# A long capillary tube of radius 0.2 mm is placed vertically inside a beaker of water. If the surface tension of water is 7.2xx10^(-2)N//m the angl

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A long capillary tube of radius 0.2 mm is placed vertically inside a beaker of water.
If the surface tension of water is 7.2xx10^(-2)N//m the angle of contact between glass and water is zero, then determine the height of the water column in the tube.
A. 3cm
B. 9cm
C. 7cm
D. 5cm

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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^@