The radius of the Earth is 6400 ,km and the a | vedclass

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(C) At the equator,the effective acceleration due to gravity $g'$ is given by $g' = g - R\omega^2$.For the weight of the body to be zero,the effective

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$\frac{1}{800}$

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(C) At the equator,the effective acceleration due to gravity $g'$ is given by $g' = g - R\omega^2$.For the weight of the body to be zero,the effective gravity must be zero,which means $g' = 0$.Therefore,$g - R\omega^2 = 0$.This implies $R\omega^2 = g$.Solving for $\omega$:$\omega^2 = \frac{g}{R}$$\omega = \sqrt{\frac{g}{R}}$Given $g = 10 \,ms^{-2}$ and $R = 6400 \,km = 6.4 	imes 10^6 \,m$.$\omega = \sqrt{\frac{10}{6.4 	imes 10^6}}$$\omega = \sqrt{\frac{10}{64 	imes 10^5}} = \sqrt{\frac{1}{64 	imes 10^4}}$$\omega = \frac{1}{8 	imes 10^2} = \frac{1}{800} \,rad/s$.

The radius of the Earth is $6400 \,km$ and the acceleration due to gravity is $g=10 \,ms^{-2}$. For the weight of a body of mass $5 \,kg$ to be zero at the equator,the rotational angular velocity of the Earth must be (in $rad/s$):

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