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A glass rod of radius ' $r$ ' ' is inserted symmetrically into a vertical capillary tube of radius ' $\mathrm{r}_2$ ' $\left(\mathrm{r}_1 < \mathrm{r}_2\right)$ such that their lower ends are at same level. The arrangement is dipped in water. The height to which water will rise into the tube will be ( $\rho=$ density of water, $\mathrm{T}=$ surface tension in water, $\mathrm{g}$ = acceleration due to gravity)
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Verified Answer
The correct answer is:
$\frac{2 \mathrm{~T}}{\left(\mathrm{r}_2-\mathrm{r}_1\right) \rho g}$
Vertical component of total force of surface tension
$$
\mathrm{F}=\left(\mathrm{r}_2+\mathrm{r}_1\right) 2 \pi \mathrm{T} \cos \theta
$$
Weight of the liquid in the capillary
$$
\mathrm{W}=\pi\left(\mathrm{r}_2^2-\mathrm{r}_1^2\right) \mathrm{h} \rho \mathrm{g}
$$
This is balanced by the vertical component of the face due to the surface tension
$$
\therefore \pi\left(\mathrm{r}_2^2-\mathrm{r}_1^2\right) \mathrm{h} \rho \mathrm{g}=\left(\mathrm{r}_2+\mathrm{r}_1\right) \times 2 \pi \mathrm{T} \cos \theta
$$
Simplifying and solving for $\mathrm{h}$ we get,
$$
\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\left(\mathrm{r}_2-\mathrm{r}_1\right) \rho g}=\frac{2 \mathrm{~T}}{\left(\mathrm{r}_2-\mathrm{r}_1\right) \rho g}
$$
(For pure water, $\theta=0^{\circ}, \cos \theta=1$ )
$$
\mathrm{F}=\left(\mathrm{r}_2+\mathrm{r}_1\right) 2 \pi \mathrm{T} \cos \theta
$$
Weight of the liquid in the capillary
$$
\mathrm{W}=\pi\left(\mathrm{r}_2^2-\mathrm{r}_1^2\right) \mathrm{h} \rho \mathrm{g}
$$
This is balanced by the vertical component of the face due to the surface tension
$$
\therefore \pi\left(\mathrm{r}_2^2-\mathrm{r}_1^2\right) \mathrm{h} \rho \mathrm{g}=\left(\mathrm{r}_2+\mathrm{r}_1\right) \times 2 \pi \mathrm{T} \cos \theta
$$
Simplifying and solving for $\mathrm{h}$ we get,
$$
\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\left(\mathrm{r}_2-\mathrm{r}_1\right) \rho g}=\frac{2 \mathrm{~T}}{\left(\mathrm{r}_2-\mathrm{r}_1\right) \rho g}
$$
(For pure water, $\theta=0^{\circ}, \cos \theta=1$ )
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