Capillary Tube and Capillarity Questions in English

(A) The height of the water column in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
Since $T$,$\rho$,$g$,and the contact angle $\theta$ are constant for a given liquid and tube,we have $h \cdot r = \text{constant}$.
Initially,$h_1 = 8 \ cm$. When the tube is pushed down such that the height above the water level is $h_2 = 5 \ cm$,the water cannot overflow because the radius of curvature $R$ of the meniscus will adjust to satisfy the equilibrium condition.
Specifically,$h_1 r_1 = h_2 R_2$,where $R_2$ is the new radius of the meniscus.
As $h$ decreases from $8 \ cm$ to $5 \ cm$,the radius of the meniscus $R$ increases to maintain the pressure balance.
Therefore,the water does not overflow; it simply adjusts its meniscus shape.

(C) The height of the liquid column in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{rdg}$,where $T$ is the surface tension,$\theta$ is the angle of contact,$r$ is the radius of the tube,$d$ is the density of the liquid,and $g$ is the acceleration due to gravity.
For water,the density $d$ is maximum at $4^{\circ}C$.
Since the height $h$ is inversely proportional to the density $d$ $(h \propto \frac{1}{d})$,the height $h$ will be minimum when the density $d$ is maximum.
Therefore,the height up to which water will rise in a capillary tube is minimum at $4^{\circ}C$.

(C) The height of liquid in a capillary tube is given by $h = \frac{2T \cos \theta}{rdg}$,where $T$ is surface tension,$r$ is the radius,$d$ is density,$g$ is acceleration due to gravity,and $\theta$ is the angle of contact.
Rearranging for surface tension: $T = \frac{hrdg}{2 \cos \theta}$.
For water $(1)$ and mercury $(2)$: $\frac{T_1}{T_2} = \frac{h_1}{h_2} \times \frac{d_1}{d_2} \times \frac{\cos \theta_2}{\cos \theta_1}$.
Given: $h_1 = 10 \text{ cm}$,$h_2 = -3.112 \text{ cm}$ (depression),$d_1 = 1 \text{ g/cm}^3$,$d_2 = 13.6 \text{ g/cm}^3$,$\theta_1 = 0^o$,$\theta_2 = 135^o$.
Note: For capillary rise/fall,we consider the magnitude of height: $|h_2| = 3.112 \text{ cm}$.
$\frac{T_1}{T_2} = \frac{10}{3.112} \times \frac{1}{13.6} \times \frac{\cos(135^o)}{\cos(0^o)} = \frac{10}{3.112 \times 13.6} \times \frac{1}{\sqrt{2}} \approx \frac{10}{42.32} \times 0.707 \approx \frac{1}{6}$.

(C) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{rdg}$.
Here,$T$ is surface tension,$\theta$ is the angle of contact,$r$ is the radius of the capillary,$d$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Since the capillary tube is the same,$r$ and $g$ are constant. Thus,$h \propto \frac{T \cos \theta}{d}$.
For the first liquid (water): $h_1 = 6 \text{ cm}$,$T_1 = 70 \text{ dynes/cm}$,$\theta_1 = 0^\circ$,$d_1 = 1 \text{ g/cm}^3$.
For the second liquid: $T_2 = 140 \text{ dynes/cm}$,$\theta_2 = 60^\circ$,$d_2 = 2 \text{ g/cm}^3$.
Using the ratio: $\frac{h_2}{h_1} = \frac{T_2}{T_1} \times \frac{\cos \theta_2}{\cos \theta_1} \times \frac{d_1}{d_2}$.
Substituting the values: $\frac{h_2}{6} = \frac{140}{70} \times \frac{\cos 60^\circ}{\cos 0^\circ} \times \frac{1}{2}$.
$\frac{h_2}{6} = 2 \times \frac{0.5}{1} \times 0.5 = 2 \times 0.25 = 0.5$.
Therefore,$h_2 = 6 \times 0.5 = 3 \text{ cm}$.

(D) The height of the liquid column in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
From this relation,we can see that $h \propto \frac{1}{r}$.
If the radius $r$ is reduced to half $(r' = r/2)$,the new height $h'$ will be $h' = \frac{2T \cos \theta}{(r/2) \rho g} = 2h$.
Therefore,the rise of the liquid column becomes two times,not four times.
Thus,statement $(d)$ is false.

(B) The capillary rise $h$ is given by $h = \frac{2T \cos \theta}{r \rho g}$. When the tube is cut at a height $h' = 12 \,cm$,which is less than the natural rise of $16.3 \,cm$,the water will not overflow. Instead,the radius of curvature $R$ of the meniscus will adjust itself such that $h' = \frac{2T \cos \theta'}{R \rho g} = 12 \,cm$. Thus,the water will simply rise to the top of the cut tube and stay there,forming a meniscus with a larger radius of curvature.

