| Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become. [NEET 2013] |
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A) \[{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}\]
B) \[\left( \frac{r}{\sqrt[3]{2}} \right)\]
C) \[\left( \frac{2r}{\sqrt{3}} \right)\]
D) \[\left( \frac{2r}{3} \right)\]
Correct Answer: B
Solution :
| [b] As \[Fe=mg\tan \theta \] |
| We have |
| \[Fe=\tan {{\theta }_{1}}\] |
| and \[F_{e}^{'}=\tan \theta \] |
| \[\therefore \] \[\frac{F_{e}^{'}}{{{F}_{e}}}=\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{1}}}\] |
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