Two pith balls, each carrying charge $+q$ are hung from a hook by two strings. It is found that when each charge is tripled, angle between the strings double. What was the initial angle between the strings?

• A

  $30^{\circ}$
• Correct answer
  B

  $60^{\circ}$
• C

  $45^{\circ}$
• D

  $90^{\circ}$

Solution

Hint: $\mathrm{T} \sin \theta=\mathrm{Fe}, \mathrm{T} \cos \theta=\mathrm{mg}$ $\tan \theta=\frac{\mathrm{Fe}}{\mathrm{mg}}=\frac{\mathrm{k} \cdot \mathrm{q}^{2}}{(2 \mathrm{~L} \sin \theta)^{2} \mathrm{mg}}$ In 2 nd case $\begin{array}{l} \tan 2 \theta=\frac{\mathrm{kq} \mathrm{q}^{2}}{(2 \mathrm{~L} \sin 2 \theta)^{2} \cdot \mathrm{mg}} \\ \frac{\tan 2 \theta}{\tan \theta}=\frac{9 \sin ^{2} \theta}{\sin ^{2} 2 \theta} \\ \frac{2}{1-\tan ^{2} \theta}=\frac{9}{4 \cos ^{2} \theta}=\frac{9}{4} \sec ^{2} \theta \\ \frac{2}{1-\tan ^{2} \theta}=\frac{9}{4}\left(1-\tan ^{2} \theta\right) \end{array}$ $\tan ^{2} \theta=x, \frac{2}{1-x}=\frac{9}{4}(1+x)$ $\Rightarrow 8=9-9 x^{2} \Rightarrow x=\frac{1}{3} \Rightarrow \tan ^{2} \theta=\frac{1}{3} \Rightarrow \theta=30^{\circ}$ Angle between the strings $=2 \theta=60^{\circ}$

Hence, the correct answer is option: 'B'.