Question

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

• A. $3.28\times {10}^{-4}rad/sec$
• B. $7.28\times {10}^{-3}rad/sec$
• C. $3.28\times {10}^{-3}rad/sec$
• D. $7.826\times {10}^{-4}rad/sec$

Answer 1

The correct option is C $3.28\times {10}^{-3}rad/sec$

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