Capillary Tube and Capillarity Questions in English

(A) The rise of liquid in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Since $T$,$\theta$,$r$,and $\rho$ are constant,the height of the liquid column is inversely proportional to the acceleration due to gravity: $h \propto \frac{1}{g}$.
At the surface of the Earth,the height is $X = \frac{k}{g}$,where $k$ is a constant.
At a depth '$d$' in a mine,the acceleration due to gravity is given by $g_d = g \left(1 - \frac{d}{R}\right)$.
The new height is $Y = \frac{k}{g_d} = \frac{k}{g \left(1 - \frac{d}{R}\right)}$.
Therefore,the ratio $\frac{Y}{X} = \frac{k / [g(1 - d/R)]}{k/g} = \frac{1}{1 - d/R} = \left(1 - \frac{d}{R}\right)^{-1}$.

(A) The height $h$ of a liquid column in a capillary tube is given by the formula:
$h = \frac{2T \cos \theta}{r \rho g}$
Where $T$ is the surface tension,$\theta$ is the angle of contact,$r$ is the radius of the capillary tube,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Assuming $T$,$\theta$,$\rho$,and $g$ are constants,we have:
$h \propto \frac{1}{r}$
This represents an inverse relationship,which is graphically depicted as a rectangular hyperbola.
Among the given options,graph $(ii)$ correctly shows this inverse relationship between $h$ and $r$.

(B) The vertical height $h$ of the water column in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Here,$T = 7 \times 10^{-2} \ N/m$,$r = 0.35 \ mm = 0.35 \times 10^{-3} \ m$,$\theta = 0^{\circ}$ (for glass-water contact),$\rho = 10^3 \ kg/m^3$,and $g = 10 \ m/s^2$.
Substituting the values:
$h = \frac{2 \times (7 \times 10^{-2}) \times \cos 0^{\circ}}{(0.35 \times 10^{-3}) \times 10^3 \times 10} = \frac{14 \times 10^{-2}}{3.5} = 4 \times 10^{-2} \ m = 0.04 \ m = 4 \ cm$.
When the capillary is inclined at an angle $\phi = 60^{\circ}$ with the vertical,the length $l$ of the water column along the capillary is given by $l = \frac{h}{\cos \phi}$.
$l = \frac{0.04}{\cos 60^{\circ}} = \frac{0.04}{0.5} = 0.08 \ m = 8 \ cm$.

(B) The height of water rising in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
For water,$\cos \theta \approx 1$,so $h \propto \frac{1}{r}$.
If the new radius is $r' = r/3$,the new height $h'$ will be $h' = \frac{h}{r/3} \cdot r = 3h$.
The mass of water in the capillary is given by $m = V \rho = (\pi r^2 h) \rho$.
For the new capillary,the mass $m'$ is $m' = \pi (r')^2 h' \rho$.
Substituting $r' = r/3$ and $h' = 3h$:
$m' = \pi (r/3)^2 (3h) \rho = \pi (r^2/9) (3h) \rho = \frac{1}{3} (\pi r^2 h \rho) = \frac{m}{3}$.

(A) The height $h$ of a liquid column in a capillary tube is given by the formula:
$h = \frac{2T \cos \theta}{\rho g r}$
where $T$ is the surface tension,$\theta$ is the angle of contact,$\rho$ is the density of the liquid,$g$ is the acceleration due to gravity,and $r$ is the radius of the capillary tube.
Given that the angle of contact $\theta = 90^{\circ}$,we substitute this into the formula:
$h = \frac{2T \cos 90^{\circ}}{\rho g r}$
Since $\cos 90^{\circ} = 0$,we get:
$h = 0$
Therefore,the liquid will neither rise nor fall in the capillary tube.

(B) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Given that the internal diameters are the same,the radii are equal: $r_A = r_B$.
Assuming the angles of contact are nearly equal: $\theta_A = \theta_B$.
Thus,the height $h$ is proportional to $\frac{T}{\rho}$,where $T$ is surface tension and $\rho$ is density.
Therefore,the ratio of the heights is: $\frac{h_A}{h_B} = \frac{T_A}{T_B} \times \frac{\rho_B}{\rho_A}$.
Given $\frac{\rho_A}{\rho_B} = \frac{4}{3}$ and $\frac{T_A}{T_B} = \frac{6}{5}$.
Substituting these values: $\frac{h_A}{h_B} = \frac{6}{5} \times \frac{3}{4} = \frac{18}{20} = \frac{9}{10}$.
So,the ratio is $9:10$.

