Question

An infinitely long cylindrical water filament of radius \( R \) is surrounded by air. Assume water and air to be static. The pressure outside the filament is \( P_{\text{out}} \) and the pressure inside is \( P_{\text{in}} \). If \( \gamma \) is the surface tension of the water-air interface, then \( P_{\text{in}} - P_{\text{out}} \) is:

• A. \( \frac{2\gamma}{R} \)
• B. \( 0 \)
• C. \( \frac{\gamma}{R} \)
• D. \( \frac{4\gamma}{R} \)

Hint:

For Laplace pressure problems, identify whether the interface is spherical or cylindrical. A cylindrical interface has a pressure difference of \( \frac{\gamma}{R} \), while a spherical interface has \( \frac{2\gamma}{R} \).

Answer 1

The Correct Option is C

Step 1: Use the Laplace pressure equation. The Laplace pressure equation for a cylindrical interface is given by: \[ P_{\text{in}} - P_{\text{out}} = \frac{\gamma}{R}. \] Here: - \( P_{\text{in}} \) is the pressure inside the water filament, - \( P_{\text{out}} \) is the pressure outside the water filament, - \( \gamma \) is the surface tension, - \( R \) is the radius of the cylindrical water filament. Step 2: Validate the equation for a cylindrical surface. For a cylindrical interface, the Laplace pressure is derived based on the curvature of the interface. Unlike a spherical interface where the pressure difference is \( \frac{2\gamma}{R} \), a cylindrical surface has a single radius of curvature, resulting in \( \frac{\gamma}{R} \). Step 3: Conclusion. The pressure difference \( P_{\text{in}} - P_{\text{out}} \) is \( \frac{\gamma}{R} \).