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The speed of earth's rotation about its axis is $\omega$ . Its speed increases to $\mathrm{x}$ times to make the effective acceleration due to gravity equal to zero at the equator. Then $\mathrm{\mathrm{x}}$ is

Option: 1
$1$

Option: 2
$8.5$

Option: 3
$17$

Option: 4
$34$

Answers (1)

best_answer

$\mathrm{ \mathrm{g}_{\mathrm{e}}^{\prime}=\mathrm{g}-\mathrm{R} \omega^2 }$

Let $\mathrm{ \omega_1 }$ be the angular speed of earth's rotation so that $\mathrm{ \mathrm{g}_{\mathrm{e}}^{\prime}=0 }$

$\mathrm{ 0=\mathrm{g}-\mathrm{R} \omega_1^2, \frac{\mathrm{R} \omega_1^2}{\mathrm{~g}}=1 }$ ............(1)

$\mathrm{ \text { also } \frac{\mathrm{R} \omega^2}{\mathrm{~g}}=\frac{6.4 \times 10^6}{9.8} \times\left(\frac{2 \pi}{24 \times 60 \times 60}\right)^2 }$

$\mathrm{ =\frac{1}{291} }$ ...........(2)

$\mathrm{ \frac{(1)}{(2)} ; \frac{\omega_1^2}{\omega^2}=291 (17.02)^2 }$

$\mathrm{ \omega_1 \simeq \omega \times 17=\omega \times \mathrm{x} }$

$\mathrm{ \therefore \quad \mathrm{x}=17 \text {. } }$

Hence option 3 is correct.

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vinayak

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