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Two metal balls of same radii are charged to 5 and -25 units of electricity . They are brought in contact with each other and then again seperated to original distance. Ratio of magnitude of force b/w two balls after and before contact is

$(A)\;5:4\\ (B)\;4:5 \\ (C)\; 1:5 \\ (D)\;5:1 $

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$F_{\large before }=\large\frac{1}{4 \pi \in _0} \frac{(5) \times (-25)}{r^2}$
Charge after $=q_1=q_2=\large\frac{5+ (-25)}{2}$
$\qquad=-10$
$F_{\large after }=\large\frac{1}{4 \pi \in _0} \frac{(-10) \times (-10)}{2}$
$\large\frac{|F_{\Large after}|}{|F_{\Large before}|}=\bigg| \large\frac{(-10) \times (-10)}{(5) \times (-25)}\bigg|$
$\qquad= \large\frac{4}{5}$
$\qquad=4:5$
Hence B is the correct answer.
answered Jan 3, 2014 by meena.p
 

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