(C) The height of liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
For water: $h_1 = 6 \ cm$,$\theta_1 = 0^{\circ}$,$\rho_1 = 1 \ g/cm^3$,$T_1 = T$.
$6 = \frac{2T \cos 0^{\circ}}{r \cdot 1 \cdot g} \implies 6 = \frac{2T}{rg} \implies rg = \frac{2T}{6} = \frac{T}{3}$.
For the other liquid: $T_2 = 2T$,$\theta_2 = 60^{\circ}$,$\rho_2 = 2 \ g/cm^3$.
$h_2 = \frac{2T_2 \cos \theta_2}{r \rho_2 g} = \frac{2(2T) \cos 60^{\circ}}{r \cdot 2 \cdot g} = \frac{4T \cdot 0.5}{2rg} = \frac{2T}{2rg} = \frac{T}{rg}$.
Substituting $rg = \frac{T}{3}$ into the equation for $h_2$:
$h_2 = \frac{T}{T/3} = 3 \ cm$.

(B) The height of water rise in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
Since the surface tension $T$,radius $r$,density $\rho$,and acceleration due to gravity $g$ remain constant,the product $h \cos \theta$ must be constant.
Initially,$h \cos 0^{\circ} = h(1) = h$.
When the tube is depressed such that the height above the surface is $h' = \frac{h}{3}$,the new angle of contact $\theta'$ satisfies:
$h' \cos \theta' = h \cos 0^{\circ}$
$\frac{h}{3} \cos \theta' = h(1)$
$\cos \theta' = \frac{1}{3}$
$\theta' = \cos ^{-1}\left(\frac{1}{3}\right)$.

(C) The height of the liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Given that $h, T, r$,and $g$ are constant for all three liquids.
Therefore,$\frac{\cos \theta}{\rho} = \text{constant}$.
This implies: $\frac{\cos \theta_1}{\rho_1} = \frac{\cos \theta_2}{\rho_2} = \frac{\cos \theta_3}{\rho_3}$.
Since $\rho_1 > \rho_2 > \rho_3$,it follows that $\cos \theta_1 > \cos \theta_2 > \cos \theta_3$.
Because the cosine function is a decreasing function in the range $[0, \pi/2]$,a larger cosine value corresponds to a smaller angle.
Therefore,$\theta_1 < \theta_2 < \theta_3$.

(D) The height of liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Given that the diameter $(2r)$,angle of contact $(\theta)$,and acceleration due to gravity $(g)$ are constant,the height is proportional to the ratio of surface tension to density: $h \propto \frac{T}{\rho}$.
Therefore,the ratio of the heights is given by: $\frac{h_1}{h_2} = \frac{T_1}{\rho_1} \times \frac{\rho_2}{T_2} = \left(\frac{T_1}{T_2}\right) \times \left(\frac{\rho_2}{\rho_1}\right)$.
Given $\frac{T_1}{T_2} = \frac{6}{5}$ and $\frac{\rho_1}{\rho_2} = \frac{4}{3}$,we have $\frac{\rho_2}{\rho_1} = \frac{3}{4}$.
Substituting these values: $\frac{h_1}{h_2} = \frac{6}{5} \times \frac{3}{4} = \frac{18}{20} = 0.9$.

(B) The height $h$ to which water rises in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$,which implies $h \propto \frac{1}{r}$.
The mass of water $m$ in the capillary is given by $m = \text{Volume} \times \text{Density} = (\pi r^2 h) \rho$.
Substituting $h \propto \frac{1}{r}$ into the mass equation,we get $m \propto r^2 \times \frac{1}{r}$,which simplifies to $m \propto r$.
For a new radius $r' = \frac{r}{3}$,the new mass $m'$ will be $m' = m \times \frac{r'}{r} = m \times \frac{r/3}{r} = \frac{m}{3}$.

(D) The height of water rising in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$, which implies $h \propto \frac{1}{r}$.
Thus, $h_1 r_1 = h_2 r_2$.
Given $r_2 = 2r_1$, we have $h_2 = \frac{h_1 r_1}{2r_1} = \frac{h_1}{2}$.
The mass of water in the capillary is $m = \pi r^2 h \rho$.
Let $m_1 = \pi r_1^2 h_1 \rho$ and $m_2 = \pi r_2^2 h_2 \rho$.
Taking the ratio: $\frac{m_2}{m_1} = \frac{\pi (2r_1)^2 h_2 \rho}{\pi r_1^2 h_1 \rho} = 4 \times \frac{h_2}{h_1} = 4 \times \frac{1}{2} = 2$.
Therefore, $m_2 = 2m$.

