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Two small spheres of masses, M₁ and M₂ are suspended by weightless insulating threads of lengths L₁ and L₂. the sphere carry charges
Q₁ and Q₂ respectively. The spheres are suspended such that they are in level with another and the threads are inclined to the vertical at angles of $θ_{1}$ and $θ_{2}$ as shown below, which one of the following conditions is essential , if $θ_{1} = θ_{2}$ .

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$M_{1} = M_{2}$

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

$Q_{1} = Q_{2}$

$L_{1} = L_{2}$

answer is B.

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Detailed Solution

There are three forces acting on each sphere are (i) tension (ii) weight(w) (iii) electrostatic force of repulsion for sphere

In equailibrium, from figure

$\tan θ_{1} = F_{1} / M_{1} g$

From sphere 2, in equilibirum from figure
$\tan θ_{2} = F_{2} / M_{2} g$

$or θ_{1} = θ_{2} only for \frac{F_{1}}{M_{1} g} = \frac{F_{2}}{M_{2} g}$

$But, F_{1} = F_{2} and then M_{1} = M_{2}$

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