An open capillary tube contains a drop of water. When the tube is in its vertical position, the drop forms a column with a length of a `2cm`, b. `4cm`, and c. `2.98cm`. The internal diameter of the capillary tube is `1mm`. Determine the radii of curvature of the upper and lower meniscuses in each case. Consider the wetting to be complete. Surface tension of water `=0.0075N//m`

Explanation

When capillary is in vertical position, the upper meniscus is concave and pressure due to surface tension
`P_(1)=(2T)/R_(1)`
is directed vertically upward `R_(1)` being radius of curvature of upper meniscus. When wetting is complete.
`P_(1)=(2T)/r`
`r` being radus of capillary tube `=D/2=1.0/2mm`
`R_(1)=r=0.5mm` in each case
`R_(1)=r=0.5mm` in each case
we have `P_(1)=(2T)/r=(2xx75xx10^(-3))/(0.5xx10^(-3))=300N//m^(2)`
The hydrostatic pressure `p_(2)=hrhog` is always directed downwaads.
a. If `h=2 cm` figure i
`p_(2)=hrhog=2xx10^(-2)xx10xx10^(3)=200N//m^(2)`
Clearly `p_(1)gtp_(2)`. That is resulting pressure is directed upward. For equilibrium the pressure due to lower meniscus should be downwards. This makes the lower meniscus concave dowwards. Then
`p_(1)-p_(2)=(2T)/(R_(2))`
or `300-200=(2xx75xx10^(-3))/(R_(2))`
`R_(2)=(2xx75xx10^(-3))/1001.50xx10^(-3)m=1.50mm`
b. If `h=4cm` fig ii
`p_(2)=hrhogh=4x10^(-2)xx10^(-3)xx10=392N//m^(2)`
In this case `p_(1)gtp_(2)`, hence the pressure due to lower meniscus should be directed upwards. For this meniscus should be concave upwards.
Then `p_(2)-p_(1)=(2T)/(R_(2))`
`400-300=(2xx75xx10^(-3))/(R_(2))`

`R_(2)=(2xx75xx10^(-3))/100=1.50xx10^(-3)m=1.50mm`
c. If `h=3 cm` fig iii
`p_(2)=hrhog=3.00xx10^(-2)xx10^(3)xx10=300N//m^(2)`
In this case `p_(1)=p_(2)`
`p_(1)-p_(2)=(2T)/(R_(2))`
`(2T)/(R_(2))=0` or `R_(2)=oo`
i.e., lower surface will be flat.