Question

A straw is placed in a cup of water. The straw has a diameter of 5 mm, the contact angle is 60 degrees and the surface tension of water is 0.073 Nm. Calculate the change in elevation between the water surface in the cup and that in the straw.

Answer 1

Answer:

The change in elevation is 0.0029mm

Explanation:

This problem is on the capillary action of water and tube

we are going use the formula below for the rise in liquid level

h= 2Tcos∅/rgρ

where h= height of liquid level in the capillary tube

          T= surface tension = 0.073 Nm

          ∅= angle of contact = 60 degrees

          r= radius of bore of capillary tube = 5/2= 2.5 mm to m= 2.5/1000=  

          2.5*10^-3

          g= acceleration due to gravity = 9.81 m/s^2

         ρ= density of the liquid (assume water 1000 kg/m³)

we can now substitute to solve for h

$h=\frac{2*0.073cos60}{ 2.5*10^-^3*9.81*10^3} \\\\h=\frac{2*0.073*0.5}{ 2.5*9.81}\\\\h=\frac{0.073}{ 2.5*9.81}\\\\h=\frac{0.073}{ 24.525}\\\\h=\frac{0.073}{ 24.525}\\\\h= 0.0029\\h= 2.9*10^-^3mm$