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
Correct Answer: D
Solution :
\[Y=\frac{F}{A}\ \frac{L}{A}\Rightarrow l\propto \frac{L}{A}\propto \frac{L}{\pi {{d}^{2}}}\] \ \[l\propto \frac{L}{{{d}^{2}}}\] [As F and Y are constant] The ratio of \[\frac{L}{{{d}^{2}}}\]is maximum for case
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