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Two spheres A and B of radius 4 cm and 6 cm are given charges of $80 µ C a n d 40 µ C$ respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is

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$16 μ C$ from $B$ to $A$

$20 μ C$ from $A t o B$

$32 μ C$ from $B$ to $A$

$32 μ C$ from $A$ to $B$

answer is D.

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Detailed Solution

Total charge $Q = 80 + 40 = 120 μ C .$

By using the formula $Q_{1}^{'} = Q [\frac{r_{1}}{r_{1} + r_{2}}]$

New charge on sphere $A is Q_{A}^{'} = Q [\frac{r_{A}}{r_{A} + r_{B}}]$

$= 120 [\frac{4}{4 + 6}] = 48 μ C$

Initially it was $80 µ C, i . e ., 32 µ C$ charge flows from A to B.

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