Water rises to a height of $15 \mathrm{~mm}$ in a capillary tube having cross-sectional area 'A'. If…

Water rises to a height of $15 \mathrm{~mm}$ in a capillary tube having cross-sectional area 'A'. If cross-sectional area of the tube is made ' $\frac{A^{\prime}}{3}$ then the water will rise to a height of
  1. $15 \sqrt{3} \times 10^{-3} \mathrm{~m}$
  2. $20 \sqrt{3} \times 10^{-3} \mathrm{~m}$
  3. $5 \sqrt{3} \times 10^{-3} \mathrm{~m}$
  4. $10 \sqrt{3} \times 10^{-3} \mathrm{~m}$

Solution

$\mathrm{h} \propto \frac{1}{\mathrm{r}}$ since $\mathrm{A}=\pi \mathrm{r}^{2}, \quad \sqrt{\mathrm{A}} \propto \mathrm{r}$ $\therefore \mathrm{h} \propto \frac{1}{\sqrt{\mathrm{A}}}$ $\therefore \frac{\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\sqrt{\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}}=\sqrt{3} \quad \because \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=3$ $\therefore \mathrm{h}_{2}=\sqrt{3} \mathrm{~h}_{1}=\sqrt{3} \times 15 \times 10^{-3} \mathrm{~m} \quad=15 \sqrt{3} \times 10^{-3} \mathrm{~m}$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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