MCQ

Figure shows a capillary tube of radius r dipped into water. If the atmospheric pressure is P_{0}, the pressure at point A is

#### Options

*P*_{0}\[P_0 + \frac{2S}{r}\]

\[P_0 - \frac{2S}{r}\]

\[P_0 - \frac{4S}{r}\]

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#### Solution

\[\text{ Here }: \]

\[\text{ Radius of the tube = r }\]

\[\text{ Net upward force due to surface tension = S }\text{ cos }\theta \times 2\pi r\]

\[\text{ Upward pressure }= \frac{\text{ S } cos \theta \times 2\pi r}{\pi r^2} = \frac{2Scos\theta}{r}\]

\[\text{ Net downward pressure due to atmosphere }= P_o \]

\[ \Rightarrow \text{ Net pressure at A }= P_o - \frac{2Scos\theta}{r}\]

\[\text{ Since }\theta \text{ is small, } \]

\[\text{ cos }\theta \approx 1 . \]

\[ \Rightarrow \text{ Net pressure }= P_o - \frac{2S}{r}\]

Is there an error in this question or solution?

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