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 $= 6400 \,km$]

• 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
$w'=w-m R \omega^{2}$
Here, $w'=(3 / 5) w=(3 / 5) \,m g $
${[\because w=m g]} $
$\therefore (3 / 5) m g=m g-m R \omega^{2}$
Or $m R \omega^{2}=m g-(3 / 5) m g\left(\frac{2}{5}\right) m g $
Or $\omega^{2}=\frac{2 g}{5 R} $
$\therefore \omega=\sqrt{\frac{2 g}{5 R}}$
Here, $g=9.8\, ms ^{-2}$ and
$R=6400\, km $
$=6400 \times 10^{3}\, m $
$\therefore \omega=\sqrt{\left(\frac{2}{5} \times \frac{9.8}{6400 \times 10^{3}}\right)} rads ^{-1} $
$=7.82 \times 10^{-4}\, rad\,s ^{-1}$