If the radius of the earth shrinks by 1 %,its | vedclass

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(C) The acceleration due to gravity on the surface of the Earth is given by $g = \frac{GM}{R^2}$.Since the mass $M$ remains constant,we have $g \propt

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increase by $2 \%$

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(C) The acceleration due to gravity on the surface of the Earth is given by $g = \frac{GM}{R^2}$.Since the mass $M$ remains constant,we have $g \propto \frac{1}{R^2}$.Taking the logarithmic differentiation,we get $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$.Given that the radius shrinks by $1 \%$,we have $\frac{\Delta R}{R} = -0.01$.Substituting this value,we get $\frac{\Delta g}{g} = -2 	imes (-0.01) = 0.02$.Therefore,the percentage change in $g$ is $0.02 	imes 100 = 2 \%$.Since the value is positive,the acceleration due to gravity increases by $2 \%$.

If the radius of the earth shrinks by $1 \%$,its mass remaining the same,then the acceleration due to gravity on the earth's surface would

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