Question

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:

Answer 1

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

Step-by-step 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 = (4 * \sigma * cos (\theta ))/( \rho * g * d)`

substituting value

`y = (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`