Question

Q.10.A capillary tube of radius 0.05 cm is immersed in water, find the value of
rise of water in capillary if value of surface tension is 0.073 N/m and angle
of contact 0° (take g = 10 m/s2 & p = 103 kg/m²)​

Answer 1

Answer:

0.0292 m.

Explanation:

Attached picture shows the explanation of the problem.

Answer 2

Given- radius(r) = 0.05 cm

            surface tension(T) = 0.073 N/m

            angle of contact(θ) = 0°

            acc. due to gravity(g) = 10m/s²

            density(p) = 10³ kg/m²

To find- The height of rise of capillary

Formula used-  H = $(\frac{2Tcos\theta}{rpg} )$

Step-by-step explanation:

                          H = $(\frac{2\times0.073\times cos0\textdegree}{0.0005\times10^{3}\times10 } )$  

                          H = $(\frac{2\times0.073\times 1}{0.0005\times10^{3}\times10 } )$   (since cos 0° = 1)

                          H = 0.029 m or 2.9 cm

Therefore, the value of rise of water in capillary = 2.9 cm