Question

Two smalI 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 $Q_1$ and $Q_2$ 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 \ne \, M_2, $ but $Q_1 = Q_2$
• B. $ M_1= M_2$
• C. $ Q_1 = Q_2$
• D. $L_1 = L_2$

Answer 1

The Correct Option is B

For sphere 1 , in equilibrium

$T_{1} \cos \theta_{1}=M_{1} g$
and $T_{1} \sin \theta_{1}=F_{1}$
$\therefore \tan \theta_{1}=\frac{F_{1}}{M_{1} g}$
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.