Two identical balls each of mass $\frac{40}{3}$ g of are charged by rubbing against each other. They are suspended from a horizontal support using two silk threads of length 1 m each with a separation of 1.5 m. The equilibrium separation between the balls become 30 cm, then charges on spheres are.

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$+ 10^{- 8} C, - 10^{- 4} C$

$+ 10^{- 5} C, + 10^{- 5} C$

$+ 10^{- 6} C, - 10^{- 6} C$

$+ 10^{- 5} C, - 10^{- 7} C$

answer is A.

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Detailed Solution

$\sin θ = \frac{0.6}{1} \cos θ = \sqrt{1 - 0 . 6^{2}} = 0.8 \Rightarrow \tan θ = \frac{0.6}{0.8} = \frac{3}{4} From equation (i) and (ii), we get : \begin{aligned} \tan θ = \frac{F}{mg} \\ ∴ F = mgtan θ \end{aligned} \Rightarrow \frac{kQQ}{r^{2}} = mgtan θ \Rightarrow \frac{9 \times 10^{9} Q^{2}}{0 . 3^{2}} = (\frac{40}{3} \times 10^{- 3}) \times 10 \times \frac{3}{4} \Rightarrow Q = + 10^{- 6} C, and Q = - 10^{- 6} C$