A) \[\upsilon \propto x\]
B) \[\upsilon \propto {{x}^{-\frac{1}{2}}}\]
C) \[\upsilon \propto {{x}^{-1}}\]
D) \[\upsilon \propto {{x}^{\frac{1}{2}}}\]
Correct Answer: B
Solution :
[b] \[\tan \theta =\frac{F}{mg}\] |
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, \[q\propto {{x}^{3/2}}\] |
\[\Rightarrow \,\,\frac{dx}{dt}\propto \frac{{{x}^{3/2}}}{{{x}^{2}}},i.e.,\,\upsilon \propto {{x}^{-1/2}}\] |
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