If the radius of the earth were to shrink by one percent its mass remaining the same, the acceleration due to greavity on the earth's surface would

Explanation

Consider the case of a body of mass m placed on the earth's surface (mass of the earth M and radius R).
`g_(s) = (GM_(e ))/(R_(e )^(2))` ….(i)
Now, when the radius reduced by `1%` i.e., become `0.99 R`, let acceleration due to gravity be g' then
`g' = (GM)/((0.99)^(2))` ....(ii)
From equation (i) and (ii), we get
`(g')/(g) = (R^(2))/((0.99)R^(2)) = (1)/((0.99)^(2))`
or `g' = gxx((1)/((0.99)))^(2)`
or `g' = 1.02 g`
Thus the value of g is increased by `2%`