(A) When the angle of contact is zero,the radius of the meniscus $(r)$ equals the radius of the tube $(d/2)$.
Excess pressure in the first tube is $P_1 = \frac{2T}{r_1} = \frac{2T}{d_1/2} = \frac{4T}{d_1}$.
Excess pressure in the second tube is $P_2 = \frac{2T}{r_2} = \frac{2T}{d_2/2} = \frac{4T}{d_2}$.
The pressure difference between the two limbs is balanced by the hydrostatic pressure of the water column of height $h$,where $h$ is the difference in levels.
$\Delta P = P_1 - P_2 = h \rho g$.
Substituting the values of $P_1$ and $P_2$:
$h \rho g = \frac{4T}{d_1} - \frac{4T}{d_2} = 4T \left( \frac{1}{d_1} - \frac{1}{d_2} \right)$.
$h \rho g = 4T \left( \frac{d_2 - d_1}{d_1 d_2} \right)$.
Therefore,$h = \frac{4T}{\rho g} \left[ \frac{d_2 - d_1}{d_1 d_2} \right]$.

(C) The height of water rise in a capillary tube is given by the formula: $h = \frac{2 T \cos \theta}{r \rho g}$.
From this,we can see that $h \propto \frac{1}{r}$.
The area of cross-section is $A = \pi r^2$,which implies $r = \sqrt{\frac{A}{\pi}}$,so $r \propto \sqrt{A}$.
Substituting this into the proportionality for height,we get $h \propto \frac{1}{\sqrt{A}}$.
Given the initial area is $A_1 = A$ and the final area is $A_2 = \frac{A}{9}$,we have the ratio:
$\frac{h_2}{h_1} = \sqrt{\frac{A_1}{A_2}} = \sqrt{\frac{A}{A/9}} = \sqrt{9} = 3$.
Therefore,the new height $h_2 = 3 h$.

(B) The height $h$ of a water column in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Since $T, \theta, \rho,$ and $g$ are constant,we have $h \propto \frac{1}{r}$.
The mass $m$ of the liquid in the capillary is given by $m = V \rho = (\pi r^2 h) \rho$.
Substituting $h \propto \frac{1}{r}$ into the mass equation,we get $m \propto r^2 \times \frac{1}{r}$,which simplifies to $m \propto r$.
If the radius changes from $r$ to $r' = \frac{r}{4}$,the new mass $m'$ will be $m' = m \times \frac{r'}{r} = m \times \frac{r/4}{r} = \frac{m}{4}$.

(B) The height of water in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$,where $T$ is surface tension,$r$ is the radius of the tube,$\rho$ is the density of water,and $g$ is the acceleration due to gravity.
On the surface of the earth,$h = \frac{2T \cos \theta}{r \rho g}$.
At a depth $d$ inside a mine,the acceleration due to gravity becomes $g^{\prime} = g(1 - \frac{d}{R})$.
The new height $h^{\prime}$ at depth $d$ is $h^{\prime} = \frac{2T \cos \theta}{r \rho g^{\prime}} = \frac{2T \cos \theta}{r \rho g(1 - \frac{d}{R})}$.
Therefore,the ratio $\frac{h}{h^{\prime}}$ is given by:
$\frac{h}{h^{\prime}} = \frac{\frac{2T \cos \theta}{r \rho g}}{\frac{2T \cos \theta}{r \rho g(1 - \frac{d}{R})}} = 1 - \frac{d}{R}$.

(A) The height $h$ to which a liquid rises in a capillary tube is given by the formula:
$h = \frac{2T \cos \theta}{r \rho g}$
From this expression, we can see that the height of the liquid column is inversely proportional to the acceleration due to gravity $(g)$:
$h \propto \frac{1}{g}$
To make $h$ much greater than $10 \,cm$, the value of $g$ must be significantly smaller than the value of $g$ on Earth.
Among the given options, the acceleration due to gravity is minimum on the surface of the Moon $(g_{moon} \approx \frac{g_{earth}}{6})$.
Therefore, the water will rise to a greater height on the Moon.

(B) The height $h$ to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Since $T$, $\theta$, $\rho$, and $g$ are constant, we have $h \propto \frac{1}{r}$, which implies $h_1 r_1 = h_2 r_2$.
Given the cross-sectional area $A = \pi r^2$, we have $A \propto r^2$, or $r \propto \sqrt{A}$.
Therefore, $\frac{r_1}{r_2} = \sqrt{\frac{A_1}{A_2}}$.
Given that the new area $A_2 = \frac{1}{16} A_1$, we have $\frac{A_1}{A_2} = 16$.
Thus, $\frac{r_1}{r_2} = \sqrt{16} = 4$, which means $r_1 = 4 r_2$.
Using the relation $h_2 = h_1 \left( \frac{r_1}{r_2} \right)$:
$h_2 = 2 \,cm \times 4 = 8 \,cm$.

