When the radius of earth is reduced by 2% without changing the mass and acceleration due to gravity on surface will?

If we consider the Earth to be a perfect sphere, then the acceleration due to gravity at its surface is given by \[g = G\frac{M}{R^2}\]Here, M is the mass of Earth; R is the radius of the Earth and G is universal gravitational constant.

If the radius of the earth is decreased by 1%, then the new radius becomes

\[R' = R - \frac{R}{100} = \frac{99}{100}R\]

\[ \Rightarrow R' = 0 . 99R\] 

New acceleration due to gravity will be given by

\[g' = G\frac{M}{R '^2} = G\frac{M}{(0 . 99R )^2}\]

\[ \Rightarrow g' = 1 . 02 \times \left( G\frac{M}{R^2} \right) = 1 . 02g\] 

Hence, the value of the acceleration due to gravity increases when the radius is decreased.
Percentage increase in the acceleration due to gravity is given by

\[\frac{g' - g}{g} \times 100\]

\[ = \frac{0 . 02g}{g} \times 100\]

= 2 %