension.
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