Let `M,R` be tha mass and redius of the Earth. The acceleration due to gravity on the surface of Earth is given by
`g = GM//R^(2)` ..(i)
Differentiating it, we have,
`dg = - (2GM)/(R^(3))dR` ..(ii) Dividing (iii) by (i), we have,
`(dg)/(g) = - (2dR)/(R )` ..(iii)
When redius of earth shrinks by `2%` then,
`(dR)/(R ) = - (2)/(100)` ..(iv)
From (iii) & (iv), the `%` increase in the value of `g` is
`= (dg)/(g) xx 100 = - (2dR)/(R ) xx 100 = - [(-2)/(100)] xx 100`
`= 4 %`