Two blocks of masses m each are hung from a balance. The scale pan A is at height H1 whereas scale pan B is at height H2 The error in weighing when H1 > H2 and R being the radius of earth is

The correct option is 2mg((H1-H2)/R).

Let us find the error in weighing step-by-step.
We are given two blocks, each of mass $m$, hung from a balance.
The scale pan A is at height $H_1$ above the surface of the Earth, and the scale pan B is at height $H_2$ above the surface of the Earth.
Let $R$ be the radius of the Earth, and $g$ be the acceleration due to gravity at the Earth's surface.

The acceleration due to gravity $g_h$ at a height $h$ above the Earth's surface (where $h\ll R$) is given by the formula:
$g_h=g\left(1-\frac{2h}{R}\right)$

Thus, the effective weight of the block in scale pan A (at height $H_1$) is:
$W_1=m{g}_{H_1}=mg\left(1-\frac{2H_1}{R}\right)$

Similarly, the effective weight of the block in scale pan B (at height $H_2$) is:
$W_2=m{g}_{H_2}=mg\left(1-\frac{2H_2}{R}\right)$

The error in weighing is the difference between the two weights:
$\Delta W=W_2-W_1$
Since $H_1>H_2$, $W_2$ is greater than $W_1$.
Substituting the expressions for $W_1$ and $W_2$:
$\Delta W=mg\left(1-\frac{2H_2}{R}\right)-mg\left(1-\frac{2H_1}{R}\right)$

Simplifying the terms:
$\Delta W=mg\left[\left(1-\frac{2H_2}{R}\right)-\left(1-\frac{2H_1}{R}\right)\right]$
$\Delta W=mg\left[1-\frac{2H_2}{R}-1+\frac{2H_1}{R}\right]$
$\Delta W=mg\left[\frac{2H_1}{R}-\frac{2H_2}{R}\right]$
$\Delta W=\frac{2mg}{R}\left(H_1-H_2\right)$

Therefore, the error in weighing is:
$2mg\left(\frac{H_1-H_2}{R}\right)$