Two possitively charged spheres of masses $m_{1}$ , and $m_{2}$ , are suspended from a common point at the ceiling by identical insulating mass-less strings of length I. Charges on the two spheres are $q_{1}$ and $q_{2}$ , respectively. At equilibrium both strings make the same angle $θ$ with the vertical. Then :

q_{1} m_{1} = q_{2} m_{2}

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m_{1} = m_{2}

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m_{1} = m_{2} sin θ

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q_{2} m_{1} = q_{1} m_{2}

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Solution

The correct option is B $m_{1} = m_{2}$
For equilibrium of $m_{1}$
$T_{1} cos θ = m_{1} g; T_{1} sin θ = F$
$tan θ = \frac{F}{m_{1} g}$
For equilibrium of $m_{2}$
$T_{2} cos θ = m_{2} g; T_{2} sin θ = F$
$tan θ = \frac{F}{m_{2} g}$
Thus comparing 2 equations $m_{1} = m_{2}$

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Two positively charged spheres of masses $m_{1}$ and $m_{2}$ , are suspended from a common point at the ceiling by identical insulating massless strings of length l. Charges on the two spheres are $q_{1}$ and $q_{2}$ , respectively. At equilibrium both strings make the same angle $θ$ with the vertical. Then