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A capillary tube of radius $\mathrm{r}$ is immersed vertically in a liquid such that liquid rises in it to height $\mathrm{h}$ (less than the length of the tube). Mass of liquid in the capillary tube is $\mathrm{m}$. If radius of the capillary tube is increased by $50 \%$, then mass of liquid that will rise in the tube, is
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The correct answer is:
$\frac{3}{2} \mathrm{~m}$
$\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \Rightarrow \mathrm{h} \propto \frac{1}{\mathrm{r}} \Rightarrow \frac{\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{2}{3}$
$$
\begin{aligned}
\left(\because \mathrm{r}_{1}=\mathrm{r}, \quad \mathrm{r}_{2}\right.&=\mathrm{r}+50 \% \text { of } \left.\mathrm{r}=\frac{3}{2} \mathrm{r}\right) \\
\text { New mass } \mathrm{m}_{2}=\pi \mathrm{r}_{2}^{2} \mathrm{~h}_{2} \rho &=\pi\left(\frac{3}{2} \mathrm{r}_{1}\right)^{2}\left(\frac{2}{3} \mathrm{~h}_{1}\right) \rho \\
&=\frac{3}{2}\left(\pi \mathrm{r}_{1}^{2} \mathrm{~h}_{1}\right) \rho=\frac{3}{2} \mathrm{~m}
\end{aligned}
$$
$$
\begin{aligned}
\left(\because \mathrm{r}_{1}=\mathrm{r}, \quad \mathrm{r}_{2}\right.&=\mathrm{r}+50 \% \text { of } \left.\mathrm{r}=\frac{3}{2} \mathrm{r}\right) \\
\text { New mass } \mathrm{m}_{2}=\pi \mathrm{r}_{2}^{2} \mathrm{~h}_{2} \rho &=\pi\left(\frac{3}{2} \mathrm{r}_{1}\right)^{2}\left(\frac{2}{3} \mathrm{~h}_{1}\right) \rho \\
&=\frac{3}{2}\left(\pi \mathrm{r}_{1}^{2} \mathrm{~h}_{1}\right) \rho=\frac{3}{2} \mathrm{~m}
\end{aligned}
$$
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