B] As \[Fe=mg\tan \THETA \] We have \[Fe=\tan {{\theta }_{1}}\] and \[F_{E}^{'}=\tan \theta \] \[\therefore \] \[\frac{F_{e}^{'}}{{{F}_{e}}}=\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{1}}}\]

"> 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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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]

NEET Physics in NEET . 8 months ago

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[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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Posted on 21 Aug 2024, this text provides information on NEET related to Physics in NEET. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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