A glass capillary sealed at the upper end is of length `0.11 m` and internal diameter `2xx10^(5)`m. The tube is immersed vertically into a liquid of surface tension `5.06xx10^(-2)N//m`. To what length the capillary has to be immersed so that the liquid level inside and outside the capillary becomes the same. What will happen to water level inside the capillary if the seal is now broken?

Options

A. 5 cm
B. 3 cm
C. 1 cm
D. 7 cm

Answer

C

Explanation

Let `P_(0)` be atmospheric pressure and `P_(1)` the pressure of air within the sealed tube. Therefore equilibrium pressure just below the meniscus should be equal to atmospheric pressure because levels of water inside and outside the tube is same.
i.e., `(P_(1)-(2T)/(r))=p_(0)` or `P_(1)=p_(0)+(2T)/(r)`
If `L=0.11` m is the length of tube and `x` the
length of immersed part, then from Boyle's law
`p_(1)V_(1)=p_(2)V_(2),P_(0)La=(P_(0)+(2T)/(r))(L-x)a`
Where `a` is the cross-sectional area of tube.
i.e., `P_(0)xxL=(P_(0)+(2T)/(r))(L-x)`
`P_(0)xxL=P_(0)(L-x)+(2T)/(r)(L-x)`
`x=(4TL)/(P_(0)d+4T)`