(A) The water rises in the annular space between the glass rod and the capillary tube. The total length of the contact line is the sum of the inner circumference of the tube and the outer circumference of the rod,which is $L = 2\pi r_1 + 2\pi r_2 = 2\pi(r_1 + r_2)$.
The upward force due to surface tension is $F = L \cdot T \cos \theta = 2\pi(r_1 + r_2) T \cos \theta$.
The weight of the water column in the annular space is $W = \text{Volume} \times \rho \times g = \pi(r_2^2 - r_1^2) h \rho g$.
Equating the upward force to the weight of the liquid column: $\pi(r_2^2 - r_1^2) h \rho g = 2\pi(r_1 + r_2) T \cos \theta$.
Using the identity $r_2^2 - r_1^2 = (r_2 - r_1)(r_2 + r_1)$,we get $\pi(r_2 - r_1)(r_2 + r_1) h \rho g = 2\pi(r_1 + r_2) T \cos \theta$.
For pure water,$\theta = 0^{\circ}$,so $\cos \theta = 1$. Simplifying for $h$:
$h = \frac{2T}{(r_2 - r_1)\rho g}$.

(D) The pressure at the lower end of the capillary tube is the sum of the hydrostatic pressure due to the depth of the tube and the pressure due to the capillary rise.
Given that the depth of the lower end is $8 \,cm$ and the capillary rise is $4 \,cm$.
The total pressure $P$ at the lower end is given by $P = h_{depth} + h_{rise}$.
Substituting the values,$P = 8 \,cm + 4 \,cm = 12 \,cm$ of water.
Therefore,the pressure required to blow an air bubble at the lower end is $12 \,cm$ of water.
Thus,$X = 12$.

(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 $r$ is the radius of the tube.
This implies $h \propto \frac{1}{r}$.
Since the cross-sectional area $A = \pi r^2$, we have $r = \sqrt{\frac{A}{\pi}}$, which means $r \propto \sqrt{A}$.
Substituting this into the proportionality, we get $h \propto \frac{1}{\sqrt{A}}$.
Given $h_1 = 15 \,mm = 15 \times 10^{-3} \,m$ for area $A_1 = A$.
For the new area $A_2 = A / 3$, we have the ratio $\frac{A_1}{A_2} = 3$.
Using the relation $\frac{h_2}{h_1} = \sqrt{\frac{A_1}{A_2}}$, we get $\frac{h_2}{h_1} = \sqrt{3}$.
Therefore, $h_2 = h_1 \times \sqrt{3} = 15 \times 10^{-3} \times \sqrt{3} \,m = 15 \sqrt{3} \times 10^{-3} \,m$.

(D) The height of the liquid column in a capillary tube is given by the formula: $h = \frac{2 T \cos \theta}{r \rho g}$.
From this expression,it is clear that $h \propto \frac{1}{g}$.
When a lift moves downward with an acceleration '$a$',the effective acceleration due to gravity becomes $g_{eff} = g - a$.
Since $a < g$,the effective gravity $g_{eff}$ is less than the actual gravity $g$.
As $g_{eff} < g$,the value of '$h$' increases because '$h$' is inversely proportional to the acceleration due to gravity.

(B) The height of the liquid column in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
From this expression,we can see that $h \propto \frac{1}{g}$.
Therefore,the ratio of the heights is $\frac{h_{1}}{h_{2}} = \frac{g_{2}}{g_{1}}$.
At a depth $d$ below the surface of the Earth,the acceleration due to gravity is given by $g_{2} = g_{1} \left(1 - \frac{d}{R}\right)$,where $g_{1}$ is the acceleration due to gravity at the surface.
Substituting this into the ratio,we get $\frac{h_{1}}{h_{2}} = \frac{g_{1}(1 - d/R)}{g_{1}} = 1 - \frac{d}{R}$.

(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 $r$ is the radius of the tube.
Since $h \propto \frac{1}{r}$, and the area of cross-section $A = \pi r^2$, we have $r = \sqrt{\frac{A}{\pi}}$, which implies $r \propto \sqrt{A}$.
Therefore, $h \propto \frac{1}{\sqrt{A}}$, or $h_1 \sqrt{A_1} = h_2 \sqrt{A_2}$.
Given $h_1 = 2.2 \text{ cm}$ and $A_2 = \frac{1}{4} A_1$, we have $\sqrt{A_2} = \frac{1}{2} \sqrt{A_1}$.
Substituting these values: $2.2 \times \sqrt{A_1} = h_2 \times \frac{1}{2} \sqrt{A_1}$.
Solving for $h_2$: $h_2 = 2.2 \times 2 = 4.4 \text{ cm}$.

