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 \[{{P}_{A}}={{P}_{0}}-\rho gh\] \[\therefore \,\,P={{P}_{A}}+\frac{2T}{r}\] \[\therefore \,\,\,\frac{{{P}_{0}}T}{L-h}={{P}_{0}}-\rho gh+\frac{2T}{r}\] \[\Rightarrow \,\,\,{{P}_{0}}\left( \frac{L}{L-h}-1 \right)=\frac{2T}{r}-\rho gh\] \[\Rightarrow \,\,\,\frac{{{P}_{0}}h}{L-h}=\frac{2T}{r}-\rho gh\Rightarrow L-h=\frac{{{P}_{0}}h}{\frac{2T}{r}-\rho gh}\] \[\Rightarrow \,\,\,\,\,\,\,L=h+\frac{{{P}_{0}}rh}{2T-\rho ghr}\]You need to login to perform this action.
You will be redirected in
3 sec