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The speed of earth's rotation about its axis is $\omega$. Its speed is increased to $x$ times to make the effective acceleration due to gravity equal to zero at the equator. Then $x$ is
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$17$
Earth rotates about its own axis in 24 h .
So, $\quad T=24 \times 60 \times 60 \mathrm{~s}$
So, angular speed of earth about its own axis
$\omega=\frac{2 \pi}{T}=\frac{2 \pi}{24 \times 60 \times 60} \mathrm{rad} / \mathrm{s}$
At equator, $g^{\prime}=g-R_e \omega^{\prime 2}$
or $\quad 0=g-R_e \omega^{\prime 2}$
or $\omega^{\prime}=\sqrt{\frac{g}{R_e}}=\frac{2 \pi}{84.6 \times 60}$
$\therefore \quad \frac{\omega^{\prime}}{\omega}=\frac{24 \times 60 \times 60}{84.6 \times 60} \approx 17$
$\Rightarrow \quad \omega^{\prime}=17 \omega$
So, value of $x$ is 17 .
So, $\quad T=24 \times 60 \times 60 \mathrm{~s}$
So, angular speed of earth about its own axis
$\omega=\frac{2 \pi}{T}=\frac{2 \pi}{24 \times 60 \times 60} \mathrm{rad} / \mathrm{s}$
At equator, $g^{\prime}=g-R_e \omega^{\prime 2}$
or $\quad 0=g-R_e \omega^{\prime 2}$
or $\omega^{\prime}=\sqrt{\frac{g}{R_e}}=\frac{2 \pi}{84.6 \times 60}$
$\therefore \quad \frac{\omega^{\prime}}{\omega}=\frac{24 \times 60 \times 60}{84.6 \times 60} \approx 17$
$\Rightarrow \quad \omega^{\prime}=17 \omega$
So, value of $x$ is 17 .
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