(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 contact angle, $r$ is the radius of the tube, $\rho$ is the density, and $g$ is acceleration due to gravity.
From this, we see that $h \propto \frac{1}{r}$.
The cross-sectional area $A$ is given by $A = \pi r^2$, which implies $r \propto \sqrt{A}$.
Substituting this into the height relation, we get $h \propto \frac{1}{\sqrt{A}}$, or $h_1 \sqrt{A_1} = h_2 \sqrt{A_2}$.
Given $h_1 = 3 \,cm$ and $A_2 = \frac{1}{9} A_1$, we have $\sqrt{A_2} = \frac{1}{3} \sqrt{A_1}$.
Substituting these values: $3 \times \sqrt{A_1} = h_2 \times \frac{1}{3} \sqrt{A_1}$.
Solving for $h_2$: $h_2 = 3 \times 3 = 9 \,cm$.

(A) The height of water in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
Since $T, \theta, \rho,$ and $g$ are constant,$h \propto \frac{1}{r}$,which implies $hr = \text{constant}$.
For a capillary of radius $r_1 = r$,the height is $h_1 = h$. For a capillary of radius $r_2 = \frac{r}{4}$,the new height $h_2$ is given by $h_1 r_1 = h_2 r_2$.
$h \times r = h_2 \times \frac{r}{4} \implies h_2 = 4h$.
The mass of water in a capillary is given by $m = \text{Volume} \times \text{density} = (\pi r^2 h) \rho$.
For the new capillary,the mass $m'$ is $m' = \pi (r_2)^2 h_2 \rho$.
Substituting $r_2 = \frac{r}{4}$ and $h_2 = 4h$:
$m' = \pi \left(\frac{r}{4}\right)^2 (4h) \rho = \pi \left(\frac{r^2}{16}\right) (4h) \rho = \frac{1}{4} (\pi r^2 h \rho) = \frac{m}{4}$.

(D) The height $h$ to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
From this relation,we can see that $h \propto \frac{1}{g}$.
For a lift going down with an acceleration $a$,the effective acceleration due to gravity is $g' = g - a$. Since $g' < g$,the height $h$ will increase.
However,in a satellite orbiting the earth,the effective gravity $g'$ becomes $0$ (weightlessness). As $g' \to 0$,$h \to \infty$. Thus,the height $h$ increases significantly in a satellite compared to the other options where $g$ is only partially reduced.

(D) The height $h$ to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Since $T$,$\theta$,$\rho$,and $g$ are constants for the given liquid and tube material,we have the relation $h \propto \frac{1}{r}$,which implies $h_{1} r_{1} = h_{2} r_{2}$.
Given $h_{1} = h$,$r_{1} = r$,and $r_{2} = \frac{r}{3}$.
Substituting these values into the equation: $h \cdot r = h_{2} \cdot \frac{r}{3}$.
Solving for $h_{2}$: $h_{2} = 3h$.

(D) The rise of a liquid in a capillary tube is given by the formula: $h = \frac{2 T \cos \theta}{r \rho g}$.
Rearranging for the angle of contact,we get: $\cos \theta = \frac{h r \rho g}{2 T}$.
Here,$h$ (rise),$r$ (radius),and $T$ (surface tension) are constants for all three liquids.
Thus,$\cos \theta \propto \rho$.
Given the densities are $\varrho_{1} > \varrho_{2} > \varrho_{3}$,it follows that $\cos \theta_{1} > \cos \theta_{2} > \cos \theta_{3}$.
Since the cosine function is a decreasing function for angles between $0$ and $\frac{\pi}{2}$,a larger cosine value corresponds to a smaller angle.
Therefore,$\theta_{1} < \theta_{2} < \theta_{3}$.

(D) The height of the liquid column in a capillary tube is given by the formula: $h = \frac{2 T \cos \theta}{\rho g r}$,where $T$ is the surface tension,$\theta$ is the contact angle,$\rho$ is the density of the liquid,$g$ is the acceleration due to gravity,and $r$ is the radius of the capillary tube.
From the formula,we can see that $h \propto \frac{1}{r}$.
This implies that the height of the water rise is inversely proportional to the radius (or diameter) of the capillary tube.
Therefore,the rise of water will be greater in the tube with the smaller diameter.

(B) 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,$\theta$ 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.
Since $T, \theta, \rho,$ and $g$ are constant,we have $h \propto \frac{1}{r}$,which means $rh = \text{constant}$.
The area of cross-section is $A = \pi r^2$,so $r = \sqrt{\frac{A}{\pi}}$,which implies $r \propto \sqrt{A}$.
Given the initial area $A_1 = A$ and final area $A_2 = \frac{A}{9}$.
The ratio of radii is $\frac{r_2}{r_1} = \sqrt{\frac{A_2}{A_1}} = \sqrt{\frac{A/9}{A}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Using the relation $r_1 h_1 = r_2 h_2$,we get $h_2 = h_1 \left( \frac{r_1}{r_2} \right)$.
Substituting the values,$h_2 = h \times 3 = 3h$.

