If the radius of the sphere having charge q at its centre is doubled,…

The correct answer is Option (3) → \( \phi=\displaystyle\frac{q}{\varepsilon_{0}} \)

By Gauss’s law, the total electric flux through any closed surface equals the charge enclosed divided by \( \varepsilon_0 \):

    \( \displaystyle \phi = \oint \vec{E}\cdot d\vec{A} = \frac{q_{\mathrm{enclosed}}}{\varepsilon_0} \)

When the radius is doubled, the field on the surface falls to one-fourth (\( E \propto 1/r^{2} \)), but the area becomes four times (\( A \propto r^{2} \)). The two changes cancel:

    \( \displaystyle \phi = \frac{1}{4\pi\varepsilon_0}\frac{q}{(2r)^{2}}\times4\pi(2r)^{2} = \frac{q}{\varepsilon_0} \)

So the flux stays \( q/\varepsilon_0 \); it is independent of the size and shape of the surface.

NCERT (2026-27):
Gauss’s Law – “… We state Gauss’s law without proof:

    \( \displaystyle \text{Electric flux through a closed surface } S = q/\varepsilon_0 \qquad (1.31) \)

q = total charge enclosed by S.”

Gauss’s Law – “… (i) Gauss’s law is true for any closed surface, no matter what its shape or size.”