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]

Question

TuteeHUB · Accepted Answer

[left( fracrsqrt[3]2 right)].

AnotherUser · Answer

[left( frac1sqrt2 right)^2] [B] As [Fe=mgtan THETA ] We have [Fe=tan theta _1] and [F_E^=tan theta ] [therefore ] [fracF_e^F_e=fractan theta _1tan theta _1]

TuteeHUB · Answer

[left( fracrsqrt[3]2 right)]

TuteeHUB · Answer

[left( frac2rsqrt3 right)]

TuteeHUB · Answer

[left( frac2r3 right)]

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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]

\[{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}\]

\[\left( \frac{r}{\sqrt[3]{2}} \right)\]

\[\left( \frac{2r}{\sqrt{3}} \right)\]

\[\left( \frac{2r}{3} \right)\]

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

Popular Categories

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]

\[{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}\]

\[\left( \frac{r}{\sqrt[3]{2}} \right)\]

\[\left( \frac{2r}{\sqrt{3}} \right)\]

\[\left( \frac{2r}{3} \right)\]

Report

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