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(C) The solid angle subtended by a cone of half-angle $	heta$ is given by $\Omega = 2\pi(1 - \cos 	heta)$.Since there are two such cones (top and bo

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$3$

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(C) The solid angle subtended by a cone of half-angle $	heta$ is given by $\Omega = 2\pi(1 - \cos 	heta)$.Since there are two such cones (top and bottom) subtending the plane surfaces of the cylinder, the total solid angle subtended by the two plane surfaces is $\Omega_{total} = 2 	imes 2\pi(1 - \cos 	heta) = 4\pi(1 - \cos 	heta)$.The solid angle subtended by the curved surface is the total solid angle $4\pi$ minus the solid angle of the plane surfaces:$\Omega_{curved} = 4\pi - 4\pi(1 - \cos 	heta) = 4\pi \cos 	heta$.The electric flux $\Phi$ through a surface is given by $\Phi = \frac{q \Omega}{4\pi \epsilon_0}$.Thus, the flux through the curved surface is $\Phi(	heta) = \frac{q}{4\pi \epsilon_0} (4\pi \cos 	heta) = \frac{q}{\epsilon_0} \cos 	heta$.For $	heta = 30^{\circ}$, $\Phi = \frac{q}{\epsilon_0} \cos 30^{\circ}$.For $	heta = 60^{\circ}$, $\Phi' = \frac{q}{\epsilon_0} \cos 60^{\circ}$.Taking the ratio: $\frac{\Phi}{\Phi'} = \frac{\cos 30^{\circ}}{\cos 60^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.Therefore, $\Phi' = \frac{\Phi}{\sqrt{3}}$.Comparing this with $\Phi / \sqrt{n}$, we get $n = 3$.

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