(C) The pressure difference across a curved liquid surface is given by the Young-Laplace equation: $\Delta P = \frac{2T}{R}$,where $T$ is the surface tension and $R$ is the radius of curvature of the meniscus.
When the capillary tube is raised by a height $h$ from its initial position,the pressure balance at the meniscus level is given by: $P_0 - \frac{2T}{R} = P_0 - \rho gh$,where $P_0$ is the atmospheric pressure,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Simplifying this,we get: $\frac{2T}{R} = \rho gh$,which implies $R = \frac{2T}{\rho gh}$.
This shows that $R$ is inversely proportional to $h$,i.e.,$R \propto \frac{1}{h}$.
As the tube is raised ($h$ increases),the radius of curvature $R$ of the meniscus increases until it reaches the radius of the tube $r$,at which point the liquid meniscus detaches or the height reaches the equilibrium capillary rise. The graph representing $R \propto \frac{1}{h}$ is a rectangular hyperbola. Among the given options,the graph that shows $R$ decreasing as $h$ increases towards a limiting value is represented by option $C$.

(B) Given the relation $T = \frac{rhg}{2} \times 10^3$.
Since $r = D/2$,the relative error in $r$ is the same as the relative error in $D$,i.e.,$\frac{\Delta r}{r} = \frac{\Delta D}{D}$.
The least count for both $D$ and $h$ is $\Delta D = \Delta h = 0.01 \times 10^{-2} \; m$.
The relative error in $T$ is given by $\frac{\Delta T}{T} = \frac{\Delta r}{r} + \frac{\Delta h}{h}$.
Substituting the values: $\frac{\Delta T}{T} = \frac{0.01 \times 10^{-2}}{1.25 \times 10^{-2}} + \frac{0.01 \times 10^{-2}}{1.45 \times 10^{-2}} = \frac{0.01}{1.25} + \frac{0.01}{1.45}$.
Percentage error $= \left( \frac{0.01}{1.25} + \frac{0.01}{1.45} \right) \times 100 = \frac{1}{1.25} + \frac{1}{1.45} = 0.8 + 0.6896 \approx 1.5 \%$.
Thus,the possible error in surface tension is $1.5 \%$.

(B) The height of the water 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.
When the entire arrangement is placed in a freely falling elevator,the effective acceleration due to gravity becomes $g_{eff} = g - a$. Since the elevator is in free fall,$a = g$,so $g_{eff} = 0$.
As $g_{eff} \to 0$,the height $h$ tends to infinity $(h \propto 1/g_{eff})$.
However,the water column cannot exceed the physical length of the capillary tube.
Therefore,the water will rise to fill the entire length of the capillary tube,which is $20 \ cm$.

(B) The behavior of a liquid in a capillary tube depends on the angle of contact $\theta$ between the liquid and the solid surface of the tube.
For liquids that wet the surface (e.g.,water),the angle of contact is acute $(\theta < 90^{\circ})$,resulting in capillary rise.
For liquids that do not wet the surface (e.g.,mercury),the angle of contact is obtuse $(\theta > 90^{\circ})$.
Since mercury has an obtuse angle of contact with glass,the cohesive forces between mercury molecules are stronger than the adhesive forces between mercury and glass.
Consequently,the mercury level in the capillary tube will be lower than the level of the mercury in the container,which is known as capillary depression.

(D) The height of capillary rise or depression is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
For water: $h_w = 10 \ cm$,$\theta_w = 0^o$,$\rho_w = 1 \ g/cc$,$T_w$ is surface tension of water.
For mercury: $h_m = -3.42 \ cm$ (depression),$\theta_m = 135^o$,$\rho_m = 13.6 \ g/cc$,$T_m$ is surface tension of mercury.
Taking the ratio: $\frac{h_w}{h_m} = \frac{T_w \cos \theta_w}{T_m \cos \theta_m} \times \frac{\rho_m}{\rho_w}$.
Substituting the values: $\frac{10}{-3.42} = \frac{T_w \cos 0^o}{T_m \cos 135^o} \times \frac{13.6}{1}$.
Since $\cos 0^o = 1$ and $\cos 135^o = -\frac{1}{\sqrt{2}} \approx -0.707$,we have:
$\frac{10}{-3.42} = \frac{T_w}{T_m \times (-0.707)} \times 13.6$.
$\frac{T_w}{T_m} = \frac{10 \times (-0.707)}{-3.42 \times 13.6} = \frac{-7.07}{-46.512} \approx 0.152$.
$0.152 \approx \frac{1}{6.58}$. Thus,the ratio is approximately $1:6.5$.

(B) The height to which water rises in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
When the tube is cut at a height $h' = 12 \, cm$ (which is less than the original rise $h = 16.3 \, cm$),the water will rise to the top of the cut tube.
It will not overflow because the radius of curvature of the meniscus will adjust to satisfy the condition $h \cdot R = \text{constant}$.
Specifically,the meniscus will become more convex (the radius of curvature $R$ increases) such that the pressure balance is maintained at the new height of $12 \, cm$.
Therefore,the water will stay at the top of the tube at a height of $12 \, cm$ without overflowing.

(B) The height $h$ to which water rises in a vertical capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
When the capillary tube is inclined at an angle $\alpha$ with the vertical,the vertical height $h$ remains constant because it depends on the surface tension and the radius of the tube.
If $l$ is the length of the water column along the inclined tube,then the vertical height is given by $h = l \cos \alpha$.
Given $h = 2.0\ cm$ and $\alpha = 60^o$,we have:
$l = \frac{h}{\cos \alpha} = \frac{2.0}{\cos 60^o} = \frac{2.0}{0.5} = 4.0\ cm$.

