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

Explanation

`g=(Gm)/(R^2)` if R decreases by 1% it becomes `(99)/(100)R`
`g'= (GM)/((99R)^2)=1.02(Gm)/(R^2)= (1+0.02)(Gm)/(R^2)`
g' increases by `0.02 (Gm)/(R^2)`, therefore increases by 2%.