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                 \[\frac{{{x}^{2}}}{4}<<{{l}^{2}}\,\,\,\Rightarrow \,\,{{x}^{3}}=\frac{2k{{q}^{2}}}{mg}{{({{l}^{2}})}^{1/2}}\]                 \[x=\frac{2k{{q}^{2}}}{mg}{{l}^{1/3}}\]                 \[x\propto {{l}^{1/3}}\]