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Question
Figure shows a capillary tube of radius r dipped into water. If the atmospheric pressure is P0, the pressure at point A is
Options
P0
\[P_0 + \frac{2S}{r}\]
\[P_0 - \frac{2S}{r}\]
\[P_0 - \frac{4S}{r}\]
Solution
\[\text{ Here }: \]
\[\text{ Radius of the tube = r }\]
\[\text{ Net upward force due to surface tension = S }\text{ cos }\theta \times 2\pi r\]
\[\text{ Upward pressure }= \frac{\text{ S } cos \theta \times 2\pi r}{\pi r^2} = \frac{2Scos\theta}{r}\]
\[\text{ Net downward pressure due to atmosphere }= P_o \]
\[ \Rightarrow \text{ Net pressure at A }= P_o - \frac{2Scos\theta}{r}\]
\[\text{ Since }\theta \text{ is small, } \]
\[\text{ cos }\theta \approx 1 . \]
\[ \Rightarrow \text{ Net pressure }= P_o - \frac{2S}{r}\]
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