A) \[v\propto x\]
B) \[v\propto {{x}^{-\frac{1}{2}}}\]
C) \[v\propto {{x}^{-1}}\]
D) \[v\propto {{x}^{\frac{1}{2}}}\]
Correct Answer: B
Solution :
\[\tan \theta =\frac{F}{mg}\] or \[\frac{x}{2l}=\frac{k{{q}^{2}}}{mg{{x}^{2}}}\] \[\frac{{{x}^{3}}}{2l}=\frac{k{{q}^{2}}}{mg}\] \[\frac{3{{x}^{2}}\frac{dx}{dt}}{2l}=\frac{2kq\frac{dq}{dt}}{mg}\] Also, \[q\propto {{x}^{3/2}}\] \[\Rightarrow \,\,\frac{dx}{dt}\propto \frac{{{x}^{3/2}}}{{{x}^{2}}},i.e.,\,v\propto {{x}^{-1/2}}\]You need to login to perform this action.
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