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(C) Let the radii of the two spheres be $R_1 = 4R$ and $R_2 = 7R$. When they are in contact,they reach the same potential $V$. Since $V = \frac{Kq}{R}

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$48$

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(C) Let the radii of the two spheres be $R_1 = 4R$ and $R_2 = 7R$. When they are in contact,they reach the same potential $V$. Since $V = \frac{Kq}{R}$,we have $\frac{Kq_1}{R_1} = \frac{Kq_2}{R_2}$. This implies $\frac{q_1}{4R} = \frac{q_2}{7R}$,so $\frac{q_1}{q_2} = \frac{4}{7}$. The total charge $q = q_1 + q_2 = 8.8 	imes 10^{-7} 	ext{ C}$. Substituting $q_2 = \frac{7}{4}q_1$,we get $q_1 + \frac{7}{4}q_1 = 8.8 	imes 10^{-7} 	ext{ C}$. $\frac{11}{4}q_1 = 8.8 	imes 10^{-7} 	ext{ C} \Rightarrow q_1 = \frac{4}{11} 	imes 8.8 	imes 10^{-7} = 3.2 	imes 10^{-7} 	ext{ C}$. The potential at a distance $r = 60 	ext{ m}$ from the smaller sphere is $V = \frac{Kq_1}{r}$. $V = \frac{9 	imes 10^9 	imes 3.2 	imes 10^{-7}}{60} = \frac{28.8 	imes 10^2}{60} = 48 	ext{ V}$.

Two metal spheres have their radii in the ratio of $4:7$. They are put in contact and a charge $8.8 \times 10^{-7} \text{ C}$ is given to the system. Then they are separated so that each can exert no influence on the other. The potential due to the smaller sphere at $60 \text{ m}$ from it in Volt is

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