Capillary Tube and Capillarity Questions in English

(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 $p^{\prime}$: $\frac{p_{0}l}{l-x} - p_{0} = \frac{2\gamma}{r}$.
$p_{0} \left( \frac{l}{l-x} - 1 \right) = \frac{2\gamma}{r} \implies p_{0} \left( \frac{l - l + x}{l-x} \right) = \frac{2\gamma}{r}$.
$\frac{p_{0}x}{l-x} = \frac{2\gamma}{r} \implies p_{0}xr = 2\gamma l - 2\gamma x$.
$x(p_{0}r + 2\gamma) = 2\gamma l$.
$x = \frac{2\gamma l}{p_{0}r + 2\gamma} = \frac{l}{\frac{p_{0}r}{2\gamma} + 1} = \frac{l}{1 + \frac{p_{0}r}{2\gamma}}$.

(D) The height of the liquid column in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
In a freely falling platform,the effective acceleration due to gravity $g_{eff}$ becomes $0$ because the system is in a state of weightlessness.
As $g_{eff} \to 0$,the height $h$ tends to infinity $(h \propto \frac{1}{g_{eff}})$.
However,the liquid cannot rise beyond the physical length of the capillary tube.
Therefore,the water will rise to fill the entire length of the capillary tube,which is $20 cm$.

(A) The formula for capillary rise is given by $h = \frac{2T \cos \theta}{\rho g r}$.
Assuming the contact angle $\theta$ is the same for both liquids and the densities $\rho$ are equal,we have $h \propto \frac{1}{r}$.
Given $r_1 > r_2$,it follows that $\frac{1}{r_1} < \frac{1}{r_2}$.
Therefore,$h_1 < h_2$.
Since the problem specifies $T_1 = T_2$,the correct relation is $h_1 < h_2$ and $T_1 = T_2$.

(D) The height $h$ of a liquid in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{\rho gr}$.
Assuming the angle of contact $\theta$ and acceleration due to gravity $g$ remain constant, the relative change is given by: $\frac{\Delta h}{h} = \frac{\Delta T}{T} - \frac{\Delta \rho}{\rho} - \frac{\Delta r}{r}$.
Given that $T$, $\rho$, and $r$ each decrease by $1\%$, we have $\frac{\Delta T}{T} = -0.01$, $\frac{\Delta \rho}{\rho} = -0.01$, and $\frac{\Delta r}{r} = -0.01$.
Substituting these values: $\frac{\Delta h}{h} = (-0.01) - (-0.01) - (-0.01) = -0.01 + 0.01 + 0.01 = +0.01$.
Therefore, the height will change by $+1\%$.