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Two metal spheres of masses $m_{1}$ and $m_{2}$ are suspended from a common point by a light insulating string of same length. The length of each string is same. The spheres are given positive charges $q_{1}$ and $q_{2}$ . Figures A shows angles made by the strings with vertical are different whereas for figure B, the angles are same. Then, which of the following is possible?

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For figure B, $m_{1} = m_{2}$ , $q_{1} \neq q_{2}$

For figure A, $m_{1} < m_{2}$ and $q_{1} = q_{2}$

For figure A, $m_{1} > m_{2}$ and $q_{1} = q_{2}$

For figure B, $m_{1} = m_{2}$ and $q_{1} = q_{2}$

answer is A, B, C.

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Detailed Solution

When $m_{1} = m_{2}$ and $q_{1} = q_{2}$ , by symmetry, both angles will be equal. Option (C) is correct.

Now, if $m_{1}$ is made slightly greater than $m_{2}$ , then $m_{1}$ will descend and we expect $θ_{1}$ to become less than $θ_{2}$ . Option (A) is correct. Option (D) is wrong.

If masses are equal, then regardless of charges, both angles are equal. Option (B) is correct.

$\tan θ = \frac{F}{m g}$

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