A) \[v\propto {{x}^{\frac{1}{2}}}\]
B) \[v\propto x\]
C) \[v\propto {{x}^{-\frac{1}{2}}}\]
D) \[v\propto {{x}^{-1}}\]
Correct Answer: C
Solution :
\[\tan \theta =\frac{{{F}_{e}}}{mg}\simeq \theta \] \[\frac{K{{q}^{2}}}{{{x}^{2}}mg}=\frac{x}{2\ell }\] or \[\] ?..(1) or \[{{x}^{3/2}}\propto q\] ?..(2) differentiate eq.(i) w. r .t. time \[3{{x}^{2}}\frac{dx}{dt}\propto 2q\frac{dq}{dt}\]but \[\frac{dq}{dt}\] is constant so x\[{{x}^{2}}(v)\propto q\] replace q from eq. (2) \[{{x}^{2}}(v)\propto {{x}^{3/2}}\] or \[\]You need to login to perform this action.
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