(B) The pressure inside the liquid column between the plates is less than the atmospheric pressure $P_0$ due to the concave meniscus.
The pressure at a depth $y$ below the meniscus is $P(y) = P_0 - \frac{2T}{d} + \rho gy$.
The average pressure difference between the outside and inside is $\Delta P_{avg} = \frac{1}{h} \int_0^h (P_0 - P(y)) dy = \frac{1}{h} \int_0^h (\frac{2T}{d} - \rho gy) dy = \frac{2T}{d} - \frac{\rho gh}{2}$.
Since the height of the water rise is $h = \frac{2T}{\rho gd}$,we substitute this into the expression for $\Delta P_{avg}$:
$\Delta P_{avg} = \frac{2T}{d} - \frac{\rho g}{2} (\frac{2T}{\rho gd}) = \frac{2T}{d} - \frac{T}{d} = \frac{T}{d}$.
The force of attraction $F$ is the product of the average pressure difference and the area of the plates submerged $(l \times h)$:
$F = \Delta P_{avg} \times l \times h = (\frac{T}{d}) \times l \times (\frac{2T}{\rho gd}) = \frac{2T^2l}{\rho gd^2}$.

(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 $R = 2.5 \, mm = 2.5 \times 10^{-3} \, m$ and $r = 1 \, mm = 1 \times 10^{-3} \, m$,the pressures at the same horizontal level $E$ in the liquid are equal.
Let $h$ be the difference in the liquid levels. The pressure at the level of the meniscus in the wider limb is $P_B = P_0 - \frac{2T}{R}$.
The pressure at the same level in the narrower limb is $P_E = P_0 - \frac{2T}{r} + h \rho g$.
Equating the pressures: $P_0 - \frac{2T}{R} = P_0 - \frac{2T}{r} + h \rho g$.
Rearranging for $h$: $h = \frac{2T}{\rho g} \left( \frac{1}{r} - \frac{1}{R} \right)$.
Substituting the given values: $h = \frac{2 \times 7 \times 10^{-2}}{10^3 \times 10} \left( \frac{1}{1 \times 10^{-3}} - \frac{1}{2.5 \times 10^{-3}} \right)$.
$h = \frac{14 \times 10^{-2}}{10^4} \times 10^3 \left( 1 - \frac{1}{2.5} \right) = 14 \times 10^{-3} \times (1 - 0.4) = 14 \times 10^{-3} \times 0.6 = 8.4 \times 10^{-3} \, m = 8.4 \, mm$.

(A) The vertical height $h$ to which a liquid rises in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
This vertical height $h$ depends only on the properties of the liquid and the radius of the capillary tube,remaining constant regardless of the tilt of the tube.
When the tube is tilted at an angle $\alpha = 60^{\circ}$ from the vertical,the length $l$ of the water column along the tube is related to the vertical height $h$ by $h = l \cos \alpha$.
Given $h = 2\, cm$ and $\alpha = 60^{\circ}$,we have $2 = l \cos 60^{\circ}$.
Since $\cos 60^{\circ} = 0.5$,we get $2 = l \times 0.5$.
Therefore,$l = \frac{2}{0.5} = 4\, cm$.

(D) The formula for the capillary rise is given by $h = \frac{2T \cos \theta}{rdg}$.
Here,$h = 5\,cm$,$r = 0.2\,cm$,$\theta = 60^o$,$d = 1\,gm/cm^3$,and $g = 980\,cm/s^2$.
Rearranging the formula for surface tension $T$,we get $T = \frac{hrdg}{2 \cos \theta}$.
Substituting the values: $T = \frac{5 \times 0.2 \times 1 \times 980}{2 \times \cos 60^o}$.
Since $\cos 60^o = 0.5$,we have $T = \frac{5 \times 0.2 \times 980}{2 \times 0.5}$.
$T = \frac{1 \times 980}{1} = 980\,dynes/cm$.

(C) The effective radius of the annular space between the glass rod and the capillary tube is given by the difference in their radii: $r_{eff} = r_2 - r_1 = 2.0\,mm - 1.0\,mm = 1.0\,mm = 1.0 \times 10^{-3}\,m$.
Using the formula for capillary rise in an annular space: $h = \frac{2T}{\rho g r_{eff}}$.
Substituting the given values: $T = 75 \times 10^{-3}\,N/m$,$\rho = 10^3\,kg/m^3$,$g = 10\,m/s^2$,and $r_{eff} = 1.0 \times 10^{-3}\,m$.
$h = \frac{2 \times 75 \times 10^{-3}}{10^3 \times 10 \times 1.0 \times 10^{-3}} = \frac{150 \times 10^{-3}}{10} = 15 \times 10^{-3}\,m$.
Converting to millimeters: $h = 15\,mm$.

