What is the height from the surface of earth,where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth (in $km$)? $(R_E = 6400 \ km)$

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 variation of acceleration due to gravity $(g)$ with distance $(r)$ from the center of the earth is correctly represented by ... (Given $R =$ radius of earth)

If the density of the Earth increases $4$ times and its radius becomes half of its present value,our weight will

If the radius of the earth is increased by a factor of $5$,by what factor must its density be changed to keep the value of $g$ the same?

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth).