Water rises to a height of \(2 \mathrm{~cm}\) in a capillary tube. If crosssectional area of the tube is reduced to \(\frac{1}{16}^{\text {th }}\) of initial area, then water will rise to a height of

\(\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \\ & \therefore \frac{\mathrm{h}_2}{\mathrm{~h}_1}=\frac{\mathrm{r}_1}{\mathrm{r}_2} \\ & \text { Area } \mathrm{A}=\pi \mathrm{r}^2 \therefore \mathrm{A} \propto \mathrm{r}^2 \\ & \therefore \frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\mathrm{r}_1^2}{\mathrm{r}_2^2} \text { or } \frac{\mathrm{r}_1}{\mathrm{r}_2}=\sqrt{\frac{\mathrm{A}_1}{\mathrm{~A}_2}}=\sqrt{16} \\ & \frac{\mathrm{r}_1}{\mathrm{r}_2}=4 \\ & \therefore \frac{\mathrm{h}_2}{\mathrm{~h}_1}=4 \text { or } \mathrm{h}_2=4 \mathrm{~h}_1=4 \times 2=8 \mathrm{~cm}\end{aligned}\)