Question

Two equal masses 'm' are hung from balances whose scale level differ in vertical height by h. Calculate the error in weighing, if any, in terms of density of earth

• A. $\frac{2}{3}\pi \rho R^{3}Gm$
• B. $\frac{8}{3}\pi \rho Gmh$
• C. $\frac{8}{3}\pi \rho R^{3}Gm$
• D. $\frac{8}{3}\pi \rho G{m}^{2}h$

Answer 1

Answer: (B)

$\frac{8}{3}\pi \rho Gmh$

$\begin{array}{c}{m}_2=m\\ \text{Let}{g}_2=g\\ g_1=g{(1+\frac{h}{R})}^{-2}\end{array}$

$\begin{array}{c}{m}_1{g}_1=m_2{g}_2\\ m_1=m_2\frac{g_2}{g_1}\\ m_1=m_2{(1+\frac{h}{R})}^2\\ \Delta m_{1}g=\mid m_2\left(1{(1+\frac{h}{R})}^2\right)\mid g\end{array}$

$\begin{array}{c}=m_2(1-(1+\frac{2h}{R}))\\ =\frac{2h{m}_{2}g}{R}=\frac{2hmg}{R}\end{array}$

$\begin{array}{cc}\frac{GM}{R^2}& =g\\ \text{also,}\rho & =\frac{M}{V}\end{array}$

$\rho =$$\frac{M}{\frac{4}{3}\pi R^3}$

$\begin{array}{c}\therefore \frac{GMR}{R^3}=g\\ \therefore GR\times \frac{4}{3}\pi f=g\\ \frac{4}{3}GR\rho \pi =g\\ \Delta mg=\frac{2hm4}{R}GR{\rho }^{-}\\ =\frac{8}{3}\pi hG\cdot \rho m\end{array}$