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
Correct Answer: A
Solution :
\[\Delta L=\frac{FL}{AY}\] or \[\Delta L\propto \frac{L}{{{d}^{2}}}\] \[\left[ \because \,A=\frac{\pi {{d}^{2}}}{4} \right]\] Therefore, \[\Delta L\] will be maximum for that wire which \[\frac{L}{A}\]is maximum.
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