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A Person Stands on a Spring Balance at the Equator. If the Speed of Earth'S Rotation is Increased by Such an Amount that the Balance Reading is Half the True Weight, What Will Be the Length

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प्रश्न

A person stands on a spring balance at the equator. If the speed of earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case?

संख्यात्मक
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उत्तर

 When the balance reading is half, we have :

\[\text { True weight }= \frac{\text{mg - m }\omega^2 r}{\text{mg}} = \frac{1}{2}\] 

\[\Rightarrow \omega^2 r = \frac{g}{2}\]

\[ \Rightarrow \omega = \sqrt{\frac{g}{2r}}\]

\[ = \sqrt{\frac{10}{2 \times 6400 \times {10}^3}} \text{rad/s}\]

\[\therefore \text { Duration of the day }= 2\pi \times \sqrt{\frac{2 \times 6400 \times {10}^3}{9 . 8}}\]s

\[ = 2\pi\sqrt{\frac{6 . 4 \times {10}^7}{49}}\]s

\[ = \frac{2\pi \times 8000}{7 \times 3600} \text{ h = 2 h}\]

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अध्याय 7: Circular Motion - Exercise [पृष्ठ ११५]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 7 Circular Motion
Exercise | Q 17.2 | पृष्ठ ११५

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