A body weighs $49 \mathrm{~N}$ on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?
(Use $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}=9.8 \mathrm{~ms}^{-2}$ and radius of earth, $\mathrm{R}=6400 \mathrm{~km} .]$
Correct Option: , 2
Weight of pole $=\mathrm{mg}=49 \mathrm{~N}$
At equator due to rotation $=\mathrm{g}_{\mathrm{e}}=\mathrm{g}-\mathrm{R} \omega^{2}$
so $\mathrm{W}=\mathrm{mg} \mathrm{e}_{e}=\mathrm{m}\left(\mathrm{g}-\mathrm{R} \omega^{2}\right)$
$\therefore \mathrm{W}_{\mathrm{P}}>\mathrm{W}_{\mathrm{e}} \quad \mathrm{W}_{\mathrm{P}}=49 \mathrm{~N}$
So, $\mathrm{W}_{\mathrm{e}}=48.83 \mathrm{~N} . \quad \mathrm{W}_{\mathrm{e}}<49 \mathrm{~N}$
Option (2) is correct.
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