If a man at the equator would weigh (3/5)^{th | vedclass

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(A) At the equator,the effective acceleration due to gravity $g'$ is given by $g' = g - \omega^2 R$,where $g$ is the acceleration due to gravity at th

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$\sqrt {\frac{2}{5} \frac{g}{R}} $

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(A) At the equator,the effective acceleration due to gravity $g'$ is given by $g' = g - \omega^2 R$,where $g$ is the acceleration due to gravity at the poles,$\omega$ is the angular speed of the Earth,and $R$ is the radius of the Earth.Given that the man's weight at the equator is $(3/5)$ of his actual weight,we have $W' = (3/5)W$.Since $W = mg$,this implies $mg' = (3/5)mg$,which simplifies to $g' = (3/5)g$.Substituting this into the equation for effective gravity: $(3/5)g = g - \omega^2 R$.Rearranging the terms: $\omega^2 R = g - (3/5)g = (2/5)g$.Solving for $\omega$: $\omega^2 = (2/5)(g/R)$,therefore $\omega = \sqrt{(2/5)(g/R)}$.

If a man at the equator would weigh $(3/5)^{th}$ of his weight,the angular speed of the earth is

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