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

The correct option is Increase by 2%.

To understand why this is the case, we can look at the formula for the acceleration due to gravity on the surface of a spherical body like the Earth.

The acceleration due to gravity $g$ is given by the formula:
$g=\frac{GM}{R^2}$
where:
• $G$ is the universal gravitational constant,
• $M$ is the mass of the Earth, and
• $R$ is the radius of the Earth.

Since the mass $M$ of the Earth remains constant and $G$ is a constant, the acceleration due to gravity is inversely proportional to the square of the radius:
$g\propto R^{-2}$

For small percentage changes, we can find the relation between the relative change in $g$ and the relative change in $R$ by taking the natural logarithm on both sides and differentiating:
$\ln \left(g\right)=\ln \left(GM\right)-2\ln \left(R\right)$

Differentiating both sides gives the fractional change:
$\frac{dg}{g}=-2\frac{dR}{R}$

Expressing this relation in terms of percentage change by multiplying both sides by 100:
$(\frac{dg}{g}\times 100)=-2\times (\frac{dR}{R}\times 100)$

The problem states that the radius of the Earth shrinks by 1%. Since it is shrinking, this is a negative change:
$\frac{dR}{R}\times 100=-1\%$

Substituting this value back into the percentage change equation:
$\text{Percentage change in}g=-2\times (-1\%)=+2\%$

Since the result is positive, the acceleration due to gravity on the Earth's surface would increase by 2%.