JEE Main & Advanced
JEE Main Online Paper (Held on 9 April 2013)
question_answer
Two balls of same mass and carrying equal charge are hung from a fixed support of length \[l\]. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, \[x\] between the balls is proportional to :
JEE Main Online Paper (Held On 09 April 2013)
A) \[l\]
B) \[{{l}^{2}}\]
C) \[{{l}^{2/3}}\]
D) \[{{l}^{1/3}}\]
Correct Answer:
C
Solution :
\[T\,\cos \theta =mg\] \[T\,\sin \theta =\frac{k{{q}^{2}}}{{{x}^{2}}}\] \[\tan \theta =\frac{k{{q}^{2}}}{mg\,{{x}^{2}}}\] ?(i) From OBC \[\Rightarrow \] \[\tan \theta =\frac{CB}{OB}=\frac{X/2}{{{\left( {{l}^{2}}-\frac{{{x}^{2}}}{4} \right)}^{1/2}}}\] ?(ii) Form Eqs. (i) and (ii) \[\frac{x}{2{{\left( {{l}^{2}}-\frac{{{x}^{2}}}{4} \right)}^{1/2}}}=\frac{k{{q}^{2}}}{mg\,{{x}^{2}}}\] \[{{x}^{3}}=\frac{2k{{q}^{2}}}{mg}\,{{\left( {{l}^{2}}-\frac{{{x}^{2}}}{4} \right)}^{1/2}}\] If \[\theta \] is small, then \[\frac{{{x}^{2}}}{4}<<{{l}^{2}}\,\,\,\Rightarrow \,\,{{x}^{3}}=\frac{2k{{q}^{2}}}{mg}{{({{l}^{2}})}^{1/2}}\] \[x=\frac{2k{{q}^{2}}}{mg}{{l}^{1/3}}\] \[x\propto {{l}^{1/3}}\]