# A body weighs

Question:

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, $\mathrm{R}=6400 \mathrm{~km} .]$

1. $49 \mathrm{~N}$

2. $48.83 \mathrm{~N}$

3. $49.83 \mathrm{~N}$

4. $49.17 \mathrm{~N}$

Correct Option: , 2

Solution:

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.