A) \[l\]
B) \[{{l}^{2}}\]
C) \[{{l}^{2/3}}\]
D) \[{{l}^{1/3}}\]
Correct Answer: D
Solution :
[d] |
\[T\,\cos \theta =mg\] |
\[T\,\sin \theta =\frac{k{{q}^{2}}}{{{x}^{2}}}\] |
\[\tan \theta =\frac{k{{q}^{2}}}{mg\,{{x}^{2}}}\] (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 |
\[\frac{{{x}^{2}}}{4}<<{{l}^{2}}\,\,\,\Rightarrow \,\,{{x}^{3}}=\frac{2k{{q}^{2}}}{mg}{{({{l}^{2}})}^{1/2}}\] |
\[x=\frac{2k{{q}^{2}}}{mg}{{l}^{1/3}}\] |
\[x\propto {{l}^{1/3}}\] |
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