Two equal masses m and m are hung from a bala | vedclass

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(C) The acceleration due to gravity at a height $h$ above the surface of the Earth is given by $g_h = g(1 - \frac{2h}{R})$.Let the two pans be at heig

Vedclass · Accepted Answer

$\frac{8}{3}\pi G\rho mh$

Vedclass · Answer

(C) The acceleration due to gravity at a height $h$ above the surface of the Earth is given by $g_h = g(1 - \frac{2h}{R})$.Let the two pans be at heights $h_1$ and $h_2$ such that $h_1 - h_2 = h$. The weights measured are $W_1 = mg_1$ and $W_2 = mg_2$.The difference in weight (error) is $\Delta W = mg_1 - mg_2 = mg(1 - \frac{2h_1}{R}) - mg(1 - \frac{2h_2}{R}) = mg(\frac{2h_2}{R} - \frac{2h_1}{R}) = \frac{2mg}{R}(h_2 - h_1)$.Since $h_1 - h_2 = h$,we have $\Delta W = \frac{2mgh}{R}$.Substituting $g = \frac{GM}{R^2}$ and $M = \frac{4}{3}\pi R^3 \rho$,we get $g = \frac{G}{R^2} \cdot \frac{4}{3}\pi R^3 \rho = \frac{4}{3}\pi G R \rho$.Substituting this into the error formula: $\Delta W = \frac{2mh}{R} \cdot (\frac{4}{3}\pi G R \rho) = \frac{8}{3}\pi G \rho mh$.

Two equal masses $m$ and $m$ are hung from a balance whose scale pans differ in vertical height by $h$. The error in weighing in terms of density of the earth $\rho$ is:

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