Question
Find the height to which water at $4^{\circ} \mathrm{C}$ will rise in a capillary tube of $10^{-3} \mathrm{~m}$ diameter. Take $\mathrm{g}=9.8 \mathrm{~ms}^{-2}$. Angle of contact, $\theta=0$ and $\mathrm{T}=0.072 \mathrm{Nm}^{-1}$.
✓
Answer
Here, $D = 10^{-3}m$, $\text{r}=\frac{\text{D}}{2}=0.5\times10^{-3}\text{m}$$\text{g}=9.8\text{ms}^{-1},\theta=0,$
$\text{T}=72\times10^{-3}\text{Nm}^{-1}$
$\cos\theta=1$
Density of water at $4^\circ C = 10^3kg m^{-3}$ Using the relation, $\text{h}=\frac{2\text{T}\cos\theta}{\text{r}\rho\text{g}}$$=\frac{2\times72\times10^{-3}\times1}{5\times10^{-4}\times10^3\times9.8}$
$=2.939\times10^{-2}\text{m}$
$\text{T}=72\times10^{-3}\text{Nm}^{-1}$
$\cos\theta=1$
Density of water at $4^\circ C = 10^3kg m^{-3}$ Using the relation, $\text{h}=\frac{2\text{T}\cos\theta}{\text{r}\rho\text{g}}$$=\frac{2\times72\times10^{-3}\times1}{5\times10^{-4}\times10^3\times9.8}$
$=2.939\times10^{-2}\text{m}$
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