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Step-by-Step Solution
Step 1: Understand the Problem
When a narrow glass capillary tube is dipped into a liquid that wets the tube (contact angle nearly 0°), the liquid rises in the tube to a certain height. We are given:
Radius of the capillary tube, r = 0.015 \text{ cm} = 0.015 \times 10^{-2} \text{ m} = 1.5 \times 10^{-4} \text{ m}
Height of the liquid column, h = 15 \text{ cm } = 0.15 \text{ m}
Density of the liquid, \rho = 900 \text{ kg m}^{-3}
Acceleration due to gravity, g = 10 \text{ m s}^{-2}
Contact angle, \theta \approx 0^\circ \;\Rightarrow\; \cos \theta \approx 1
We need to find the surface tension T of the liquid.
Step 2: Recall the Capillary Rise Formula
The height to which the liquid rises in a capillary tube (for small contact angles) is given by:
h = \dfrac{2T \cos \theta}{\rho \, g \, r}
Rearranging to solve for the surface tension T :
T = \dfrac{\rho \, g \, r \, h}{2 \cos \theta}
Step 3: Substitute the Known Values
Since \theta \approx 0^\circ , we have \cos \theta \approx 1 . Thus,
T = \dfrac{(900 \,\text{kg m}^{-3}) \times (10 \,\text{m s}^{-2}) \times (1.5 \times 10^{-4}\,\text{m}) \times (0.15\,\text{m})}{2 \times 1}
Step 4: Perform the Calculation
First, multiply all the numerators:
\rho \, g \, r \, h = (900) \times (10) \times (1.5 \times 10^{-4}) \times (0.15) \\
= 9000 \times (1.5 \times 10^{-4}) \times 0.15 \\
= 9000 \times 1.5 \times 0.15 \times 10^{-4} \\
= 9000 \times 0.225 \times 10^{-4} \\
= 2025 \times 10^{-4} \\
= 0.2025
Now, divide by 2:
T = \dfrac{0.2025}{2} = 0.10125 \,\text{N/m}
To express in milliNewton per meter (mN/m):
0.10125 \,\text{N/m} = 101.25 \,\text{mN/m}
Step 5: State the Final Answer
Rounding to the nearest integer:
\boxed{101 \,\text{mN/m}}
