–10μC, 40μC and q are the charges on three identical spherical conductors P, Q and R respectively. Now P and Q attract each other with a force F when they are separated by a distance d. Now P and Q are made in contact with each other and then separated. Again Q and R are touched and they are separated by a distance ‘d’. The repulsive force between Q and R is 4F. Then the charge q is

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30μC

40μC

10μC

65μC

answer is D.

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Detailed Solution

$F α \frac{10 \times 40}{d^{2}} . . . . . . . . (1)$ (Force between P and Q) P and Q touch each other, then net charge is $(- 10 + 40) = 30 μ$ and they are separated, then charge on each becomes $15 μ c, 15 μ c$ If Q and R touch each other and separated,

then charge on each becomes $= (\frac{15 + q}{2})$
but force $4 F α \frac{{(\frac{Q + 15}{2})}^{2}}{d^{2}} . . . . . . . (2)$
$\begin{array}{l} \frac{(1)}{(2)} \Rightarrow \frac{1}{4} = \frac{400}{{(\frac{15 + q}{2})}^{2}} \\ \Rightarrow \frac{15 + q}{2} = 2 \times 20 = 40 μ c \end{array}$