Two equal masses m and m are hung from a bala | vedclass

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Question

$\mathrm{g}_{1}=\frac{\mathrm{g}}{\left[1+\frac{\mathrm{h}}{\mathrm{R}}\right]^{2}}=\mathrm{g}\left[1-\frac{2 \mathrm{h}}{\mathrm{R}}\right]$

$\mat

Vedclass · Accepted Answer

$\frac{8}{3}\pi G\rho mh$

Vedclass · Answer

$\mathrm{g}_{1}=\frac{\mathrm{g}}{\left[1+\frac{\mathrm{h}}{\mathrm{R}}ight]^{2}}=\mathrm{g}\left[1-\frac{2 \mathrm{h}}{\mathrm{R}}ight]$

$\mathrm{W}_{2}-\mathrm{W}_{1}=\mathrm{mg}_{2}-\mathrm{mg}_{1}$

$=2 \mathrm{mg}\left[\frac{\mathrm{h}_{1}}{\mathrm{R}}-\frac{\mathrm{h}_{2}}{\mathrm{R}}ight]=2 \mathrm{m} \frac{\mathrm{GM}}{\mathrm{R}^{2}} \frac{\mathrm{h}}{\mathrm{R}}$

$\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} 	ext { and } \mathrm{h}_{1}-\mathrm{h}_{2}=\mathrm{h}ight]$

$	herefore \mathrm{W}_{2}-\mathrm{W}_{1}=$ error in weighing

$=2 \mathrm{mG} \cdot \frac{4}{3} \pi \mathrm{R}^{3} ho \frac{\mathrm{h}}{\mathrm{R}^{3}}=\frac{8 \pi}{3} \mathrm{Gm} ho \mathrm{h}$

Two equal masses $m$ and $m$ are hung from a balance whose scale pans differ in vertical height by $'h'$. The error in weighing in terms of density of the earth $\rho $ is

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