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Leno4ka [110]
3 years ago
8

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:
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
4 0

Answer:

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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