In capillary tube having area of cross-section ' $\mathrm{A}$ ', water rises to a height ' $h$ '. If cross…

In capillary tube having area of cross-section ' $\mathrm{A}$ ', water rises to a height ' $h$ '. If cross sectional area is reduced to $\frac{A}{9}$, the rise of water in the capillary tube is
  1. $\mathrm{h}$
  2. $4 h$
  3. $3 \mathrm{~h}$
  4. $2 \mathrm{~h}$

Solution

$\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r \rho g}} \\ & \Rightarrow \mathrm{h} \propto \frac{1}{\mathrm{r}}\end{aligned}$ And $\mathrm{A}=\pi \mathrm{r}^2 \Rightarrow \mathrm{r} \propto \sqrt{\mathrm{A}}$ $\Rightarrow \mathrm{h} \propto \frac{1}{\sqrt{\mathrm{A}}}$ Now, $\frac{h_2}{h_1}=\sqrt{\frac{A_1}{A_2}}$ $\begin{aligned} & \Rightarrow \frac{h_2}{h}=\sqrt{\frac{A}{A / 9}}=3 \\ & \Rightarrow h_2=3 h\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 1)

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