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Standard IX

Physics

Third Law of Motion

Two small sph...

Question

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 $Q_{1}$ and $Q_{2}$ 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, which one of the following conditions is essential, if $θ_{1} = θ_{2} :$

M_{1} \neq M_{2}

but

Q_{1} = Q_{2}

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M_{1} = M_{2}

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Q_{1} = Q_{2}

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L_{1} = L_{2}

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Solution

The correct option is B $M_{1} = M_{2}$
For a single change system:-
We can write from $F B D$ ,
$T cos θ_{1} = M g$ $T sin θ_{1} = F e$ .
Now for both change, $F e$ is equal in magnitude but, $θ_{1} = θ_{2}$ necessarily imply.
$M_{1} = M_{2}$
But, we can't say $Q_{1} = Q_{2}$ as both change face $F e = \frac{K Q_{1} Q_{2}}{d^{2}}$ in equal magnitude but opposite direction [Newtons third law] so $T sin θ$ component would be equal for both.

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