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?

Answer

A, B, C

Explanation

`p=(2T)/(R )=p_(0)`
`:. P=(2T)/(R )+p_(0)`
`R=(r )/(cos theta)`
For water `theta =0^(@)`
`:. R=r `or` P=(2T)/(r)+p_(0)`
As temperature is constant
`p_(1)V_(1)=p_(2)V_(2)`
`:. p_(0)(0.11)= p (0.11-h)`
Subsitituting the values, we get
`(0.1 xx 10^(5) (0.11) =[1.01 xx 10^(5)+ (2xx5.06 xx 10^(-2))/(10^(-5))] (0.11 =h)`
Solving this equation we get,
`h=0.01m` or `1 cm`.
If seal is broken, then water will rise in the capillary.