Two possitively charged spheres of masses $m_{1}$, and $m_{2}$, are suspended from a common point at the ceiling by identical insulating massless strings of length I. Charges on the two spheres are $q_{1}$ and $q_{2}$, respectively. At equilivrium both strings make the same angle $\theta$ with the vertical. Then

• A

  $q_{1} m_{1}=q_{2} m_{2}$
• Correct answer
  B

  $m_{1}=m_{2}$
• C

  $m_{1}=m_{2} \sin \theta$
• D

  $q_{2} m_{1}=q_{1} m_{2}$

Solution

$\begin{array}{l} \tan \theta=\frac{F}{m g} \quad(F \rightarrow \text { same }) \\ \tan \theta \alpha \frac{1}{m} \\ \therefore m_{1}=m_{2} \end{array}$

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