(C) The height of the liquid column in a capillary tube is given by the ascent 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 tube,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
For both water and soap solution,the meniscus is concave upwards,meaning the angle of contact $\theta$ is acute $(< 90^{\circ})$,so $\cos \theta$ is positive.
The surface tension $(T)$ of a soap solution is significantly lower than that of pure water.
Since $h \propto T$,the height of the liquid column in the capillary tube containing the soap solution $(B)$ will be less than the height of the liquid column in the capillary tube containing water $(A)$.
Therefore,the water column in tube $(A)$ will rise higher than the soap solution column in tube $(B)$,and both will have a concave meniscus.

(D) When the finger is removed,the water column is supported by the surface tension forces at both the upper and lower menisci.
The total upward force due to surface tension is $F_{up} = 2 \pi r T + 2 \pi r T = 4 \pi r T$.
The downward force due to the weight of the water column is $F_{down} = Mg = V \rho g = \pi r^2 h \rho g$.
For equilibrium,$F_{up} = F_{down}$,so $4 \pi r T = \pi r^2 h \rho g$.
Solving for height $h$,we get $h = \frac{4 T}{r \rho g}$.
Given $T = 70 \,\, dyne/cm$,$r = 1 \,\, mm = 0.1 \,\, cm$,$\rho = 1 \,\, g/cm^3$,and $g = 980 \,\, cm/sec^2$:
$h = \frac{4 \times 70}{0.1 \times 1 \times 980} = \frac{280}{98} = 2.857 \approx 2.86 \,\, cm$.
Thus,only $2.86 \,\, cm$ of water will stay in the capillary and the rest will fall down.

(B) The height of the liquid rise in a capillary tube 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,
$r$ is the radius of the capillary tube,
$\rho$ is the density of the liquid,
$g$ is the acceleration due to gravity.
From the formula,it is clear that the rise $h$ depends on $T, \theta, r, \rho,$ and $g$.
It does not depend on the atmospheric pressure.

(A) The density of the mixture $(\rho_{\text{mix}})$ is calculated as: $\rho_{\text{mix}} = \frac{\text{Total Mass}}{\text{Total Volume}} = \frac{m + m}{\frac{m}{0.8} + \frac{m}{1}} = \frac{2m}{1.25m + m} = \frac{2}{2.25} = \frac{200}{225} = \frac{8}{9} \, g/cm^3$.
The formula for capillary rise is $h = \frac{2T \cos \theta}{r \rho g}$. Assuming the contact angle $\theta = 0^\circ$ $(\cos 0^\circ = 1)$:
$T = \frac{h r \rho g}{2}$.
Given: $h = 5 \, cm$, $r = 1 \, mm = 0.1 \, cm$, $\rho = \frac{8}{9} \, g/cm^3$, and $g = 980 \, cm/s^2$.
$T = \frac{5 \times 0.1 \times (8/9) \times 980}{2}$.
$T = \frac{0.5 \times 8 \times 980}{18} = \frac{4 \times 980}{18} = \frac{3920}{18} \approx 217.77 \, dyne/cm$.
Rounding to the nearest option, the value is $217.9 \, dyne/cm$.

(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 of the liquid,and $g$ is the acceleration due to gravity.
From the formula,it is clear that $h \propto \frac{1}{r}$,which implies $h_1 r_1 = h_2 r_2$.
Therefore,the ratio of the radii is given by:
$\frac{r_1}{r_2} = \frac{h_2}{h_1}$
Given $h_1 = 6.6\,cm$ and $h_2 = 2.2\,cm$:
$\frac{r_1}{r_2} = \frac{2.2}{6.6} = \frac{1}{3}$
Thus,the ratio of the radii is $1:3$.

(A) The forces acting on the capillary tube are its weight $(mg)$ acting downwards and the upward force due to surface tension $(F_s)$ acting along the circumference of the tube at the liquid-air interface.
Given:
Mass of tube $m = \pi \,g = \pi \times 10^{-3} \,kg$.
Radius $r = 2\,mm = 2 \times 10^{-3} \,m$.
Surface tension $T = 0.1\,N/m$.
Angle of contact $\theta = 0^\circ$.
Acceleration due to gravity $g = 10\,m/s^2$.
Weight of the tube $W = mg = (\pi \times 10^{-3}) \times 10 = 10\pi \times 10^{-3} \,N = 10\pi \,mN$.
The upward force due to surface tension is $F_s = T \times (2\pi r) \times \cos(\theta)$.
Since $\theta = 0^\circ$,$\cos(0^\circ) = 1$.
$F_s = 0.1 \times 2\pi \times (2 \times 10^{-3}) = 0.4\pi \times 10^{-3} \,N = 0.4\pi \,mN$.
To hold the tube in equilibrium,the external force $F$ applied upwards must satisfy:
$F + F_s = W$
$F = W - F_s$
$F = 10\pi \,mN - 0.4\pi \,mN = 9.6\pi \,mN$.
Wait,re-evaluating the force balance: The liquid rises in the tube,exerting a downward force on the tube due to surface tension. Thus,the external force $F$ must support both the weight and the downward pull of surface tension.
$F = W + F_s = 10\pi \,mN + 0.4\pi \,mN = 10.4\pi \,mN$.

