If the radius of the earth shrinks by $1 \%$, its mass remaining the same, then the acceleration due to gravity on the earth surface would

Acceleration due to gravity, $g_1=\frac{G M}{R_1^2}$ As, $M$ remain constant and $R$ becomes $R_2=R_1-1 \% \text { of } R_1=R_1-\frac{R_1}{100}=\frac{99}{100} R_1$ Therefore, the new gravitational acceleration becomes, $g_2=\frac{G M}{R_2^2}=\frac{G M \times(100)^2}{(99)^2 R_1^2}=\frac{(100)^2}{(99)^2} g_1=1.02 g_1$ Change in the gravitational acceleration, $g_2-g_1=1.02 g_1-g_1=0.02 g_1$ Hence, the percentage increase is $\frac{g_2-g_1}{g_1} \times 100=0.02 \times 100=2 \%$ So, if the radius of earth shrinks by $1 \%$, the value of $g$ would increase by $2 \%$.