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729 drops of same size are charged to a potential of 1 V each . If they coalesce to form a single drop , its potential would be

$(a)\;V\qquad(b)\;9 V\qquad(c)\;81 V\qquad(d)\;729 V$

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Answer ; (c) 81
Explanation : Charge on final drop = 729 q
radius of final drop =$\; \sqrt[3]{729}\;r$
$= 9\;r$
Potential of final drop = $\;\large\frac{729}{9} = 81 V\;.$
answered Jan 30, 2014 by yamini.v

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