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Chapter 1: Problem 118

When a 2 -mm-diameter tube is inserted into a liquid in an open tank, the liquid is observed to rise \(10 \mathrm{mm}\) above the free surface of the liquid (see Video \(V 1.10\) ). The contact angle between the liquid and the tube is zero, and the specific weight of the liquid is \(1.2 \times 10^{4} \mathrm{N} / \mathrm{m}^{3}\). Determine the value of the surface tension for this liquid.

Short Answer

Expert verified
The value of the surface tension for the liquid is \(0.006 \, N/m\).

Step by step solution

01

Understand the given information and what needs to be found

In this case, the following information is given: the diameter of the tube \(d = 2 \, mm = 2 \times 10^{-3} \, m\), the height to which the liquid rises in the tube \(h = 10 \, mm = 10 \times 10^{-3} \, m\), the specific weight of the liquid \(w = 1.2 \times 10^{4} \, N/m^3\), and the contact angle between the liquid and the tube \(\theta = 0^{\circ}\). The value of the surface tension of the liquid \(\sigma\) needs to be found.
02

Use the formula for capillary rise

The formula for capillary rise is given by: \(h = \frac{2 \sigma \cos(\theta)}{r w}\), where: \(h\) is the capillary rise height, \(\sigma\) is the surface tension, \(\theta\) is the contact angle, \(r\) is the radius of the tube, and \(w\) is the specific gravity of the fluid. In this case, \(r = d / 2 = 1 \times 10^{-3} \, m\) and \(\cos(\theta) = \cos(0^{\circ}) = 1\). Therefore, the formula becomes \(h = \frac{2 \sigma}{r w}\).
03

Substitute the given values into the formula and solve for the unknown

Substituting the given values into the formula yields: \(10 \times 10^{-3} \, m = \frac{2 \sigma}{1 \times 10^{-3} \, m \times 1.2 \times 10^{4} \, N/m^3}\). Solving this equation for \(\sigma\) yields: \(\sigma = 10 \times 10^{-3} \, m \times 1 \times 10^{-3} \, m \times 1.2 \times 10^{4} \, N/m^3 / 2 = 0.006 \, N/m\).

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