JEE Advanced 2024 Paper 2, Physics Section 3 Q3: Gauss's Law

The solid angle subtended at $P$ by a flat circular cap (the end of a cone of half-angle $\theta$) is $\Omega =2\pi (1-\cos \theta )$.

The flux through one flat end of the cylinder is

${\varphi }_{\text{end}}=\frac{\Omega }{4\pi }\cdot \frac{Q}{{\epsilon }_0}=\frac{Q}{2{\epsilon }_0}(1-\cos \theta ).$

By symmetry, both flat ends carry the same flux, so the total flux through the two ends is $\frac{Q}{{\epsilon }_0}(1-\cos \theta )$.

Since the total flux through the closed surface is $Q/{\epsilon }_0$, the flux through the curved side is

$\Phi (\theta )=\frac{Q}{{\epsilon }_0}-\frac{Q}{{\epsilon }_0}(1-\cos \theta )=\frac{Q\cos \theta }{{\epsilon }_0}.$

For $\theta =30^{\circ }$: $\Phi =\frac{Q}{{\epsilon }_0}\cdot \frac{\sqrt{3}}{2}.$

For $\theta =60^{\circ }$: ${\Phi }^{\prime }=\frac{Q}{{\epsilon }_0}\cdot \frac{1}{2}.$

$\therefore \frac{\Phi }{{\Phi }^{\prime }}=\sqrt{3}\Rightarrow {\Phi }^{\prime }=\frac{\Phi }{\sqrt{3}}.$ Hence $n=3$.