Answer:
The capillary rise of the glycerin is most nearly  [tex]y = 0.0204 \ m[/tex]
Explanation:
From the question we are told that
 The diameter of the glass tube is  [tex]d = 1 \ mm = 0.001 \ m[/tex]
  The density of glycerin is  [tex]\rho = 1260 \ kg /m^3[/tex]
  The surface tension of the glycerin is [tex]\sigma = 6.3 *10^{-2} \ N /m[/tex]
The capillary rise of the glycerin is mathematically represented as
    [tex]y = \frac{4 * \sigma * cos (\theta )}{ \rho * g * d}[/tex]
substituting value Â
    [tex]y = \frac{4 * 6.3 *10^{-2} * cos (0 )}{ 1260 * 9.8 * 0.001}[/tex]
   [tex]y = 0.0204 \ m[/tex]
Therefore the height  of the glass tube  the glycerin was able to cover is
[tex]y = 0.0204 \ m[/tex] Â
