A) doubled
B) half
C) constant
D) None of these
Correct Answer: B
Solution :
\[Y=\frac{mgL}{\pi {{r}^{2}}.I}\] \[\Rightarrow \] \[Y\propto \frac{L}{{{r}^{2}}}\] \[\therefore \] \[Y'=\frac{2L}{4{{r}^{2}}}=\frac{Y}{2}\]
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