If surface tension of water is 0.07 N/m the weight of water supported in a capillary tube of radius 0.1 mm is -- ( Angle of contact is 0⁰)

Given,

$$\begin{gathered} \mathrm{S}=7\times {10}^{-2}\mathrm{N}/\mathrm{m} \\ \mathrm{r}=0.1\mathrm{mm} \\ \mathrm{r}=0.1\times {10}^{-3}\mathrm{m} \end{gathered}$$

We know that,

$$\begin{gathered} \mathrm{F}=\mathrm{mg} \\ \mathrm{Sl}=\mathrm{mg} \\ \mathrm{mg}=\mathrm{S}\times 2\mathrm{\pi r} \end{gathered}$$

Therefore the weight of water,

$\mathrm{Weight}=7\times {10}^{-2}\times 2\times \frac{22}{7}\times 0.1\times {10}^{-3}$

$\mathrm{Weight}=44\times {10}^{-6}\mathrm{N}$

Hence the correct answer is $44\times {10}^{6}N.$