(A) The height of the water 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 water,and $g$ is the acceleration due to gravity.
Since $T, \theta, r,$ and $\rho$ are constants,the height $h$ is inversely proportional to $g$,i.e.,$h \propto \frac{1}{g}$.
At the surface of the earth,the height is $x = \frac{k}{g}$,where $k = \frac{2T \cos \theta}{r \rho}$.
At a depth $d$ inside a mine,the acceleration due to gravity becomes $g' = g \left( 1 - \frac{d}{R} \right)$.
The new height is $y = \frac{k}{g'} = \frac{k}{g \left( 1 - \frac{d}{R} \right)}$.
Therefore,the ratio $\frac{x}{y} = \frac{k/g}{k / [g(1 - d/R)]} = 1 - \frac{d}{R}$.

(B) The height of the liquid rise in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{rdg}$,where $T$ is the surface tension,$\theta$ is the angle of contact,$r$ is the radius of the capillary,$d$ is the density,and $g$ is the acceleration due to gravity.
Statement-$II$ is true because the surface tension $T$ of a liquid generally decreases as the temperature increases.
According to the formula,since $h \propto T$,if the surface tension $T$ decreases with an increase in temperature,the height $h$ of the liquid rise must also decrease.
Therefore,Statement-$I$ is false because it claims that the height $h$ increases with temperature,whereas it actually decreases.

(C) The angle of contact $\theta$ is determined by the relation $\cos \theta = \frac{T_{SA} - T_{SL}}{T_{LA}}$,where $T_{SA}$,$T_{SL}$,and $T_{LA}$ are the surface tensions between solid-air,solid-liquid,and liquid-air interfaces respectively.
For a clean glass capillary,water wets the surface,resulting in an acute angle of contact $(\theta < 90^{\circ})$ and capillary rise $(h > 0)$.
When the inner wall is coated with wax,the surface becomes hydrophobic. For water on a waxy surface,the adhesive force between water and wax is weaker than the cohesive force of water. This makes the surface tension relation such that $\cos \theta$ becomes negative.
Consequently,the angle of contact $\theta$ increases to an obtuse angle $(90^{\circ} < \theta < 180^{\circ})$.
Since the capillary rise is given by $h = \frac{2T \cos \theta}{r \rho g}$,when $\theta > 90^{\circ}$,$\cos \theta$ becomes negative,meaning the liquid level in the capillary tube falls relative to the outside level. Thus,$h$ decreases.

(D) The height of liquid rise in a capillary tube is given by $h = \frac{2T \cos \theta}{\rho rg}$.
Since $T, \theta, \rho,$ and $g$ are constant,$h \propto \frac{1}{r}$.
When the radius becomes $2r$,the new height $h'$ becomes $h/2$.
The mass of the liquid in the capillary tube is $M = \pi r^2 h \rho$.
For the new tube,the mass $M'$ is given by $M' = \pi (2r)^2 h' \rho$.
Substituting $h' = h/2$,we get $M' = \pi (4r^2) (h/2) \rho = 2 \pi r^2 h \rho = 2M$.

(D) The height of capillary rise or depression is given by the formula $h = \frac{2S \cos \theta}{r \rho g}$.
For mercury $(1)$: $h = \frac{2S_1 \cos \theta_1}{r_1 \rho_1 g}$.
For water $(2)$: $h = \frac{2S_2 \cos \theta_2}{r_2 \rho_2 g}$.
Since the magnitude of $h$ is the same,we equate them: $\frac{2S_1 \cos \theta_1}{r_1 \rho_1 g} = \frac{2S_2 \cos \theta_2}{r_2 \rho_2 g}$.
Rearranging for the ratio $\frac{r_1}{r_2}$: $\frac{r_1}{r_2} = \frac{S_1}{S_2} \cdot \frac{\rho_2}{\rho_1} \cdot \frac{\cos \theta_1}{\cos \theta_2}$.
Given: $\frac{S_1}{S_2} = 7.5$,$\frac{\rho_1}{\rho_2} = 13.6 \Rightarrow \frac{\rho_2}{\rho_1} = \frac{1}{13.6}$,$\theta_1 = 135^o$,$\theta_2 = 0^o$.
$\cos 135^o = -\frac{1}{\sqrt{2}}$ and $\cos 0^o = 1$.
Taking the magnitude of the depression/rise: $\frac{r_1}{r_2} = 7.5 \times \frac{1}{13.6} \times \frac{1/\sqrt{2}}{1} \approx 7.5 \times 0.0735 \times 0.707 \approx 0.39$.
This is approximately equal to $2/5 = 0.4$.

