The radius of the earth is 6400 , km and g = | vedclass

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(C) The effective acceleration due to gravity at the equator is given by $g' = g - R\omega^2$. For a body to weigh zero at the equator,the effective g

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$1/800 \, rad/s$

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(C) The effective acceleration due to gravity at the equator is given by $g' = g - R\omega^2$. For a body to weigh zero at the equator,the effective gravity must be zero,so $g' = 0$. This implies $g = R\omega^2$,or $\omega = \sqrt{\frac{g}{R}}$. Given $g = 10 \, m/s^2$ and $R = 6400 \, km = 6.4 	imes 10^6 \, m$. Substituting these values: $\omega = \sqrt{\frac{10}{6.4 	imes 10^6}} = \sqrt{\frac{1}{6.4 	imes 10^5}} = \sqrt{\frac{1}{64 	imes 10^4}} = \frac{1}{8 	imes 10^2} = \frac{1}{800} \, rad/s$.

The radius of the earth is $6400 \, km$ and $g = 10 \, m/s^2$. In order that a body of $5 \, kg$ weighs zero at the equator,the angular speed of the earth is

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