A man weights 80 kg on earth surface. The height above ground where he will weigh 40kg is (radius of earth is r)

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0.414 times r

0.31 times r

0.51 times r

0.61 times r

answer is B.

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Detailed Solution

$\begin{array}{l} g_{h} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \\ {mg}_{h} = \frac{mg}{{(1 + \frac{h}{R})}^{2}} \Rightarrow 40 kgwt = \frac{80 kgwt}{{(1 + \frac{h}{R})}^{2}} \\ {(1 + \frac{h}{R})}^{2} = 2 \Rightarrow (1 + \frac{h}{R}) = \sqrt{2} \Rightarrow 1 + \frac{h}{R} = 1 .414 \\ h = 0 .414 R = 0.414 r \end{array}$

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