At what altitude will the acceleration due to gravity be $25\%$ of that at the earth's surface (given radius of earth is $R$)?

If $R$ is the radius of the Earth and $g$ is the acceleration due to gravity on the Earth's surface,then the mean density of the Earth is:

Assume that the Earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the Earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 \, g$. The time period of the motion of the particle will be (approximately) (take $g = 10 \, m/s^2$, radius of Earth $R = 6400 \, km$):

The height at which the weight of the body becomes $\frac{1}{16}$ of its weight on the surface of the earth of radius $R$ is: (in $R$)

$A$ body weighs $500 \, N$ on the surface of the earth. How much would it weigh halfway below the surface of the earth (in $, N$)?

Choose the correct alternative:
$(a)$ Acceleration due to gravity increases/decreases with increasing altitude.
$(b)$ Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to be a sphere of uniform density).
$(c)$ Acceleration due to gravity is independent of mass of the Earth/mass of the body.
$(d)$ The formula $-G M m(1 / r_{2}-1 / r_{1})$ is more/less accurate than the formula $m g(r_{2}-r_{1})$ for the difference of potential energy between two points $r_{2}$ and $r_{1}$ distance away from the centre of the Earth.