(C) The capillary rise in a tube of radius $r$ is given by $h = \frac{2T \cos \theta}{\rho g r}$.
For water,the contact angle $\theta \approx 0^\circ$,so $\cos \theta = 1$.
The difference in levels in a $U$-tube with limbs of radii $r_1$ and $r_2$ is $\Delta h = h_1 - h_2 = \frac{2T}{\rho g} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.
Given: $T = 0.07\,N/m$,$\Delta h = 3.6\,cm = 3.6 \times 10^{-2}\,m$,$\rho = 1000\,kg/m^3$,$g = 9.8\,m/s^2$,and $d_1 = 0.4\,mm = 0.4 \times 10^{-3}\,m$ (so $r_1 = 0.2 \times 10^{-3}\,m$).
Substituting the values: $3.6 \times 10^{-2} = \frac{2 \times 0.07}{1000 \times 9.8} \left( \frac{1}{0.2 \times 10^{-3}} - \frac{1}{r_2} \right)$.
$3.6 \times 10^{-2} = \frac{0.14}{9800} \left( 5000 - \frac{1}{r_2} \right)$.
$3.6 \times 10^{-2} = 1.428 \times 10^{-5} \times (5000 - \frac{1}{r_2})$.
Solving for $r_2$,we find $r_2 = 4 \times 10^{-3}\,m$.
Since $d = 2r_2$,$d = 8 \times 10^{-3}\,m$.

(D) The upward force due to surface tension is given by $F = T \times L$,where $L$ is the circumference of the capillary $(L = 2 \pi r)$.
Given that the upward force balances the weight of the water,we have:
$T \times (2 \pi r) = \text{Weight of water}$
$T \times L = 75 \times 10^{-4} \ N$
Substituting the given value of surface tension $T = 12 \times 10^{-2} \ N/m$:
$(12 \times 10^{-2}) \times L = 75 \times 10^{-4}$
$L = \frac{75 \times 10^{-4}}{12 \times 10^{-2}}$
$L = 6.25 \times 10^{-2} \ m$
Thus,the internal circumference of the capillary is $6.25 \times 10^{-2} \ m$.

(A) The capillary rise in a tube of radius $r$ is given by the formula: $h = \frac{2T \cos \theta}{\rho rg}$.
For the first arm with radius $r_1$,the capillary rise is $h_1 = \frac{2T \cos 0^o}{\rho r_1 g} = \frac{2T}{\rho r_1 g}$.
For the second arm with radius $r_2$,the capillary rise is $h_2 = \frac{2T \cos 0^o}{\rho r_2 g} = \frac{2T}{\rho r_2 g}$.
The level difference between the two arms is $h = h_1 - h_2$.
Substituting the expressions for $h_1$ and $h_2$: $h = \frac{2T}{\rho g} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.
Simplifying the term in the bracket: $h = \frac{2T}{\rho g} \left( \frac{r_2 - r_1}{r_1 r_2} \right)$.
Solving for surface tension $T$: $T = \frac{h \rho g r_1 r_2}{2(r_2 - r_1)}$.

(B) Given: Radius of capillary $r = 2.5 \times 10^{-5} \, m$,Surface tension $S = 7.28 \times 10^{-2} \, N \, m^{-1}$,Angle of contact $\theta = 0^{\circ}$,Density $\rho = 10^3 \, kg \, m^{-3}$,Acceleration due to gravity $g = 9.8 \, m \, s^{-2}$.
The formula for the height of capillary rise is given by $h = \frac{2 S \cos \theta}{r \rho g}$.
Substituting the values:
$h = \frac{2 \times (7.28 \times 10^{-2}) \times \cos 0^{\circ}}{(2.5 \times 10^{-5}) \times 10^3 \times 9.8}$.
Since $\cos 0^{\circ} = 1$:
$h = \frac{14.56 \times 10^{-2}}{2.5 \times 10^{-2} \times 9.8} = \frac{14.56}{2.5 \times 9.8} = \frac{14.56}{24.5} \approx 0.594 \, m$.
Rounding to two decimal places,the height is $0.59 \, m$.

(C) The height $h$ to which water rises in a capillary tube of radius $r$ is given by $h = \frac{2T \cos \theta}{r \rho g}$.
The mass of water $m$ in the capillary tube is given by $m = V \rho = (\pi r^2 h) \rho$.
Substituting the expression for $h$:
$m = \pi r^2 \left( \frac{2T \cos \theta}{r \rho g} \right) \rho = \frac{2 \pi r T \cos \theta}{g}$.
From this expression,we see that $m \propto r$.
For the second capillary tube with radius $r' = 2r$,the mass $m'$ will be:
$\frac{m'}{m} = \frac{r'}{r} = \frac{2r}{r} = 2$.
Therefore,$m' = 2m = 2 \times 5 \, g = 10 \, g$.

(D) The height of the water 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 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 $g_{eff}$ becomes $0$.
As $g_{eff} \to 0$,the height $h = \frac{2T \cos \theta}{r \rho g_{eff}}$ tends to infinity.
However,the water column is physically limited by the length of the capillary tube.
Therefore,the water will rise until it fills the entire length of the capillary tube,which is $20\, cm$.

(A) 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 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.
From the formula,it is clear that $h \propto \frac{1}{g}$.
On the moon,the acceleration due to gravity $g_m = \frac{g}{6}$,where $g$ is the acceleration due to gravity on Earth.
Therefore,the new height $h'$ on the moon will be:
$h' = \frac{2T \cos \theta}{r \rho (g/6)} = 6 \times \left( \frac{2T \cos \theta}{r \rho g} \right) = 6h$.
Thus,the water will rise to a height of $6h$ on the moon.

