Correct Answer - C

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^@`