A) l
B) \[{{l}^{2}}\]
C) \[{{l}^{2/3}}\]
D) \[{{l}^{1/3}}\]
Correct Answer: D
Solution :
[d] In equilibrium, \[{{F}_{e}}=T\sin \theta \] \[mg=T\cos \theta \] \[\tan \theta =\frac{{{F}_{e}}}{mg}=\frac{{{q}^{2}}}{4\pi {{\in }_{0}}{{x}^{2}}\times mg}\] \[\text{also }\tan \theta \approx \sin =\frac{x/2}{l}\] \[\text{Hence, }\frac{x}{2l}=\frac{{{q}^{2}}}{4\pi {{\in }_{0}}{{x}^{2}}kt}\] \[\Rightarrow {{x}^{3}}=\frac{2{{q}^{2}}l}{4\pi {{\in }_{0}}mg}\therefore x={{\left( \frac{{{q}^{2}}l}{2\pi {{\in }_{0}}mg} \right)}^{1/3}}\] \[\Rightarrow x\propto {{l}^{1/3}}\]You need to login to perform this action.
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