When the value of acceleration due to gravity ' g ' becomes g/3 above surface of height ' h ' then relation between ' h ' and ' R ' is ( R = radius of earth) (a) h = (R)/(√(3)-1) (b) h = √(3)/R (c) h = (√(2)-1)R (d) h = (√(3)-1)R

Answer: (d). (R/R+h)² = 1/3 R + h = √(3)R h = (√(3) - 1)R So the answer is (d) h = (√(3)-1)R.

When the value of acceleration due to gravity ' g ' becomes g/3 above… | MHT CET 2021 (21 Sep, Shift 1)

Q86MHT CET 2021 · 21 SEP · SHIFT 1Easy

Acceleration Due to Gravity

When the value of acceleration due to gravity ' $g$ ' becomes $\frac{g}{3}$ above surface of height ' $h$ ' then relation between ' $h$ ' and ' $R$ ' is ( $R =$ radius of earth)

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Answer: (d)

$\left(\frac{R}{R+h}\right)^2 = \frac{1}{3}$
$R + h = \sqrt{3}R$
$h = (\sqrt{3} - 1)R$
So the answer is (d) $h = (\sqrt{3}-1)R$ .

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