At what altitude in metres will the acceleration due to gravity be $25\%$ of that at the earth's surface? (Radius of earth $= R \ m$)

The time period of a simple pendulum is $T_1$ when on the Earth's surface and $T_2$ when taken to a height $h = 2R$ above the Earth's surface, where $R$ is the radius of the Earth. The ratio $T_1 : T_2$ is:

The magnitudes of the gravitational field at distances $r_1$ and $r_2$ from the center of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively. Then:

If the radius of earth shrinks by $2 \%$ while its mass remains the same,the acceleration due to gravity on the earth's surface will approximately:

The height $h$ from the surface of the earth at which the value of $g$ will be reduced by $64 \%$ from the value at the surface of the earth is ($R=$ radius of the earth).

If the radius of the earth were to shrink by $1\%$ while its mass remains the same,the acceleration due to gravity on the earth's surface would