Q174 / 200 · Manipal (MET) 2013 · Physics
The speed with which the earth have to rotate on its axis so that a person on the equator would weight $(3 / 5)^{\text {th }}$ as much as present. [Radius of earth $=6400 \mathrm{~km}$ ]
- A$4.83 \times 10^{-3} \mathrm{rads}^{-1}$
- B$5.41 \times 10^{-3} \mathrm{rads}^{-1}$
- Correct answerC$7.82 \times 10^{-4} \mathrm{rads}^{-1}$
- D$8.88 \times 10^{-14} \mathrm{rads}^{-1}$
Solution
The apparent weight of person on the equator (latitude $\lambda=0$ ) is given by
$\omega^{\prime}=w-m R \omega^2$
Here, $\therefore \quad \omega^{\prime}=(3 / 5) \mathrm{w}=(3 / 5) \mathrm{mg} \quad[\because \omega=m g]$
or $\quad m R \omega^2=m g-(3 / 5) m g=\left(\frac{2}{5}\right) m g$
or $\quad \omega^2=\frac{2 g}{5 R}$
$\omega=\sqrt{\frac{2 g}{5 R}}$
Here, $g=9.8 \mathrm{~ms}^{-2}$
and $\quad R=6400 \mathrm{~km}=6400 \times 10^3 \mathrm{~m}$
$\therefore \quad \omega=\sqrt{\left(\frac{2}{5} \times \frac{9.8}{6400 \times 10^3}\right)} \mathrm{rads}^{-1}$
$=7.82 \times 10^{-4} \mathrm{rads}^{-1}$
Hence, the correct answer is option: 'C'.