Two bodies each of mass $m$ are hung from a balance whose scale pans differ in a vertical height by $h$. If…
Two bodies each of mass $m$ are hung from a balance whose scale pans differ in a vertical height by $h$. If the mean density of the earth is $\rho$, the error in weighing is
$\frac{4 \pi \rho G m h}{3}$
$\frac{3 \pi \rho \mathrm{Gmh}}{4}$
$\frac{8 \pi \rho \mathrm{Gmh}}{3}$
$\frac{3 \pi \rho \mathrm{Gmh}}{8}$
Solution
Gravitational force on a body of mass $m$ at height $h$ due to earth,
$
F=\frac{G M_e m}{(R+h)^2}
$
where, $M_e$ is mass of the earth.
$
\because \text { Density of earth, } \rho=\frac{\text { mass of earth }\left(M_e\right)}{\text { volume of earth }(V)}
$
or $M_e=\rho \cdot V=\rho\left(\frac{4}{3} \pi R^3\right)$
$
\begin{gathered}
\therefore \quad F=\frac{G\left(\frac{4}{3} \pi R^3\right) \rho m}{(R+h)^2} \\
\text { or } \quad F=\frac{G\left(\frac{4}{3} \pi R\right) \rho m}{\left(1+\frac{h}{R}\right)^2} \\
\text { or } \quad F=\frac{4}{3} \pi G R \rho m\left(1+\frac{h}{R}\right)^{-2}
\end{gathered}
$
or
$
F=\frac{4}{3} \pi G R \rho m\left(1+\frac{h}{R}\right)^{-2}
$
By using Binomial expansion,
$
F=\frac{4}{3} \pi G R \rho m\left(1-\frac{2 h}{R}\right)
$
$\left[\because \quad(1+x)^n=1+n x+\frac{n(n-1)}{2 !} \cdot x^2+\ldots\right.$ and neglecting higher terms.]
Difference in weight $=\frac{4}{3} \pi G R \rho m-\frac{4}{3} \pi G \rho m R\left(\frac{2 h}{R}\right)$
Hence, error in weight $=\frac{8}{3} \pi \rho G m h$