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The acceleration due to gravity at a height ${(1 / 20)}^{t h}$ of the radius of earth above the earth’s surface is $9 m . s^{- 2}$ . Its value at an equal depth below the surface of earth is

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$9.25 m . s^{- 2}$

$9.8 m . s^{- 2}$

$9 m . s^{- 2}$

$9.5 m . s^{- 2}$

answer is C.

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Detailed Solution

$g_{h} = 9 m / s^{2}, h = R / 20$

$g_{h} = g {[1 - \frac{h}{R}]}^{- 2} \Rightarrow 9 = g {[1 - \frac{1}{20}]}^{- 2}$

$g = \frac{9}{{(19 / 20)}^{2}} = \frac{9 \times 400}{361} = 9.97$

$g_{d} = ?, d = R / 20$

$= 9.97 [\frac{19}{20}]$

$g_{d} = (0.498) (19) \approx 9.5 m / s^{2}$

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