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Gravitation Part 1

Section: Physics

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Question : 20 of 100

Marks: +1, -0

If the angular velocity of Earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately (Take

2 g = 10 \mathrm{ms}^{-2}

, the radius of Earth,

R = 6400 \times 10^{3} \mathrm{m}

\pi = 3.14)

[18 Mar 2021 Shift 2]

$60\,\mathrm{min}$
does not change
$1200\,\mathrm{min}$
$84\,\mathrm{min}$

Solution:

Condition of weightlessness,

g_{\;\text{equator }\;}=0

g_{\;\text{equazar }\;}=g-\omega^{2}R

The radius of the Earth,

R=6400\,\mathrm{km}=6.4\times10^{6}\,\mathrm{m}

\Rightarrow

\omega = \sqrt{\;\frac{g}{R}}

\omega = \sqrt{\;\frac{10}{6.4 \times 10^{6}\,\mathrm{m}}} \Rightarrow \omega = \;\frac{1}{800}\,\mathrm{rad/s}

The duration of the day,

T = \;\frac{2\pi}{\omega} \Rightarrow T = \;\frac{2\pi}{\frac{1}{800}}

\Rightarrow T = 5024\,\mathrm{s} \Rightarrow T = \;\frac{5024}{60}\,\mathrm{min}

\Rightarrow \;\; T = 84\,\mathrm{min}

Hence, the duration of the day would be 84 min.

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