A) 2.00 mm
B) 4.00 mm
C) 6.00mm
D) 8.00 mm
Correct Answer: D
Solution :
\[l=\frac{FL}{\pi {{r}^{2}}Y}\therefore l\propto \frac{1}{{{r}^{2}}}\] (F,L and Y are constant) \[\frac{{{l}_{2}}}{{{l}_{1}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}={{(2)}^{2}}\Rightarrow {{l}_{2}}=4{{l}_{1}}=4\times 2=8\ mm.\]You need to login to perform this action.
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