If the radius of earth decreases by 10%, the mass remaining unchanged, what will happen to the acceleration due to gravity?

$\mathrm{g} = \frac{\mathrm{GM}}{{\mathrm{R}}^2}$  --------(i)

          ${\mathrm{g}}^{\text{'}} = \frac{\mathrm{GM}}{{\left({\displaystyle \frac{90}{100}}\mathrm{R}\right)}^2} = \frac{100}{81}\frac{\mathrm{GM}}{{\mathrm{R}}^2}$ -------(ii)

From (i) and (ii), $$\begin{gathered} {\mathrm{g}}^{\text{'}} = \frac{100}{81}\mathrm{g} \\ \frac{{\mathrm{g}}^{\text{'}}}{\mathrm{g}} = \frac{100}{81} \end{gathered}$$

$\frac{{\mathrm{g}}^{\text{'}}}{\mathrm{g}}-1 = \frac{100}{81}-1$

$∆\mathrm{g} = \frac{19}{81}\mathrm{g} = 23\% \mathrm{of} \mathrm{g}$

So increase is more than 19% of g.