Question

Two equal masses m and m are hung from a spring balance whose scale pans differ in vertical height by h. If $\rho$ is the mean density of earth ,then what is the error in weighing?

• A. $\frac{8}{3}\pi G\rho mh$
  
• B. $\frac{4}{3}\pi G\rho mh$
  
• C. $\frac{5}{3}\pi G\rho mh$
  
• D. $\infty$

Answer 1

The correct option is A $\frac{8}{3}\pi G\rho mh$

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

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

$\Delta m_{1}g=\frac{2h{m}_{2}g}{R}$$=\frac{2hmg}{R}$

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

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

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