(C) The rise of water $h$ in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{\rho g R_{eff}}$,where $R_{eff}$ is the effective radius of the meniscus.
In this case,the water rises in the annular space between the capillary tube of radius $R$ and the wire of radius $r$.
The effective radius for the annular space is the difference between the radii,$R_{eff} = R - r$.
Assuming the contact angle $\theta = 0$ for water and glass/metal,we have $\cos \theta = 1$.
Substituting these values into the formula,we get $h = \frac{2T}{\rho g (R - r)}$.

(C) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Given for liquid '$A$': $h_1 = 5 \,cm$,surface tension = $T_1$,density = $\rho_1$.
Given for liquid '$B$': surface tension $T_2 = 2T_1$,density $\rho_2 = 2\rho_1$,and the radius $r$ and angle of contact $\theta$ remain the same.
Substituting these into the formula for liquid '$B$':
$h_2 = \frac{2T_2 \cos \theta}{r \rho_2 g} = \frac{2(2T_1) \cos \theta}{r(2\rho_1) g} = \frac{4T_1 \cos \theta}{2r \rho_1 g} = \frac{2T_1 \cos \theta}{r \rho_1 g}$.
Since $h_1 = \frac{2T_1 \cos \theta}{r \rho_1 g} = 5 \,cm$,it follows that $h_2 = h_1 = 5 \,cm$.
Converting to meters: $5 \,cm = 0.05 \,m$.

(B) The height of capillary rise $h$ of a liquid in a capillary tube is given by the formula:
$h = \frac{2 T \cos \theta}{r \rho g}$
Since $T$,$\theta$,$\rho$,and $g$ are constant for the same liquid and tube material,we have:
$h \propto \frac{1}{r}$
Since the diameter $D = 2r$,we can also write $h \propto \frac{1}{D}$.
Given that $h_P = \frac{2}{3} h_Q$,we have $\frac{h_P}{h_Q} = \frac{2}{3}$.
Using the inverse proportionality $h \propto \frac{1}{D}$,we get:
$\frac{h_P}{h_Q} = \frac{D_Q}{D_P} = \frac{2}{3}$
Therefore,the ratio of their diameters is $\frac{D_P}{D_Q} = \frac{3}{2}$ or $3: 2$.

(C) The height of water in a capillary tube is given by the formula:
$h = \frac{2T \cos \theta}{r \rho g}$
where $T$ is the surface tension,$\theta$ is the angle of contact,$r$ is the radius of the capillary tube,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
From this formula,we can see that $h \propto \frac{1}{r}$.
Since the height $h$ is inversely proportional to the radius $r$ of the tube,the rise of water will be greater in the tube with the smaller diameter (smaller radius).

(A) The rise or fall of liquids in narrow tubes or porous materials is known as capillarity.
Plant fibres act as a network of fine capillary tubes.
Therefore,water rises in plant fibres due to the phenomenon of capillarity.

(B) Given: Radius of capillary tube $r = 0.1 \,mm = 0.1 \times 10^{-3} \,m$.
Height of water rise $h = 2 \,cm = 2 \times 10^{-2} \,m$.
Surface tension $T = 0.072 \,N/m$.
Density of water $\rho = 10^3 \,kg/m^3$ and acceleration due to gravity $g = 10 \,m/s^2$.
The formula for capillary rise is $h = \frac{2T \cos \theta}{\rho g r}$.
Rearranging for $\cos \theta$: $\cos \theta = \frac{h \rho g r}{2T}$.
Substituting the values: $\cos \theta = \frac{(2 \times 10^{-2}) \times (10^3) \times (10) \times (0.1 \times 10^{-3})}{2 \times 0.072}$.
$\cos \theta = \frac{2 \times 10^{-2} \times 10^4 \times 0.1 \times 10^{-3}}{0.144} = \frac{0.02}{0.144} = \frac{20}{144} = \frac{1}{7.2}$.
Therefore, $\theta = \cos^{-1}\left(\frac{1}{7.2}\right)$.

