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 $+1\%$.

(B) The pressure difference across the meniscus of the water in the capillary is given by the Young-Laplace equation: $\Delta P = 2T/R_{eff}$.
Here, the effective radius of the capillary space is the gap between the tube and the wire, which is $R_{eff} = R - r$.
At equilibrium, the pressure difference balances the hydrostatic pressure of the water column of height '$h$': $\Delta P = h\rho g$.
Equating the two expressions: $2T/(R-r) = h\rho g$.
Solving for '$h$': $h = 2T/((R-r)\rho g)$.

(B) The formula for the capillary rise is given by $h = \frac{2T \cos \theta}{r \rho g}$.
Given that $h$, $T$, $r$, and $g$ are constant for all three liquids, we have the relation $\cos \theta \propto \rho$.
Since the densities are given as $\rho_1 > \rho_2 > \rho_3$, it follows that $\cos \theta_1 > \cos \theta_2 > \cos \theta_3$.
As the cosine function is a decreasing function for angles between $0^\circ$ and $90^\circ$, a larger cosine value corresponds to a smaller angle.
Therefore, the relationship between the angles of contact is $\theta_1 < \theta_2 < \theta_3$.

(A) The height of water rise in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$, where $T$ is surface tension, $\rho$ is density, and $g$ is acceleration due to gravity.
From this, we see that $h \propto \frac{1}{r}$.
If the radius becomes $xr$, the new height $h'$ will be $h' = \frac{h}{x}$.
The mass of water in the capillary tube is given by $m = V \cdot \rho = (\pi r^2 h) \cdot \rho$.
Substituting $h = \frac{2T \cos \theta}{r \rho g}$ into the mass formula:
$m = \pi r^2 \left( \frac{2T \cos \theta}{r \rho g} \right) \rho = \frac{2 \pi r T \cos \theta}{g}$.
Thus, $m \propto r$.
If the radius changes from $r$ to $xr$, the new mass $m'$ will be $m' = x \cdot m$.

(D) The height $h$ to which a liquid rises in a capillary tube of radius $r$ is given by the formula:
$h = \frac{2T \cos \theta}{r \rho g}$
Where:
$T$ is the surface tension of the liquid,
$\theta$ is the angle of contact,
$\rho$ is the density of the liquid,
$g$ is the acceleration due to gravity.
For a given liquid and capillary tube material, $T$, $\theta$, $\rho$, and $g$ are constants.
Therefore, $h \propto \frac{1}{r}$, which implies $h \cdot r = \text{constant}$.
Thus, for different capillary tubes, $h_n r_n = \text{constant}$.

(A) The height $h$ to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$, where $T$ is surface tension, $\theta$ is the angle of contact, $r$ is the radius of the capillary, $\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity.
From this formula, we see that $h \propto \frac{1}{r}$.
The cross-sectional area $A$ of the capillary tube is given by $A = \pi r^2$, which implies $r = \sqrt{\frac{A}{\pi}}$, so $r \propto \sqrt{A}$.
Substituting this into the height relation, we get $h \propto \frac{1}{\sqrt{A}}$.
Given that the new area $A' = \frac{A}{3}$, the new height $h'$ will be:
$h' = h \times \sqrt{\frac{A}{A'}} = 3 \times \sqrt{\frac{A}{A/3}} = 3 \times \sqrt{3} = 3\sqrt{3}$ cm.

(B) The formula for the capillary rise $h$ is given by $h = \frac{2T \cos \theta}{r \rho g}$, where $T$ is the surface tension, $\theta$ is the angle of contact, $r$ is the radius of the capillary, $\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity.
Given that $h$, $T$, $r$, and $g$ are the same for all three liquids, we have the relation $\cos \theta \propto \rho$.
Since $\rho_1 < \rho_2 < \rho_3$, it follows that $\cos \theta_1 < \cos \theta_2 < \cos \theta_3$.
As the cosine function is a decreasing function for angles between $0^\circ$ and $90^\circ$, a smaller cosine value corresponds to a larger angle.
Therefore, $\theta_1 > \theta_2 > \theta_3$.

(B) The height '$h$' to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$, where '$T$' is surface tension,'$\theta$' is the angle of contact,'$r$' is the radius of the capillary tube,'$\rho$' is the density of the liquid, and '$g$' is the acceleration due to gravity.
From this formula, we see that $h \propto \frac{1}{r}$.
The area of cross-section '$a$' is given by $a = \pi r^2$, which implies $r = \sqrt{\frac{a}{\pi}}$, so $r \propto \sqrt{a}$.
Substituting this into the height relation, we get $h \propto \frac{1}{\sqrt{a}}$.
Let $h_1 = h$ for area $a_1 = a$, and $h_2$ be the height for area $a_2 = 4a$.
Then, $\frac{h_2}{h_1} = \sqrt{\frac{a_1}{a_2}} = \sqrt{\frac{a}{4a}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore, $h_2 = \frac{h}{2}$.

(C) The height of the liquid column in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$, where $T$ is surface tension, $\theta$ is the contact angle, $r$ is the radius of the tube, $\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity.
In a freely falling elevator, the effective acceleration due to gravity $g_{\text{eff}} = g - a$. Since the elevator is in free fall, $a = g$, therefore $g_{\text{eff}} = 0$.
As $h \propto \frac{1}{g_{\text{eff}}}$, when $g_{\text{eff}} = 0$, the height $h$ tends to infinity.
However, the water will rise until it reaches the top of the capillary tube. Since the tube is $30 \text{ cm}$ long, the water column will fill the entire tube.