Two equal masses m and m are hung from a balance whose scale pans differ in vertical height by '

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7.Gravitation

hard

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

$\pi G\rho mh$

$\frac{1}{3}\pi G\rho mh$

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

$\frac{4}{3}\pi G\rho mh$

Solution

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

$\mathrm{W}_{2}-\mathrm{W}_{1}=\mathrm{mg}_{2}-\mathrm{mg}_{1}$

$=2 \mathrm{mg}\left[\frac{\mathrm{h}_{1}}{\mathrm{R}}-\frac{\mathrm{h}_{2}}{\mathrm{R}}\right]=2 \mathrm{m} \frac{\mathrm{GM}}{\mathrm{R}^{2}} \frac{\mathrm{h}}{\mathrm{R}}$

$\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \text { and } \mathrm{h}_{1}-\mathrm{h}_{2}=\mathrm{h}\right]$

$\therefore \mathrm{W}_{2}-\mathrm{W}_{1}=$ error in weighing

$=2 \mathrm{mG} \cdot \frac{4}{3} \pi \mathrm{R}^{3} \rho \frac{\mathrm{h}}{\mathrm{R}^{3}}=\frac{8 \pi}{3} \mathrm{Gm} \rho \mathrm{h}$

Standard 11

Physics

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easy

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hard

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

Solution

Similar Questions

Two masses $m_1$ and $m_2\, (m_1 < m_2)$ are released from rest from a finite distance. They start under their mutual gravitational attraction

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