Two unequal masses, $m_{1} = 2 m a n d m_{2} = m$ have unequal positive charge on them. They are suspended by two mass-less threads of unequal lengths from a common point such that, in equilibrium, both the masses are on same horizontal level. The angle between the two strings is $θ = 45^{\circ}$ in this position. Then the electrostatic force applied by $m_{1}$ on $m_{2}$ in this position has magnitude $(\frac{\sqrt{17} - n}{2}) m g$ . Find $n = ?$

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answer is 3.

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Detailed Solution

F = Electrostatic force
For equilibrium
$\tan θ_{1} = \frac{F}{2 m g} a n d \tan θ_{2} = \frac{F}{m g}$ ; $\tan (θ_{1} + θ_{2}) = \frac{\tan θ_{1} + \tan θ_{2}}{1 - \tan θ_{1} \tan θ_{2}}$
$\tan 45^{\circ} = \frac{\frac{F}{2 m g} + \frac{F}{m g}}{1 - \frac{F}{2 m g} . \frac{F}{m g}}$
$\Rightarrow 2 m^{2} g^{2} - F^{2} = 3 m g F \Rightarrow F^{2} + 3 m g F - 2 m^{2} g^{2} = 0$
$F = \frac{- 3 m g \pm \sqrt{9 m^{2} g^{2} + 8 m^{2} g^{2}}}{2}$
$F = (\frac{\sqrt{17} - 3}{2}) m g$