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(B) Let the surface charge densities be $\sigma$. Since the surface charge densities on the outer surfaces are equal,we have:$\sigma = \frac{q_1}{4\pi

Vedclass · Accepted Answer

$1 : 3 : 5$

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(B) Let the surface charge densities be $\sigma$. Since the surface charge densities on the outer surfaces are equal,we have:$\sigma = \frac{q_1}{4\pi R^2} = \frac{q_2}{4\pi (2R)^2} = \frac{q_3}{4\pi (3R)^2}$,where $q_1, q_2, q_3$ are the charges on the outer surfaces of the shells.From the property of induction,the charge on the outer surface of the first shell is $q_1 = Q_1$.The charge on the outer surface of the second shell is $q_2 = Q_1 + Q_2$.The charge on the outer surface of the third shell is $q_3 = Q_1 + Q_2 + Q_3$.Equating the densities: $\frac{Q_1}{R^2} = \frac{Q_1 + Q_2}{4R^2} = \frac{Q_1 + Q_2 + Q_3}{9R^2}$.From $\frac{Q_1}{1} = \frac{Q_1 + Q_2}{4}$,we get $4Q_1 = Q_1 + Q_2 \Rightarrow Q_2 = 3Q_1$.From $\frac{Q_1}{1} = \frac{Q_1 + Q_2 + Q_3}{9}$,we get $9Q_1 = Q_1 + (3Q_1) + Q_3 \Rightarrow 9Q_1 = 4Q_1 + Q_3 \Rightarrow Q_3 = 5Q_1$.Thus,the ratio $Q_1 : Q_2 : Q_3 = Q_1 : 3Q_1 : 5Q_1 = 1 : 3 : 5$.

Three concentric metallic spherical shells of radii $R, 2R, 3R$ are given charges $Q_1, Q_2, Q_3$ respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then,the ratio of the charges given to the shells,$Q_1 : Q_2 : Q_3$ is

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