(C) The height to which a liquid rises in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$. This height $h$ is the equilibrium height determined by the balance between surface tension forces and the weight of the liquid column. If the physical length of the tube is less than $h$,the liquid will rise to the top of the tube. At the top,the radius of curvature of the meniscus will adjust such that the new height $h'$ equals the length of the tube. The liquid will not overflow because the contact angle $\theta$ will increase to satisfy the equilibrium condition $h' = \frac{2T \cos \theta'}{r \rho g}$,where $\theta' > \theta$. Thus,the water stays at the top without overflowing.

(B) The formula for the height of liquid rise in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$. Assuming the contact angle $\theta = 0^\circ$ for water,the formula becomes $h = \frac{2T}{r \rho g}$.
Rearranging for the radius $r$,we get $r = \frac{2T}{h \rho g}$.
Given values:
$T = 73 \times 10^{-3}\, N/m$
$h = 10\, cm = 0.1\, m$
$\rho = 10^3\, kg/m^3$
$g = 9.8\, m/s^2$
Substituting these values into the formula:
$r = \frac{2 \times 73 \times 10^{-3}}{0.1 \times 10^3 \times 9.8}$
$r = \frac{146 \times 10^{-3}}{98}$
$r \approx 1.489 \times 10^{-3}\, m \approx 0.149\, cm$.
Wait,checking the calculation: $r = \frac{0.146}{98} \approx 0.00149\, m = 0.149\, cm$. Given the options,let's re-evaluate the provided solution logic: $r = \frac{2 \times 73 \times 10^{-3}}{0.1 \times 10^3 \times 9.8} = 0.00149\, m = 0.149\, cm$. If we assume $g = 9.8$ and $T = 0.073$,the result is $0.149\, cm$. If the question intended $T = 0.0735$ and $g=9.8$,$r = 0.0015\, m = 0.15\, cm$. The provided option $B$ is $0.015\, cm$. Recalculating with $h=10\, cm$ and $r=0.015\, cm$ $(0.00015\, m)$: $h = \frac{2 \times 0.073}{0.00015 \times 1000 \times 9.8} \approx 0.099\, m \approx 10\, cm$. Thus,$0.015\, cm$ is the correct choice.

(B) For water in a glass capillary tube,the contact angle is acute,which results in a concave meniscus.
To balance the weight of the water column,the net force due to surface tension must act upwards.
If the meniscus at the top is concave,the surface tension force acts upwards.
If the meniscus at the bottom is also concave,the surface tension force acts downwards.
However,for a stable column of water,the net upward force must balance the weight.
In the case of a water column in a capillary tube,the top surface is concave (pulling upwards) and the bottom surface is also concave (pulling downwards).
Wait,actually,for a water column to be supported,the net surface tension force must be upward.
If both ends are concave,the net force is zero.
However,in reality,the water column will form a concave meniscus at the top and a convex meniscus at the bottom to provide a net upward force to support the weight of the liquid column.

(D) When a capillary tube is tilted at an angle $\theta$ with the vertical,the vertical height $h$ of the liquid column remains constant because it depends only on the surface tension,radius of the tube,and density of the liquid.
If $l$ is the length of the water column along the tilted capillary,then the vertical height is given by $h = l \cos \theta$.
Given $\theta = 45^o$ and the vertical height is $h$,we have:
$h = l \cos 45^o$
$h = l \times \frac{1}{\sqrt{2}}$
$l = h\sqrt{2}$
Therefore,the length of the water column in the tilted capillary becomes $h\sqrt{2}$.

(C) The height of capillary rise is given by $h = \frac{2T \cos \theta}{r \rho g}$. For a given tube and liquid,$h \propto \frac{1}{\cos \theta}$ is not the case,rather $h \cdot r = \text{constant}$ for a fixed contact angle.
When the tube length $L$ is less than the equilibrium height $h$,the water does not overflow.
Instead,the radius of curvature $R$ of the meniscus increases to adjust to the new height $L$ such that $h \cdot r = L \cdot R$.
Since $R = \frac{r}{\cos \theta'}$,where $\theta'$ is the new contact angle,we have $h \cdot r = L \cdot \frac{r}{\cos \theta'}$.
This implies $\cos \theta' = \frac{L}{h}$.
Given $h = 5\, cm$ and $L = 3\, cm$,we have $\cos \theta' = \frac{3}{5} = 0.6$.
Since the original contact angle for water-glass is nearly $0^\circ$ $(\cos 0^\circ = 1)$,and $\cos \theta' = 0.6 < 1$,the new contact angle $\theta'$ must be greater than $0^\circ$.
Therefore,the angle of contact increases.

(C) The height $h$ to which a liquid rises in a capillary tube of radius $r$ is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
Since the material of the capillaries and the liquid are the same,$T$,$\theta$,$\rho$,and $g$ are constants.
Therefore,$h \propto \frac{1}{r}$,which implies $h_1 r_1 = h_2 r_2$.
Given $h_1 = 22 \ cm$ and $h_2 = 66 \ cm$.
The ratio of the radii is $\frac{r_1}{r_2} = \frac{h_2}{h_1}$.
Substituting the values,we get $\frac{r_1}{r_2} = \frac{66}{22} = \frac{3}{1}$.
Thus,the ratio of their radii is $3 : 1$.