(C) The height of capillary rise in a tube is given by $h = \frac{2T \cos \theta}{r \rho g}$,where $T$ is surface tension,$\theta$ is the angle of contact,$r$ is the radius of the capillary tube,and $\rho$ is the density of the liquid.
Given: $T = 0.03 \ Nm^{-1}$,$\rho = 1500 \ kgm^{-3}$,$\theta = 0^\circ$ (so $\cos \theta = 1$),$g = 10 \ ms^{-2}$,$r_1 = 2 \ mm = 2 \times 10^{-3} \ m$,and $r_2 = 4 \ mm = 4 \times 10^{-3} \ m$.
The heights in the two limbs are $h_1 = \frac{2T}{r_1 \rho g}$ and $h_2 = \frac{2T}{r_2 \rho g}$.
The difference in liquid levels is $\Delta h = h_1 - h_2 = \frac{2T}{\rho g} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.
Substituting the values: $\Delta h = \frac{2 \times 0.03}{1500 \times 10} \left( \frac{1}{2 \times 10^{-3}} - \frac{1}{4 \times 10^{-3}} \right)$.
$\Delta h = \frac{0.06}{15000} \left( 500 - 250 \right) = \frac{0.06}{15000} \times 250 = \frac{0.06}{60} = 0.001 \ m$.
Therefore,$\Delta h = 1 \ mm$.

(D) The height of the rise of liquid due to surface tension in a capillary tube is given by the formula: $h = \frac{2 S \cos \theta}{r \rho g}$.
Given values are: height of liquid rise $h = 7.5 \ cm = 7.5 \times 10^{-2} \ m$,surface tension $S = 7.5 \times 10^{-2} \ N \ m^{-1}$,angle of contact $\theta = 0^{\circ}$,density of water $\rho = 1000 \ kg \ m^{-3}$,and acceleration due to gravity $g = 10 \ m \ s^{-2}$.
Rearranging the formula to solve for the radius $r$: $r = \frac{2 S \cos \theta}{h \rho g}$.
Substituting the values: $r = \frac{2 \times (7.5 \times 10^{-2}) \times \cos 0^{\circ}}{(7.5 \times 10^{-2}) \times 1000 \times 10}$.
Since $\cos 0^{\circ} = 1$,we get: $r = \frac{2 \times 7.5 \times 10^{-2}}{7.5 \times 10^{-2} \times 10^4} = \frac{2}{10^4} = 2 \times 10^{-4} \ m$.
Converting to millimeters: $r = 2 \times 10^{-4} \times 10^3 \ mm = 0.2 \ mm$.

(D) Consider a capillary tube of radius $r$ dipped in a liquid. The liquid forms a meniscus at the top of the tube.
Let $R$ be the radius of curvature of the meniscus and $\theta$ be the angle of contact.
From the geometry of the meniscus,we can form a right-angled triangle where the hypotenuse is $R$,the base is $r$,and the angle between the radius of the meniscus and the horizontal is $\theta$.
Using trigonometry,we have $\cos \theta = \frac{r}{R}$.
Therefore,the radius of curvature of the meniscus is $R = \frac{r}{\cos \theta}$.

(A) The height of water in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
Since $T$,$\theta$,$\rho$,and $g$ are constant,$h \propto \frac{1}{r}$.
Thus,$h_1 = \frac{k}{R}$ and $h_2 = \frac{k}{2R}$,where $k$ is a constant.
The mass of water in the tube is $m = \text{Volume} \times \text{Density} = (\pi r^2 h) \rho$.
Substituting $h = \frac{2T \cos \theta}{r \rho g}$,we get $m = \pi r^2 \left( \frac{2T \cos \theta}{r \rho g} \right) \rho = \frac{2 \pi T r \cos \theta}{g}$.
Since $T$,$\theta$,and $g$ are constant,$m \propto r$.
Therefore,$\frac{m_1}{m_2} = \frac{R}{2R} = \frac{1}{2}$.

(A) The pressure just below the meniscus in a capillary tube of radius $r$ is given by $P = P_0 - \frac{2T}{r}$,where $P_0$ is the atmospheric pressure and $T$ is the surface tension.
For the two limbs with radii $R_1 = 2 \ mm = 2 \times 10^{-3} \ m$ and $R_2 = 4 \ mm = 4 \times 10^{-3} \ m$,the pressures at the same horizontal level $OO'$ are equal.
Let $h_1$ and $h_2$ be the heights of the water columns in the two limbs above the level $OO'$. The pressure at level $OO'$ in the two limbs is:
$P_{OO'} = P_0 - \frac{2T}{R_1} + \rho g h_1 = P_0 - \frac{2T}{R_2} + \rho g h_2$
Since the total amount of water is constant,the difference in levels $x = h_1 - h_2$ is determined by the capillary rise difference:
$x = h_1 - h_2 = \frac{2T}{\rho g} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Substituting the values:
$x = \frac{2 \times 7.3 \times 10^{-2}}{1.0 \times 10^3 \times 10} \left( \frac{1}{2 \times 10^{-3}} - \frac{1}{4 \times 10^{-3}} \right)$
$x = \frac{14.6 \times 10^{-2}}{10^4} \left( \frac{2 - 1}{4 \times 10^{-3}} \right) = 14.6 \times 10^{-6} \times \frac{1}{4 \times 10^{-3}} = 3.65 \times 10^{-3} \ m = 3.65 \ mm$.
Thus,the difference between the levels is $3.65 \ mm$.

