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At 25∘C, how high will water rise in a glass capillary tube with an inner diameter of 0.75 mm? For water, T=71.99mNm and ρ=1.0gcm3. Your answer should have two significant figures.

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The height to which water will rise in the capillary tube is approximately \(0.014 \, \text{m}\) or \(1.4 \, \text{cm}\).

The height (\(h\)) to which water will rise in a capillary tube can be calculated using the Jurin's Law equation:

\[ h = \frac{2T \cos \theta}{r\rho g} \]

where:

- \( T \) is the surface tension of water,

- \( \theta \) is the contact angle between water and the glass,

- \( r \) is the radius of the capillary tube,

- \( \rho \) is the density of water,

- \( g \) is the acceleration due to gravity.

Assuming a full capillary rise (\( \cos \theta \approx 1 \)), the equation simplifies to:

\[ h = \frac{2T}{r\rho g} \]

Given:

- \( T = 71.99 \, \text{mN/m} \),

- \( r = 0.75 \, \text{mm} \) (convert to meters),

- \( \rho = 1.0 \, \text{g/cm}^3 \) (convert to kg/m\(^3\)),

- \( g = 9.8 \, \text{m/s}^2 \).

Convert the units and substitute the values into the formula:

\[ h = \frac{2 \times 71.99 \times 10^{-3}}{0.75 \times 10^{-3} \times 1.0 \times 10^3 \times 9.8} \]

\[ h \approx 0.014 \, \text{m} \]

The height to which water will rise in the capillary tube is approximately \(0.014 \, \text{m}\) or \(1.4 \, \text{cm}\).

User Matsolof
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