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MHT CET 2024 (4 May, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2024 (4 May, Shift 1)

Q30MHT CET 2024 · 4 MAY · SHIFT 1Easy

Acceleration Due to Gravity

The height ' $h$ ' above the earth's surface at which the value of acceleration due to gravity $(g)$ becomes $\left(\frac{g}{3}\right)$ is ( $R =$ radius of the earth)

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Answer: (b)

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

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