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A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of water in the capillary tube is $5 \mathrm{~g}$. Another capillary tube of radius $2 r$ is immersed in water. The mass of water that will rise in this tube is
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The correct answer is:
$10 \mathrm{~g}$
$: h=\frac{2 S \cos \theta}{r \rho g}$
Mass of water in the first tube,
$$
\begin{aligned}
m= & \pi r^2 h \rho=\pi r^2 \times\left(\frac{2 S \cos \theta}{r \rho g}\right) \times \rho \\
& =\frac{2 \pi r S \cos \theta}{g}
\end{aligned}
$$
$\therefore \quad m \propto r$. Hence, $\frac{m^{\prime}}{m}=\frac{2 r}{r}=2$
or $m^{\prime}=2 m=2 \times 5 g=10 g$.
Mass of water in the first tube,
$$
\begin{aligned}
m= & \pi r^2 h \rho=\pi r^2 \times\left(\frac{2 S \cos \theta}{r \rho g}\right) \times \rho \\
& =\frac{2 \pi r S \cos \theta}{g}
\end{aligned}
$$
$\therefore \quad m \propto r$. Hence, $\frac{m^{\prime}}{m}=\frac{2 r}{r}=2$
or $m^{\prime}=2 m=2 \times 5 g=10 g$.
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