A) Increased by 1%
B) Decreased by 1%
C) Increased by 2%
D) Decreased by 2%
Correct Answer: D
Solution :
\[I=\frac{L}{{{r}^{2}}}\] \[\Rightarrow \ \ \frac{dI}{I}\ =-\frac{2dr}{r}\] (\[\because \] L = constant) Þ \[\frac{dI}{I}\times 100\ =-\frac{2\times dr}{r}\times 100\] \[=-2\times 1=-2%\]You need to login to perform this action.
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