Let $R_1, R_2$ and $R_3$ be the radii of three mercury drops. A big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is ______.
Show Hint
Any time liquid drops merge or split, Volume is conserved. Therefore, $R_{\text{new}}^3 = \sum R_{\text{old}}^3$. This logic applies equally to spherical capacitors combining their charges!
Step 1: Understanding the Concept:
When multiple drops merge into one large drop, the total volume remains constant because mercury is an incompressible liquid. Step 2: Formula Application:
Volume of a sphere $V = \frac{4}{3}\pi R^3$.
Total Volume $V = V_1 + V_2 + V_3$. Step 3: Explanation:
$\frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_1^3 + \frac{4}{3}\pi R_2^3 + \frac{4}{3}\pi R_3^3$.
Canceling the common factor $\frac{4}{3}\pi$:
$R^3 = R_1^3 + R_2^3 + R_3^3 \implies R = (R_1^3 + R_2^3 + R_3^3)^{1/3}$. Step 4: Final Answer:
The radius of the resultant drop is $(R_1^3 + R_2^3 + R_3^3)^{1/3}$.