question_answer

A glass capillary tube sealed at the upper end has internal radius r. The tube is held vertical with its lower end touching the surface of water. Calculate the length (L) of such a tube for water in it to rise to a height \[h(<L)\]. Atmospheric pressure is \[{{P}_{0}}\]and surface tension of water is T. Assume that water perfectly wets glass (Density of water \[=\rho \])

A) \[h+\frac{{{P}_{0}}rh}{2T-\rho ghr}\]   

B)       \[h+\frac{{{P}_{0}}rh}{T-\rho ghr}\]

C) \[h+\frac{{{P}_{0}}rh}{2T+\rho ghr}\]   

D)       \[h+\frac{{{P}_{0}}rh}{T+\rho ghr}\]

Correct Answer: A

Solution :

[a] When the tube is brought into contact with water, it is filled with air at atmospheric pressure. When water rises to a height h, the air pressure (P) is given by \[PA(L-h)={{P}_{0}}AL.\] \[\therefore \,\,\,\,\,\,\,P=\frac{{{P}_{0}}L}{L-h}\] Radius of curvature of meniscus \[R=r,\]since contact angle is zero. Pressure at A is \[{{P}_{A}}={{P}_{0}}-\rho gh\] \[\therefore \,\,P={{P}_{A}}+\frac{2T}{r}\] \[\therefore \,\,\,\frac{{{P}_{0}}T}{L-h}={{P}_{0}}-\rho gh+\frac{2T}{r}\] \[\Rightarrow \,\,\,{{P}_{0}}\left( \frac{L}{L-h}-1 \right)=\frac{2T}{r}-\rho gh\] \[\Rightarrow \,\,\,\frac{{{P}_{0}}h}{L-h}=\frac{2T}{r}-\rho gh\Rightarrow L-h=\frac{{{P}_{0}}h}{\frac{2T}{r}-\rho gh}\] \[\Rightarrow \,\,\,\,\,\,\,L=h+\frac{{{P}_{0}}rh}{2T-\rho ghr}\]