Q.Obtain the formula for acceleration due to gravity at the depth 'd' below the Earth's surface.

[!TLDR] Only the inner sphere of radius $(R-d)$ pulls on a point at depth d (shell theorem); with uniform density this gives $g_d …

Model the Earth (radius R, uniform density $\rho$) as made of concentric uniform shells. Consider a point P at depth d below the surface, so its distance from the centre is $(R-d)$. By the shell theorem, the outer spherical shell of thickness d (from radius $R-d$ out to R) exerts ZERO net force on P, since P lies inside it; only the inner solid sphere of radius $(R-d)$ contributes, and, again by the shell theorem (now viewed from outside this smaller sphere), that inner sphere's mass acts as though concentrated at the centre. With uniform density, the inner sphere's mass is $M'=\dfrac{4}{3}\pi(R-d)^3\rho,$ so the acceleration due to gravity at P is $g_d=\dfrac{GM'}{(R-d)^2}=\dfrac{4}{3}\pi G\rho(R-d).$ The full Earth's surface gravity, similarly, is $g=\dfr …

Apply the shell theorem to isolate only the inner sphere of radius $(R-d)$ as contributing to gravity at depth d, express its mass via uniform density, and d …

Forgetting that the OUTER shell (between depth d and the surface) contributes zero force at P, and mist …