Two small spheres of masses $M_{1}$ and $M_{2}$ are suspended by weightless insulating thread lengths $L_{1}$ and $L_{2}$ . The spheres carry charges $Q_{1}$ and $Q_{2}$ , respectively. The spheres suspended such that they are in level with one another and the threads are inclined to vertical at angles of $θ_{1}$ and ${\dot{θ}}_{2}$ as shown. Which one of the following conditions is essential for $θ_{1} = θ_{2}$ ?

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$Q_{1} = Q_{2}$

$M_{1} \neq M_{2}$ , but $Q_{1} = Q_{2}$

$L_{1} \neq L_{2}$

$M_{1} = M_{2}$

answer is B.

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Detailed Solution

$For sphere 1 in equilibrium,$

$T_{1} \cos θ_{1} = M_{1} g$ and $T_{1} \sin θ_{1} = F_{1}$

$∴$ $\tan θ_{1} = \frac{F_{1}}{M_{1} g}$

Similarly for sphere $2, \tan θ_{2} = \frac{F_{2}}{H_{2}}$

$F$ is same on both the charges, 0 will 1 if their masses $M$ are equal.

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