Q.A charge of + 7 $\mu C$ is placed at the centre of two concentric spheres with radius 2.0 cm and 4.0 cm respectively. The ratio of the flux through them will be (A) 1:4 (B) 1:2 (C) 1:1 (D) 1:16

[!TLDR] By Gauss' law, flux through a closed surface depends only on the enclosed charge, never on the surface's radius -- both spheres enclose the same +7 $\mu C$, so the ratio is 1:1. [!ANSWER] (C) 1:1.

By Gauss' law, the total electric flux through any closed surface is $\phi_E=\dfrac{Q_{enclosed}}{\epsilon_0}$, a result that depends ONLY on the total charge enclosed WITHIN the surface, and not at all on the surface's size, shape, or radius. Both the 2.0 cm sphere and the 4.0 cm sphere are concentric with, and fully enclose, the SAME +7 $\mu C$ point charge at their common centre -- neither sphere's radius changes how much charge is inside it, since the charge sits at a single point well within both radii. Since $Q_{enclosed}=+7\,\mu C$ for both spheres, and $\epsilon_0$ is a universal constant, the flux through each sphere is IDENTICAL: $\phi_1=\phi_2=q/\epsilon_0$, giving a ratio of exactly 1:1. (Physically, this happens because although the electric field E itself is four times weaker on the larger sphere, having fallen off as $1/r^2$ while r doubled, the larger sphere's surface area is correspondingly four times bigger, having grown as $r^2$ -- the two effects exactly cancel in the product $E\times\text{Area}$ that gives the flux.) [!ANSWER] (C) 1:1.

Apply Gauss' law directly: flux through a closed surface depends only on the enclosed charge, so two concentric spheres enclosing the same charge always carry identical flux, whatever their radii.

Assuming flux must scale with the sphere's surface area (like a fixed number of lines crowding a bigger area) without realising E itself weakens by exactly the same factor, leaving the total flux unchanged.