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Q79 (IAS/2003) Science & Technology › Basic Science (Physics, Chemistry, Biology) › Basic mechanics concepts Answer Verified

If the radius of the earth were shrink by one per cent , its mass remaining the same, the value of ‘g’ on the earth’s surface would

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Your answer: —  ·  Correct: B
Explanation

Surface gravity g = GM/R^2, so if Earth's mass M is constant and radius R decreases to R' = 0.99R, the new gravity g' = GM/(0.99R)^2 = g/(0.99^2). Thus g'/g = 1/(0.99^2) ≈ 1.020408, an increase of about 2.04% (rounded to 2%). The inverse-square dependence and the numerical substitution give the result that g increases by ≈2% when radius shrinks by 1% with mass unchanged .

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