If the radius of earth shrinks by 1% and its mass remains the same, then acceleration due to gravity on the earth’s surface will

Acceleration due to gravity is given by:

$g=\frac{GM}{R^2}$
$g\propto \frac{1}{R^2}$
Radius is decreased by 1%,
$R_1=\frac{99}{100}R$
$\frac{g_1}{g}=\frac{R^2}{{R^2}_1}$

$\frac{g_1}{g}=\frac{{\left(100\right)}^2{R}^2}{{\left(99\right)}^2{R}^2}$=$\frac{{100}^2}{{99}^2}$
$\% Change in g= \frac{g_1-g}{g}\times 100=(\frac{g_1}{g}-1) \times 100$

$= (\frac{{100}^2}{{99}^2}-1) \times 100$
$=2.03\%$
Hence option B is the correct answer.