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A glass capillary tube of inner diameter 0.28 $\mathrm{mm}$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that water level in the tube is same as that in the vessel (in $\mathrm{N} / \mathrm{m}^2$ ) is
Surface tension of water $=0.07 \mathrm{~N} / \mathrm{m}$
Atmospheric pressure $=10^5 \mathrm{~N} / \mathrm{m}^2$
Options:
Surface tension of water $=0.07 \mathrm{~N} / \mathrm{m}$
Atmospheric pressure $=10^5 \mathrm{~N} / \mathrm{m}^2$
Solution:
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Verified Answer
The correct answer is:
$101 \times 10^3$
Diameter $d=0.28 \mathrm{~mm}=0.28 \times 10^{-3} \mathrm{~m}$
Radius $r=\frac{d}{2}=0.14 \times 10^{-3} \mathrm{~m}$
Surface tension $T=0.07 \mathrm{~N} / \mathrm{m}$
$$
T=\frac{r h d g}{2 \cos \theta}
$$
For water $\theta=0^{\circ}$
$$
\begin{aligned}
\therefore \quad \cos \theta & =1 \\
T & =\frac{r(h d g)}{2}=\frac{r P}{2}
\end{aligned}
$$
$\begin{aligned} & \qquad \begin{aligned} P & =\frac{2 T}{r}=\frac{2 \times 0.07}{0.14 \times 10^{-3}} \\ & =1 \times 10^3 \mathrm{~N} / \mathrm{m}^2 \\ \text { Total pressure } & =P+\text { atmospheric pressure } \\ & =1 \times 10^3+10^5 \\ & =1 \times 10^3+100 \times 10^3 \\ & =101 \times 10^3 \mathrm{~N} / \mathrm{m}^2\end{aligned}\end{aligned}$
Radius $r=\frac{d}{2}=0.14 \times 10^{-3} \mathrm{~m}$
Surface tension $T=0.07 \mathrm{~N} / \mathrm{m}$
$$
T=\frac{r h d g}{2 \cos \theta}
$$
For water $\theta=0^{\circ}$
$$
\begin{aligned}
\therefore \quad \cos \theta & =1 \\
T & =\frac{r(h d g)}{2}=\frac{r P}{2}
\end{aligned}
$$
$\begin{aligned} & \qquad \begin{aligned} P & =\frac{2 T}{r}=\frac{2 \times 0.07}{0.14 \times 10^{-3}} \\ & =1 \times 10^3 \mathrm{~N} / \mathrm{m}^2 \\ \text { Total pressure } & =P+\text { atmospheric pressure } \\ & =1 \times 10^3+10^5 \\ & =1 \times 10^3+100 \times 10^3 \\ & =101 \times 10^3 \mathrm{~N} / \mathrm{m}^2\end{aligned}\end{aligned}$
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