Water rises to a height $3 \mathrm{~cm}$ in a capillary tube. If cross-sectional area of capillary tube is reduced to $\frac{1}{0}$ th initial area then water will rise to a height of

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Aicharya · Accepted Answer

$\mathrm{h}_{1} \mathrm{r}_{1}=\mathrm{h}_{2} \mathrm{r}_{2}$ $\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}$ $\mathrm{~A} \propto \mathrm{r}^{2}$ $\sqrt{\mathrm{A}} \propto \mathrm{r}$ $	herefore \quad \mathrm{h}_{2}=\mathrm{h}_{1} \sqrt{\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}}=3 \sqrt{9}=3 	imes 3=9 \mathrm{~cm}$

Water rises to a height $3 \mathrm{~cm}$ in a capillary tube. If cross-sectional area of capillary tube is…

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