The diagram above illustrates a Hare's apparatus. \(\mathit{e}_1, \mathit{e}_2\) represent densities and \(\mathit{h}_1, \mathit{h}_2\) heights of columns o...

Answer Details

In a Hare's apparatus, a liquid with a higher density is placed in the taller column and a liquid with a lower density is placed in the shorter column. The pressure at the base of each column is the same. Therefore, we can equate the pressures at the base of each column: \(\text{Pressure}=\text{Height} \times \text{Density} \times \text{Gravitational acceleration}\) Thus, we can write: \(\mathit{h}_1 \mathit{e}_1 g = \mathit{h}_2 \mathit{e}_2 g\) Where g is the acceleration due to gravity. We can cancel g from both sides of the equation, which gives: \(\mathit{h}_1 \mathit{e}_1 = \mathit{h}_2 \mathit{e}_2\) Then, we can rearrange this equation to find \(\mathit{h}_1\): \(\mathit{h}_1 = \frac{\mathit{h}_2 \mathit{e}_2}{\mathit{e}_1}\) Therefore, the correct equation is: \(\mathit{h}_1 = \frac{\mathit{h}_2 \mathit{e}_2}{\mathit{e}_1}\)