Weight of a body of mass m decreases by 1%&nb | vedclass

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Question

$g^{\prime}=g\left(1-\frac{2 h}{R}\right) ; \frac{\Delta g}{g}=\frac{2 h}{R}$

$1=2 \frac{\mathrm{h}}{\mathrm{R}} \Rightarrow \frac{\mathrm{h}}{\mat

Vedclass · Accepted Answer

decrease by $0.5\%$

Vedclass · Answer

$g^{\prime}=g\left(1-\frac{2 h}{R}\right) ; \frac{\Delta g}{g}=\frac{2 h}{R}$

$1=2 \frac{\mathrm{h}}{\mathrm{R}} \Rightarrow \frac{\mathrm{h}}{\mathrm{R}}=\frac{1}{2} ; \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$

$\frac{\Delta g^{\prime}}{g}=\frac{d}{R} \Rightarrow \frac{h}{R} \quad$ $g$ decreases by $0.5 \%$

Weight of a body of mass $m$ decreases by $1\%$ when it is raised to height $h$ above the Earth's surface. If the body is taken to a depth $h$ in a mine, then its weight will

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