A) 2 mm
B) 0.5 mm
C) 4 mm
D) 0.25 mm
Correct Answer: B
Solution :
\[l=\frac{FL}{AY}\Rightarrow l\propto \frac{L}{{{r}^{2}}}\] (F and Y are same) \\[\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{L}_{2}}}{{{L}_{1}}}{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=2\times {{\left( \frac{1}{2} \right)}^{2}}=\frac{1}{2}\] Þ\[{{l}_{2}}=\frac{{{l}_{1}}}{2}=\frac{l}{2}=0.5mm.\]
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