The weight of a body on the earth is 400,N. T | vedclass

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(D) The weight of a body on the surface of the earth is given by $W = mg = 400\,N$.At a depth $d$ below the surface of the earth,the acceleration due

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$200$

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(D) The weight of a body on the surface of the earth is given by $W = mg = 400\,N$.At a depth $d$ below the surface of the earth,the acceleration due to gravity $g'$ is given by the formula $g' = g(1 - \frac{d}{R})$,where $R$ is the radius of the earth.Given that the depth $d = \frac{R}{2}$,we substitute this into the formula:$g' = g(1 - \frac{R/2}{R}) = g(1 - \frac{1}{2}) = \frac{g}{2}$.The new weight $W'$ of the body at this depth is $W' = mg' = m(\frac{g}{2}) = \frac{mg}{2}$.Since $mg = 400\,N$,we have $W' = \frac{400}{2} = 200\,N$.

The weight of a body on the earth is $400\,N$. Then weight of the body when taken to a depth half of the radius of the earth will be ............ $N$.

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