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Updated on : 2022-09-05

Solution

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Correct option is D)

Lrt $R_{e}$ be the radius of earth

Given $h=R_{e}/2$Since h is comparable to $R_{e}$

So, $g_{h}=(R_{e}+h)_{2}GM $

Putting $h=R_{e}/2$,we get

$⇒g_{h}=94 ×R_{e}GM =94g $

Where g is acceleration due to gravity at the surface of earth

Therefore weight at a height equal to half of the earth's radius $=94 ×72=32N$

Hence correct answer is option $D$

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