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 :
\[\Delta {{L}_{1}}\propto \frac{100}{{{1}^{2}}}\propto 100\] \[\Delta {{L}_{2}}\propto \frac{100}{{{2}^{2}}}\propto 50\] \[\Delta {{L}_{3}}\propto \frac{300}{{{3}^{2}}}\propto \frac{100}{3}\] \[\Delta {{L}_{3}}\propto \frac{50}{1/4}\propto 200\] \[\Delta L\propto \frac{L}{A}\]or \[\Delta L\propto \frac{L}{{{D}^{2}}}\] The ratio of\[\frac{L}{{{D}^{2}}}\] is maximum for case [a]. Hence, option [a] is correct.
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