Question

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

• A. $\frac{4 \pi \rho\, Gmh }{3}$
• B. $\frac{3 \pi \rho\, Gmh }{4}$
• C. $\frac{8 \pi \rho \,Gmh }{3}$
• D. $\frac{3 \pi \rho\, Gmh }{8}$

Answer 1

Answer: (C)

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.
$\because$ Density of earth, $\rho=\frac{\text { mass of earth }\left(M_{e}\right)}{\text { volume of earth }(V)}$
or $M_{e}=\rho \cdot V=\rho\left(\frac{4}{3} \pi R^{3}\right)$
$\therefore F=\frac{G\left(\frac{4}{3} \pi R^{3}\right) \rho\, m}{(R+h)^{2}}$
or $F=\frac{G\left(\frac{4}{3} \pi R\right) \rho \, m}{\left(1+\frac{h}{R}\right)^{2}}$
or $F=\frac{4}{3} \pi G R \rho\, m\left(1+\frac{h}{R}\right)^{-2}$
By using Binomial expansion,
$F=\frac{4}{3} \pi G R \rho \, m\left(1-\frac{2 h}{R}\right)$
$\left[\because ( l +x)^{n}=1+n x+\frac{n(n-1)}{2 !} \cdot x^{2}+\ldots\right.$ 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 $=\frac{8}{3} \pi \rho\, G m h$