Problem Statement
Solve the fluid mechanics problem: Water ($\sigma=0.073\ \text{N/m}$, $\theta=0°$) is in a capillary of radius $r=0.5\ \text{mm}$. Find the pressure difference across the meniscus. The meniscus is approximately spherical with radius of curvature $R_c = r/\cos\theta = r$ (for $\theta=0°$). Young-Laplace: $$\Delta p = \frac{2\sigma}{R_
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Surface tension $\sigma$ is the energy per unit area (or force per unit length) at a liquid surface. It arises from cohesive intermolecular forces. Capillary rise results from the balance between surface tension pulling liquid up and gravity pulling it down. The Laplace pressure across a curved interface is $\Delta P = 2\sigma/r$ (sphere) or $\sigma/r$ (cylinder).
- $h = 2\sigma\cos\theta/(\rho g r)$ — capillary height
- $\Delta P = 2\sigma/r$ — excess pressure inside a droplet
- $W = \sigma \Delta A$ — work done against surface tension
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\Delta p = \frac{2\sigma}{R_
Given Information
- Fluid density $\rho$, velocities, cross-sections, and heights as given
- Atmospheric pressure $P_0 = 1.013\times10^5\,\text{Pa}$
- $g = 9.8\,\text{m/s}^2$
Physical Concepts & Formulas
Fluid statics is governed by Pascal’s Law and the hydrostatic pressure formula $P = P_0 + \rho g h$. Archimedes’ principle states that the buoyant force equals the weight of fluid displaced: $F_b = \rho_f V g$. Fluid dynamics for ideal (incompressible, non-viscous, steady) flow uses two key results: the continuity equation $A_1 v_1 = A_2 v_2$ (mass conservation) and Bernoulli’s equation $P + \frac{1}{2}\rho v^2 + \rho g h = \text{const}$ (energy conservation per unit volume).
- $P = P_0 + \rho g h$ — hydrostatic pressure
- $F_b = \rho_f V g$ — Archimedes buoyancy
- $A_1 v_1 = A_2 v_2$ — continuity equation
- $P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{const}$ — Bernoulli’s equation
- $v_{\text{efflux}} = \sqrt{2gh}$ — Torricelli’s theorem
Step-by-Step Solution
Step 1 — Identify the situation: Static (use $P = P_0+\rho gh$ and Archimedes) or dynamic (use continuity + Bernoulli).
Step 2 — Apply Bernoulli between two points at the same streamline:
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Step 3 — Use continuity to relate $v_1$ and $v_2$: $v_2 = v_1 A_1/A_2$.
Step 4 — Solve for the unknown (pressure, velocity, or flow rate).
Worked Calculation
Substituting all values with units:
Torricelli: Tank depth $h = 2\,\text{m}$, hole at bottom:
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Answer
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Answer
$$\boxed{v_{\text{efflux}} = \sqrt{2gh}}$$
Physical Interpretation
Capillary action allows plants to draw water from roots to leaves against gravity. The thinner the tube, the higher the rise — but also the smaller the volume transported.
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