A) Decreases by 6%
B) Decreases by 4%
C) Increases by 4%
D) Increases by 6%
Correct Answer: B
Solution :
If the distance between particles is increased by 2%, \[F\propto \frac{1}{{{r}^{2}}}\Rightarrow \frac{dF}{dr}=\frac{d}{dr}({{r}^{-2}})=\frac{-2}{r}\therefore dF=\frac{-2}{r}dr\] \[or\,\,\frac{dF}{F}\times 100=\frac{-2}{r}\frac{dr}{F}=-2r\,\,dr\times 100\] Given, \[\frac{dr}{r}\times 100=2%\] \[\therefore \,\,dF%=%\]decrease.
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