Height of rise of liquid due to surface tension in a capillary tube, $h=\frac{2S\cos \theta }{r\rho g}$

Given that, height of liquid rise $h=7.5 \mathrm{cm},$surface tension of water $S=7.5\times {10}^{-2} \mathrm{N} {\mathrm{m}}^{-1}$and the angle of contact between water and glass is $\theta =0°$

$$\begin{gathered} h=\frac{2S\cos \theta }{r\rho g} \\ \Rightarrow r=\frac{2S\cos \theta }{h\rho g}=\frac{2\left(7.5\times {10}^{-2} \mathrm{N} {\mathrm{m}}^{-1}\right)\cos 0° }{\left(7.5\times {10}^{-2} \mathrm{m}\right)\left(1000 \mathrm{kg} {\mathrm{m}}^{-3}\right)\left(10 \mathrm{m} {\mathrm{s}}^{-2}\right)} \\ \Rightarrow r=0.2\times {10}^{-3} \mathrm{m} \text{or} 0.2 \mathrm{mm} \end{gathered}$$