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  • If the angular velocity of earth;s spin is increased such that the both bodies at the equator start floating, the duration of the day would be approimately: [Take g = 10 ms-2,       the radius of earth, <img alt="R=6

If the angular velocity of earth;s spin is increased such that the both bodies at the equator start floating, the duration of the day would be approimately: [Take g = 10 ms-2,       the radius of earth, R=6400 \times 10^3 m, Take \pi = 3.14]
Option: 1 does not change
Option: 2 1200 minutes
Option: 3 60 minutes
Option: 4 84 minutes

Answers (1)

best_answer

\\ \text{For objects to float}
mg = m \omega^{2} R\\ \\ \omega= \ \text{angular velocity of earth.} \\ \text{R = Radius of earth}

\omega=\sqrt{\frac{ g }{ R }} \dots(1)
\begin{aligned} &T =\frac{2 \pi}{\omega}\\ &\Rightarrow T =2 \pi \sqrt{\frac{ R }{ g }}\\\\ &=2 \pi \sqrt{\frac{6400 \times 10^{3}}{10}}\\\\ &\Rightarrow \frac{ T }{60}=83.775 \text { minutes }\\\\ &\simeq 84 \text { minutes } \end{aligned}

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Deependra Verma

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