Q.
Two small spheres of masses $M_{1}$ and $M_{2}$ are suspended by weightless insulating threads of lengths $L_{1}$and $L_{2}$. The spheres carry charges $\mathcal{Q}_{1}$ and $\mathcal{Q}_{2}$ respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of $\theta_{1}$ and $\theta_{2}$as shown. Which one of the following conditions is essential, if $\theta_{1} = \theta_{2}$ ?
A) $M_{1} \neq M_{2},$ but $\mathcal{Q}_{1} = \mathcal{Q}_{2}$
B) $M_{1} = M_{2}$
C) $\mathcal{Q}_{1} = \mathcal{Q}_{2}$
D) $L_{1} = L_{1}$
Explanation
(b) For sphere 1, in equilibrium
[image:1_image258.emf]
and $T_{1}\cos{\theta_{1} = M_{1}g}$
$T_{1}\ \ \sin\ \theta_{1} = F_{1}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \tan\ \theta_{1} = \frac{F_{1}}{M_{1}g}$
Similarly for sphere 2, tan $\theta_{2} = \frac{F_{2}}{M_{2}g}$
$F$is same on both the charges, $\ \theta$ will be same only if their masses $M$ are equal.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.