Two isolated conducting spheres 1 and 2 each carry a charge +Q. Sphere 1 has radius a and sphere B has radius 4a. If the two spheres are connected by a conducting wire, the charge on sphere 1 is $Q_{1}$ and on sphere 2 is $Q_{2}$ . Then

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$Q_{1} = \frac{2 Q}{3}, Q_{2} = \frac{4 Q}{3}$

$Q_{1} = \frac{2 Q}{5}, Q_{2} = \frac{8 Q}{5}$

$Q_{1} = \frac{Q}{2}, Q_{2} = \frac{3 Q}{2}$

$Q_{1} = Q, Q_{2} = Q$

answer is C.

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Detailed Solution

The moment the two spheres are connected by a conducting wire, they quickly form an equipotential surface. Charge will flow from sphere 1 $(w h i c h w a s a t h i g h e r p o t e n t i a l = \frac{Q}{4 π \in_{0} a})$ to sphere 2 $(w h i c h w a s a t l o w e r p o t e n t i a l = \frac{Q}{4 π \in_{0} (4 a)})$ until their potentials are equalized. It $Q_{1}$ is charge on sphere 1 and $Q_{2}$ on sphere 2, then
$\frac{Q_{1}}{4 π ϵ_{0} (a)} = \frac{Q_{2}}{4 π ϵ_{0} (4 a)} \Rightarrow Q_{2} = 4 Q_{1} (1) But Q_{1} + Q_{2} = Q + Q = 2 Q (2) From (1) and (2) we get 5 Q_{1} = 2 Q or Q_{1} = \frac{2 Q}{5} and$