Question

The speed with which the earth have to rotate on its axis so that a person on the equator would weigh ${(3/5)}^{th}$ as much as present.
[Radius of earth $=6400km$]

• A. $4.83\times {10}^{-3}rad{s}^{-1}$
• B. $5.41\times {10}^{-3}rad{s}^{-1}$
• C. $7.82\times {10}^{-4}rad{s}^{-1}$
• D. $8.88\times {10}^{-14}rad{s}^{-1}$

Answer 1

Answer: (C)

The apparent weight of person on the equator (latitude $\lambda =0$ ) is given by
$w^{\prime }=w-mR{\omega }^2$
Here, $w^{\prime }=(3/5)w=(3/5)mg$
$[\because w=mg]$
$\therefore (3/5)mg=mg-mR{\omega }^2$
Or $mR{\omega }^2=mg-(3/5)mg\left(\frac{2}{5}\right)mg$
Or ${\omega }^2=\frac{2g}{5R}$
$\therefore \omega =\sqrt{\frac{2g}{5R}}$
Here, $g=9.8m{s}^{-2}$ and
$R=6400km$
$=6400\times 10^{3}m$
$\therefore \omega =\sqrt{\left(\frac{2}{5}\times \frac{9.8}{6400\times 10^3}\right)}rad{s}^{-1}$
$=7.82\times 10^{-4}rad{s}^{-1}$