A long capillary tube of radius `0.2 mm` is placed vertically inside a beaker of water.
If the tube is now pushed into water so that only `5.0 cm` of its length is above the surface, then determine the angle of contact between the liquid and glass surface.

Options

A. `cos^(-1)(4/5)`
B. `cos^(-1)(5/7)`
C. `cos^(-1)(3/5)`
D. `cos^(-1)(5/4)`

Answer

B

Explanation

Here `s=7.0x10^(-2) N//m,r=0.2xx10^(-3)m,r=10^(3)kg//m^(3)`
we know that the hieght of liquid in the tube is
`h=2sigmacostheta//rhoRg`
where `R` is the radius of the menisus, and `theta` is the angle of contact.
Since `theta=0` (given), radius of the meniscus is equal to the radius of the capillary tube i.e. `R=r`
`:. h=(2(7.0xx10^(-2)))/((10^(3))(0.2xx10^(-3))(10))=0.07m`
or `h=7.0cm`
When the length of the capillary tube above the free surface of the liquid is less than the height of liquid that rise in the tube, radius `R` of the free surface is not equal to the radius of the tube. It is greater than `r` as the surface tennis to be flatter.
According to the equation `p_(1)-p_(2)=(4sigma//R)`, the pressure difference across te surface is given by
`/_\p=(2sigmacostheta)/R`
If `p_(1)` and `p_(2)` are the pressure just above and below the mensisus, respectively then `p_(1)-p_(2)=rhogh_(0)`
`:. rhogh_(0)=(2sigma)/r`...........i
In part a we have seen than when `h_(0)=h,theta=0, R=r`, and `rhogh=(2sigma)/r`.......ii
Therefore dividing eqn i by eqn ii we have `r/R=(h_(0))/h`
From the figure, it is clear that `costheta=r/R`
Therefore the angle of contact is
`theta=(cos^(-1)h_(0))/h=cos^(-1)(5/7)~~44^@`