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(B) Let the surface charge densities on the shells be $\sigma_1, \sigma_2, \sigma_3$ respectively.
Given that $\sigma_1 = \sigma_2 = \sigma_3 = \sigm

Vedclass · Accepted Answer

$1 : 4 : 9$

Vedclass · Answer

(B) Let the surface charge densities on the shells be $\sigma_1, \sigma_2, \sigma_3$ respectively.
Given that $\sigma_1 = \sigma_2 = \sigma_3 = \sigma$.
The surface charge density is defined as $\sigma = \frac{Q}{A}$,where $A = 4\pi r^2$ is the surface area of the sphere.
For the first shell: $Q_1 = \sigma \cdot 4\pi R^2$.
For the second shell: $Q_2 = \sigma \cdot 4\pi (2R)^2 = \sigma \cdot 16\pi R^2$.
For the third shell: $Q_3 = \sigma \cdot 4\pi (3R)^2 = \sigma \cdot 36\pi R^2$.
Now,the ratio $Q_1 : Q_2 : Q_3$ is:
$Q_1 : Q_2 : Q_3 = (\sigma \cdot 4\pi R^2) : (\sigma \cdot 16\pi R^2) : (\sigma \cdot 36\pi R^2)$.
Dividing by $4\pi R^2$,we get:
$Q_1 : Q_2 : Q_3 = 1 : 4 : 9$.

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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