If the radius of earth shrinks by 2 % while its mass remains same. The acceleration due to gravity on the earth's surface will approximately :

Solution

Physics sir · 4 steps

Mass fixed, so g goes as 1/R²: a 2% smaller radius means about a 4% bigger g.

1. What g depends on. On the surface, g = GM/R². The mass M stays the same, so g changes only through R, as g ∝ R⁻².
2. Small change rule. For a small change, the percentage change in Rⁿ is n times the percentage change in R. Here n = -2 and the radius changes by -2%.

   (Δ g)/g× 100 = -2((Δ R)/R× 100) = -2 × (-2%) = +4%

3. Exact check. With R' = 0.98R, g'/g = 1/(0.98)² = 1/0.9604 ≈ 1.041. That is an increase of about 4.1%, so 4% is right.
4. So. A smaller Earth with the same mass pulls harder at its surface. The acceleration due to gravity increases by about 4%, which is option (d).

The idea to keep. If y ∝ xⁿ, a small percentage change in x gives n times that percentage change in y. Watch the sign of n: an inverse square flips decrease into increase.

Answer from the NTA final answer key. Solution by PrepWiser.