What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value that at the surface of earth? (radius of earth $=R$ )

The acceleration due to gravity about the earth's surface would be half of its value on the surface of the earth at an altitude of ......... $mile$. ($R = 4000$ mile)

If the radius of the earth shrinks by $1.5\%$ (mass remaining same), then the value of acceleration due to gravity changes by ....... $\%$.

At a certain depth "$d$" below surface of earth. value of acceleration due to gravity becomes four times that of its value at a height $3\,R$ above earth surface. Where $R$ is Radius of earth (Take $R =6400\,km$ ). The depth $d$ is equal to $............\,km$

Spot the wrong statement :The acceleration due to gravity $‘g’$ decreases if

Assuming the earth to be a sphere of uniform density, the acceleration due to gravity inside the earth at a distance of $r$ from the centre is proportional to