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NEET PHYSICSEasy

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

A

length = 100 cm, diameter = 1 mm

B

length = 200 cm, diameter = 2 mm

C

length = 300 cm, diameter = 3 mm

D

length = 50 cm, diameter = 0.5 mm

Step-by-Step Solution

  1. Formula: The extension (Δl\Delta l) of a wire under tension (FF) is derived from Young's Modulus (Y=FLAΔlY = \frac{FL}{A\Delta l}), giving Δl=FLAY\Delta l = \frac{FL}{AY} .
  2. Dependencies: Since the material (same YY) and tension (FF) are constant, the extension is proportional to the ratio of length to cross-sectional area: Δl∝LA\Delta l \propto \frac{L}{A}.
  3. Area: The cross-sectional area A=πr2=π(d/2)2∝d2A = \pi r^2 = \pi (d/2)^2 \propto d^2. Therefore, Δl∝Ld2\Delta l \propto \frac{L}{d^2}.
  4. Comparison: We calculate the ratio Ld2\frac{L}{d^2} for each option (keeping units consistent is not strictly necessary for comparison if we treat cm and mm consistently as relative units, but converting to standard units prevents errors. Here we simply compare the numerical values of Lcm/dmm2L_{cm}/d_{mm}^2):
  • Option A: 10012=100\frac{100}{1^2} = 100
  • Option B: 20022=2004=50\frac{200}{2^2} = \frac{200}{4} = 50
  • Option C: 30032=3009≈33.3\frac{300}{3^2} = \frac{300}{9} \approx 33.3
  • Option D: 50(0.5)2=500.25=200\frac{50}{(0.5)^2} = \frac{50}{0.25} = 200
  1. Conclusion: The wire in Option D has the highest ratio, meaning it will experience the largest extension.
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