Question
An object weights $72 \mathrm{~N}$ on the earth. Its weight at a height $(\mathrm{R} / 2)$ from earth is $=\ldots \ldots \ldots \ldots \ldots \mathrm{N}$(A) 32(B) 56(C) 72(D) zero
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On the surface of the Earth, $g$ is approximately $9.8 \, m/s^2$. Show more…
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An object weights 72 N on earth. Its weight at a height of R/2 from earth is (a) 32 N (b) 56 N (c) 72 N (d) Zero
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At surface of earth weight of a person is $72 N$ then his weight at height $R / 2$ from surface of earth is ( $R$ = radius of earth) (a) $28 N$ (b) $16 N$ (c) $3^{2} N$ (d) $72 N$
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