(A) The height of a liquid column in a capillary tube is given by $h = \frac{2T \cos \theta}{rdg}$.
For water: $h_1 = \frac{2T_1 \cos \theta_1}{rdg}$. Given $T_1 = T$,$\theta_1 = 0^{\circ}$ (so $\cos \theta_1 = 1$),and mass $m_1 = 5 \times 10^{-3} \ kg$.
Since $m = \pi r^2 h d$,we have $h_1 = \frac{m_1}{\pi r^2 d}$.
For the second liquid: $T_2 = \sqrt{2}T$ and $\theta_2 = 45^{\circ}$.
The new height $h_2 = \frac{2T_2 \cos \theta_2}{rdg} = \frac{2(\sqrt{2}T) \cos 45^{\circ}}{rdg} = \frac{2\sqrt{2}T \times (1/\sqrt{2})}{rdg} = \frac{2T}{rdg} = h_1$.
Since the radius '$r$' and density '$d$' (assuming same density for simplicity or as implied by the problem context) remain constant,and $h_2 = h_1$,the mass of the liquid $m_2 = \pi r^2 h_2 d = m_1 = 5 \times 10^{-3} \ kg$.

(B) The height of capillary rise is given by the formula: $h = \frac{2 S \cos \theta}{r \rho g}$,where $r$ is the radius of the tube.
For liquid $A$: $h_A = 10 \ cm$,$\rho_A = 1 \ g/cm^3$,$\theta_A = 0^{\circ}$.
$10 = \frac{2 S_A \cos(0^{\circ})}{r \times 1 \times g} = \frac{2 S_A}{r g}$ --- $(1)$
For liquid $B$: $h_B = -2 \ cm$ (fall),$\rho_B = 10 \ g/cm^3$,$\theta_B = 135^{\circ}$.
$-2 = \frac{2 S_B \cos(135^{\circ})}{r \times 10 \times g} = \frac{2 S_B (-1/\sqrt{2})}{10 r g} = -\frac{S_B}{5 \sqrt{2} r g}$
$2 = \frac{S_B}{5 \sqrt{2} r g}$ --- $(2)$
Dividing $(2)$ by $(1)$:
$\frac{2}{10} = \frac{S_B / (5 \sqrt{2} r g)}{2 S_A / (r g)} = \frac{S_B}{5 \sqrt{2} r g} \times \frac{r g}{2 S_A} = \frac{S_B}{10 \sqrt{2} S_A}$
$\frac{1}{5} = \frac{S_B}{10 \sqrt{2} S_A} \Rightarrow \frac{S_B}{S_A} = \frac{10 \sqrt{2}}{5} = 2 \sqrt{2}$.

(D) For a capillary tube open at both ends,the water column is supported by surface tension forces at the meniscus.
When a water column of length $h$ is held in a capillary tube,the pressure difference due to surface tension must balance the hydrostatic pressure of the water column.
For a tube open at both ends,the water will not be able to be supported if the tube is vertical because the pressure at the top and bottom meniscus would be atmospheric,and gravity would pull the water down.
However,if we consider the capillary rise formula $h = \frac{2T \cos \theta}{r \rho g}$,this applies to a tube dipping in a reservoir.
In a tube open at both ends,the water column cannot be supported against gravity unless there is a pressure difference.
Since the tube is open at both ends,the pressure at both ends is atmospheric pressure $(P_0)$.
Thus,the water column will fall out due to gravity.
Therefore,the maximum length of the water column that can stay without falling is $0 \ cm$.

(C) The height to which water rises in a capillary tube is given by $h = \frac{2T}{\rho g r}$.
The potential energy of the water column of mass $m$ is $U = \frac{mgh}{2}$.
Since $m = \pi r^2 h \rho$,we have $U = \frac{(\pi r^2 h \rho) g h}{2} = \frac{\pi r^2 \rho g h^2}{2}$.
Substituting $h = \frac{2T}{\rho g r}$,we get $U = \frac{\pi r^2 \rho g}{2} \left( \frac{2T}{\rho g r} \right)^2 = \frac{2 \pi T^2}{\rho g}$.
The work done by the surface tension force is $W = (2 \pi r T) h = 2 \pi r T \left( \frac{2T}{\rho g r} \right) = \frac{4 \pi T^2}{\rho g}$.
From the principle of conservation of energy,the heat evolved $Q$ is the difference between the work done and the potential energy gained:
$Q = W - U = \frac{4 \pi T^2}{\rho g} - \frac{2 \pi T^2}{\rho g} = \frac{2 \pi T^2}{\rho g}$.