$\text{A charge is kept at the central point}P\text{of a cylindrical region. }\text{The two edges subtend a half-angle}\mathrm{ \theta }\text{at}\mathrm{ P},$

$\text{as shown in the figure. }\text{When}\mathrm{ \theta }=30°,\text{then the electric flux through the curved surface of the cylinder}$

$\text{is }\Phi .\text{If }\theta =60°,\text{then the electric flux through the curved surface becomes}\frac{\varphi }{\sqrt{n}},\text{where the value of}\mathrm{ n}\text{ is \_\_\_\_}$

The correct answer is 3.

Step 1: Understand the Total Flux and Solid Angle Concept

According to Gauss's Law, the total electric flux originating from a point charge $q$ kept inside a closed surface enclosing it completely (which subtends a total solid angle of $4\pi$ steradians) is:

${\Phi }_{\text{total}}=\frac{q}{{\epsilon }_0}$

From the given image, a cylinder has two flat circular end-caps and one curved surface. A charge is kept at the central point $P$ inside the cylinder. The line joining the point $P$ to the edges of the top and bottom circular bases subtends a semi-vertical angle (half-angle) $\theta$ with the central axis.

Step 2: Flux through the Flat Surfaces and Curved Surface

The solid angle subtended by a circular disc of semi-vertical angle $\theta$ at a point on its axis is given by:

$\Omega =2\pi (1-\cos \theta )$

Since the cylinder has two symmetrical circular end-caps (top and bottom), the total solid angle subtended by both flat caps together at point $P$ is:

${\Omega }_{\text{caps}}=2\times 2\pi (1-\cos \theta )=4\pi (1-\cos \theta )$

The electric flux passing through the two flat end-caps is therefore:

${\Phi }_{\text{caps}}=\frac{q}{{\epsilon }_0}·\frac{{\Omega }_{\text{caps}}}{4\pi }=\frac{q}{{\epsilon }_0}(1-\cos \theta )$

Since the total electric flux through the entire cylinder is ${\Phi }_{\text{total}}=\frac{q}{{\epsilon }_0}$, the flux through the curved surface of the cylinder, ${\Phi }_{\text{curved}}$, is:

${\Phi }_{\text{curved}}={\Phi }_{\text{total}}-{\Phi }_{\text{caps}}$

${\Phi }_{\text{curved}}=\frac{q}{{\epsilon }_0}-\frac{q}{{\epsilon }_0}(1-\cos \theta )=\frac{q}{{\epsilon }_0}\cos \theta$

Step 3: Calculating Flux for the Two Angles

1. When $\theta =30°$, the flux through the curved surface is given as $\Phi$:

$\Phi =\frac{q}{{\epsilon }_0}\cos (30°)=\frac{q}{{\epsilon }_0}·\frac{\sqrt{3}}{2}$

2. When $\theta =60°$, let the new flux through the curved surface be $\Phi \text{'}$:

$\Phi \text{'}=\frac{q}{{\epsilon }_0}\cos (60°)=\frac{q}{{\epsilon }_0}·\frac{1}{2}$

Step 4: Finding the Ratio and Value of n

Dividing $\Phi \text{'}$ by $\Phi$:

$\frac{\Phi \text{'}}{\Phi }=\frac{\frac{q}{{\epsilon }_0}·\frac{1}{2}}{\frac{q}{{\epsilon }_0}·\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$

$\Phi \text{'}=\frac{\Phi }{\sqrt{3}}$

Comparing this with the given expression $\frac{\varphi }{\sqrt{n}}$, we get:

$n=3$