## Step 1: Find the Pressures in Each Droplet The pressure difference across interface of a droplet to surface tension is given by the **Young-Laplace equation**: \[ \Delta P = \gamma \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \] For a barrel-shaped droplet on a fiber, assume the radii of curvature are dominated by the maximum radius of the droplet. Let the surface tension be \(\gamma\). - For droplet 1 (radius \(R_1\)): \[ \Delta P_1 = \frac{2\gamma}{R_1} \] - For droplet 2 (radius \(R_2 > R_1\)): \[ \Delta P_2 = \frac{2\gamma}{R_2} \] **Explanation:** The pressure inside each droplet is higher than the outside pressure by an amount set by the curvature (radius) of the droplet. The smaller the droplet (smaller \(R\)), the higher the internal pressure. --- ## Step 2: Is This Drop Tandem Stable? To check stability, consider the pressure difference: Since \(R_2 > R_1\), \[ \Delta P_2 < \Delta P_1 \] This means the **smaller droplet has higher internal pressure than the bigger droplet**. If they are connected by a thin film, liquid will flow from high pressure to low pressure, i.e., **from the smaller to the larger droplet**. **Explanation:** This configuration is **not stable**: over time, the smaller droplet will shrink and the larger droplet will grow (Ostwald ripening). --- ## **Final Answers** **A)** - Pressure in droplet 1: \(\Delta P_1 = \frac{2\gamma}{R_1}\) - Pressure in droplet 2: \(\Delta P_2 = \frac{2\gamma}{R_2}\) **B)** - **No, this drop tandem is not stable.** Liquid will flow from the smaller (higher pressure) droplet to the larger (lower pressure) droplet.