Physics: A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/m, and the contact angle is zero. The capillary rise of the glycerin is most nearly:

Explanation:P.E=MGHSO 100×40×40=160000...

A glass tube 1 mm in diameter is dipped into glycerin.

ctyrector

29.01.2022

The capillary rise of the glycerin is most nearly $y = 0.0204 \ m$

Explanation:

From the question we are told that

The diameter of the glass tube is $d = 1 \ mm = 0.001 \ m$

The density of glycerin is $\rho = 1260 \ kg /m^3$

The surface tension of the glycerin is $\sigma = 6.3 *10^{-2} \ N /m$

The capillary rise of the glycerin is mathematically represented as

$y = \frac{4 * \sigma * cos (\theta )}{ \rho * g * d}$

substituting value

$y = \frac{4 * 6.3 *10^{-2} * cos (0 )}{ 1260 * 9.8 * 0.001}$

$y = 0.0204 \ m$

Therefore the height of the glass tube the glycerin was able to cover is

$y = 0.0204 \ m$

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