If radius of the earth were to shrink by 1%, its mass remaining the sa

If radius of the earth were to shrink by 1%, its mass remaining the same, g would decrease by nearly

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2015

The correct option is B. g would increase by nearly 2%. (Note: The question contains a likely typo stating “decrease” instead of “increase” or “change”).

The acceleration due to gravity (g) on the surface of a sphere of mass M and radius R is given by g = GM/R², where G is the gravitational constant.
If the radius R shrinks by 1%, the new radius R’ = R – 0.01R = 0.99R. The mass M remains the same.
The new gravity g’ on the surface would be g’ = GM/(R’)² = GM/(0.99R)² = GM/(0.9801R²) = (1/0.9801) * (GM/R²) ≈ 1.01928 * g.
The change in gravity is g’ – g = 1.01928g – g = 0.01928g.
The percentage change in gravity is ((g’ – g)/g) * 100% = (0.01928g / g) * 100% ≈ 1.93%.
Using the approximation for small changes, if R changes by a small fraction ε (R’ = R(1-ε)), then g changes by approximately 2ε (g’ ≈ g(1+2ε)). Here ε = 0.01 (1% decrease in R). So g changes by approximately 2 * 0.01 = 0.02, which is a 2% increase.

A decrease in Earth’s radius while keeping the mass constant would result in a stronger gravitational pull at the surface, hence an *increase* in ‘g’. The question states “g would decrease by nearly”, which contradicts the physics. Assuming the question intends to ask about the magnitude of the percentage change calculated from the given change in radius, the change is approximately 2% (an increase). Option B (2%) is the closest value to the calculated change. It is highly probable that “decrease” is a typo.

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