Question

Consider a large drop of liquid on a glass plate. Contact angle is $\theta$ and density of liquid is $\rho$. Surface tension is S

• A. H = $\sqrt{\frac{2S}{\rho g}(1-sin \theta)}$
• B. H = $\sqrt{\frac{2S}{\rho g}(1+sin \theta)}$
• C. H = $\sqrt{\frac{2S}{\rho g}(1+cos \theta)}$
• D. H = $\sqrt{\frac{2S}{\rho g}(1-cos \theta)}$

Answer 1

Answer: (A)

H = $\sqrt{\frac{2S}{\rho g}(1-sin \theta)}$

Answer 2

For a sessile drop (a drop resting on a surface), the height H is related to surface tension S, density $\rho$, and contact angle $\theta$. Assuming the drop is a spherical cap and balancing the hydrostatic pressure ($\rho g H$) with the capillary pressure ($\frac{2S}{R}$), where R is the radius of curvature, we have $\rho g H = \frac{2S}{R}$. The relationship between the height H and the radius of curvature R for a spherical cap is $H = R(1 - \sin \theta)$. Substituting R from this equation into the pressure balance gives $\rho g H = \frac{2S}{\frac{H}{1-\sin \theta}}$. Rearranging this equation yields $H^2 = \frac{2S(1-\sin \theta)}{\rho g}$, and thus $H = \sqrt{\frac{2S(1-\sin \theta)}{\rho g}}$.