MCQ
The surface tension for pure water in a capillary tube experiment is
- A$\frac{{\rho g}}{{2hr}}$
- B$\frac{2}{{hr\rho g}}$
- C$\frac{{r\rho g}}{{2h}}$
- ✓$\frac{{hr\rho g}}{2}$
✓
Answer
Correct option: D.
$\frac{{hr\rho g}}{2}$
d
(d)$T = \frac{{rhdg}}{{2\cos \theta }}.$ For pure water $\theta = {0^o}$ so $T = \frac{{rhdg}}{2}$
(d)$T = \frac{{rhdg}}{{2\cos \theta }}.$ For pure water $\theta = {0^o}$ so $T = \frac{{rhdg}}{2}$
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