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(A) The weight of a body at height $h$ is given by $W_h = W \left(1 - \frac{2h}{R}\right)$,where $W = mg$ is the weight at the surface.Percentage decr

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$0.75 \%$ decrease

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(A) The weight of a body at height $h$ is given by $W_h = W \left(1 - \frac{2h}{R}ight)$,where $W = mg$ is the weight at the surface.Percentage decrease in weight at height $h$ is $\frac{W - W_h}{W} 	imes 100 = \frac{2h}{R} 	imes 100 = 1.5 \%$.From this,we get $\frac{h}{R} = \frac{1.5}{2 	imes 100} = 0.75 \%$.The weight of a body at depth $h$ is given by $W_d = W \left(1 - \frac{h}{R}ight)$.Percentage decrease in weight at depth $h$ is $\frac{W - W_d}{W} 	imes 100 = \frac{h}{R} 	imes 100$.Substituting the value of $\frac{h}{R}$,we get $\frac{h}{R} 	imes 100 = 0.75 \%$.Thus,the weight decreases by $0.75 \%$.

Weight of a body decreases by $1.5 \%$ when it is raised to a height $h$ above the surface of the Earth. When the same body is taken to the same depth $h$ in a mine,its weight will show ........

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