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 $F=\frac{4}{3}\pi GR\rho m{\left(1+\frac{h}{R}\right)}^{-2}$
By using Binomial expansion,
$F=\frac{4}{3}\pi GR\rho m\left(1-\frac{2h}{R}\right)$
$[\because (l+x{)}^n=1+nx+\frac{n(n-1)}{2!}\cdot x^2+\dots$ and
neglecting higher terms. ]
Difference in weight $=\frac{4}{3}\pi GR\rho m-\frac{4}{3}\pi G\rho mR\left(\frac{2h}{R}\right)$
Hence, error in weight $=\frac{8}{3}\pi \rho Gmh$