A uniform capillary tube of length l and inne | vedclass

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(D) Let $A$ be the cross-sectional area of the capillary tube.Initially,the tube is filled with air at pressure $p_{0}$ and volume $V = lA$.When the t

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$\frac{l}{\left(1+\frac{p_{0} r}{2 \gamma}\right)}$

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(D) Let $A$ be the cross-sectional area of the capillary tube.Initially,the tube is filled with air at pressure $p_{0}$ and volume $V = lA$.When the tube is submerged by a length $x$,the air is compressed into a length $(l-x)$. Let the new pressure be $p^{\prime}$.Using Boyle's Law: $p_{0}(lA) = p^{\prime}(l-x)A$,which gives $p^{\prime} = \frac{p_{0}l}{l-x}$.Since the water levels inside and outside coincide,the pressure difference across the meniscus is given by the Young-Laplace equation: $p^{\prime} - p_{0} = \frac{2\gamma}{r}$.Substituting $p^{\prime}$: $\frac{p_{0}l}{l-x} - p_{0} = \frac{2\gamma}{r}$.$p_{0} \left( \frac{l}{l-x} - 1 \right) = \frac{2\gamma}{r} \implies p_{0} \left( \frac{l - l + x}{l-x} \right) = \frac{2\gamma}{r}$.$\frac{p_{0}x}{l-x} = \frac{2\gamma}{r} \implies p_{0}xr = 2\gamma l - 2\gamma x$.$x(p_{0}r + 2\gamma) = 2\gamma l$.$x = \frac{2\gamma l}{p_{0}r + 2\gamma} = \frac{l}{\frac{p_{0}r}{2\gamma} + 1} = \frac{l}{1 + \frac{p_{0}r}{2\gamma}}$.

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