Solution 13: Ratio of Masses for Similar Drops
The problem states that the drops are similar in shape. Let $L$ be a characteristic linear dimension (e.g., radius $r$).
- Surface Energy ($U_S$): Proportional to Area $\times \sigma \propto L^2 \sigma$.
- Gravitational Potential Energy ($U_G$): Proportional to $mgh$. Since $m \propto \rho L^3$ and $h \propto L$, $U_G \propto (\rho L^3) g L = \rho g L^4$.
We are given that the ratio of Surface Energy to Gravitational Potential Energy is a constant definite ratio for this specific shape family. $$ \frac{U_S}{U_G} = \text{constant} $$ $$ \frac{\sigma L^2}{\rho g L^4} = k \implies \frac{\sigma}{\rho L^2} = k’ $$ From this, we can find the scaling relationship for the length $L$: $$ L^2 \propto \frac{\sigma}{\rho} \implies L \propto \sqrt{\frac{\sigma}{\rho}} $$
3. Mass Ratio Calculation
Mass is given by $m = \text{Density} \times \text{Volume} \propto \rho L^3$.
Substituting the scaling for $L$:
$$ m \propto \rho \left( \sqrt{\frac{\sigma}{\rho}} \right)^3 = \rho \frac{\sigma^{3/2}}{\rho^{3/2}} = \frac{\sigma^{3/2}}{\rho^{1/2}} $$
Now, take the ratio for Mercury ($H$) and Water ($W$): $$ \frac{m_H}{m_W} = \frac{\sigma_H^{3/2} / \rho_H^{1/2}}{\sigma_W^{3/2} / \rho_W^{1/2}} $$ $$ \frac{m_H}{m_W} = \left( \frac{\rho_W}{\rho_H} \right)^{1/2} \left( \frac{\sigma_H}{\sigma_W} \right)^{3/2} $$