The height above the earth's surface at which the acceleration due to gravity becomes (1/n) times the value at the surface is ( R = radius of earth) (a) R/√(n) (b) R·√(n) (c) (√(n)+1)R (d) (√(n)-1)R

Answer: (d). (R/R+h)² = 1/n R + h = √(n)R h = (√(n) - 1)R So the answer is (d) (√(n)-1)R.

The height above the earth's surface at which the acceleration due to… | MHT CET 2024 (2 May, Shift 1)

Q32MHT CET 2024 · 2 MAY · SHIFT 1Easy

Acceleration Due to Gravity

The height above the earth's surface at which the acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface is ( $R =$ radius of earth)

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Answer: (d)

$\left(\frac{R}{R+h}\right)^2 = \frac{1}{n}$
$R + h = \sqrt{n}R$
$h = (\sqrt{n} - 1)R$
So the answer is (d) $(\sqrt{n}-1)R$ .

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