(C) The height of water in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
The mass of water in the capillary tube is $m = \text{Volume} \times \text{Density} = (\pi r^2 h) \rho$.
Substituting the value of $h$ into the mass equation: $m = \pi r^2 \left( \frac{2T \cos \theta}{r \rho g} \right) \rho = \frac{2 \pi r T \cos \theta}{g}$.
Since $T$,$\theta$,and $g$ are constants,we find that $m \propto r$.
For the first capillary,$m_1 = m$ and $r_1 = r$.
For the second capillary,$r_2 = 2r$. Let the mass be $m_2$.
Therefore,$\frac{m_2}{m_1} = \frac{r_2}{r_1} = \frac{2r}{r} = 2$.
Thus,$m_2 = 2m$.

(A) The vertical height $h$ of the liquid column in a capillary tube remains constant regardless of the tilt angle $\theta$ from the vertical.
Therefore,the relationship between the length $l$ of the liquid column and the vertical height $h$ is given by $h = l \cos \theta$.
For the two cases,we have $h = l_1 \cos 45^o$ and $h = l_2 \cos 60^o$.
Equating the two expressions for $h$: $l_1 \cos 45^o = l_2 \cos 60^o$.
Substituting the values: $l_1 \times (1 / \sqrt{2}) = l_2 \times (1 / 2)$.
Rearranging the terms to find the ratio $l_1 / l_2$: $l_1 / l_2 = \sqrt{2} / 2 = 1 / \sqrt{2}$.
Thus,the ratio $l_1 : l_2$ is $1 : \sqrt{2}$.

(A) When a capillary tube is tilted at an angle $\alpha$ with the vertical,the vertical height $h$ of the water column remains constant because it depends on the surface tension,radius of the tube,and density of the liquid.
The length of the water column $\ell$ along the tube is related to the vertical height $h$ by the relation:
$\cos \alpha = \frac{h}{\ell}$
Therefore,the length of the water column is:
$\ell = \frac{h}{\cos \alpha}$
Given $h = 4 \, cm$ and $\alpha = 30^{\circ}$:
$\ell = \frac{4}{\cos 30^{\circ}}$
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$:
$\ell = \frac{4}{\frac{\sqrt{3}}{2}} = \frac{8}{\sqrt{3}} \, cm$

(C) The formula for capillary rise or fall is given by $h = \frac{2 \sigma \cos \theta}{r \rho g}$.
From this,we can write $\sigma = \frac{h r \rho g}{2 \cos \theta}$.
Since the capillary tube is the same,$r$ is constant. Thus,$\sigma \propto \frac{h \rho}{\cos \theta}$.
For water: $h_w = 10 \, cm$,$\rho_w = 1 \, g/cm^3$,$\theta_w = 0^{\circ}$.
For mercury: $h_m = -3.1 \, cm$ (fall),$\rho_m = 13.6 \, g/cm^3$,$\theta_m = 135^{\circ}$.
The ratio of surface tensions is $\frac{\sigma_w}{\sigma_m} = \frac{h_w \rho_w}{\cos \theta_w} \times \frac{\cos \theta_m}{h_m \rho_m}$.
Substituting the values: $\frac{\sigma_w}{\sigma_m} = \frac{10 \times 1}{\cos 0^{\circ}} \times \frac{\cos 135^{\circ}}{-3.1 \times 13.6}$.
Since $\cos 0^{\circ} = 1$ and $\cos 135^{\circ} = -\frac{1}{\sqrt{2}} \approx -0.707$,we get:
$\frac{\sigma_w}{\sigma_m} = \frac{10 \times (-0.707)}{-3.1 \times 13.6} = \frac{-7.07}{-42.16} \approx \frac{1}{6}$.

(D) The angle of contact between mercury and soda lime glass is $\theta = 140^{\circ}$.
Radius of the narrow tube is $r = 1.00 \; mm = 1.00 \times 10^{-3} \; m$.
Surface tension of mercury is $s = 0.465 \; N m^{-1}$.
Density of mercury is $\rho = 13.6 \times 10^{3} \; kg m^{-3}$.
Acceleration due to gravity is $g = 9.8 \; m s^{-2}$.
The capillary depression $h$ is given by the formula:
$h = \frac{2s \cos \theta}{r \rho g}$
Substituting the values:
$h = \frac{2 \times 0.465 \times \cos(140^{\circ})}{1.00 \times 10^{-3} \times 13.6 \times 10^{3} \times 9.8}$
Since $\cos(140^{\circ}) \approx -0.766$:
$h = \frac{0.93 \times (-0.766)}{133.28} \approx -0.00534 \; m$
$h = -5.34 \; mm$.
The negative sign indicates a depression in the level of mercury. Thus,the mercury dips down by $5.34 \; mm$.

(A) The capillary rise $h$ in a tube of radius $r$ is given by $h = \frac{2S \cos \theta}{r \rho g}$.
Given: $S = 7.3 \times 10^{-2} \;N m^{-1}$,$\theta = 0^\circ$,$\rho = 1.0 \times 10^3 \;kg m^{-3}$,$g = 9.8 \;m s^{-2}$.
Radii are $r_1 = 1.5 \times 10^{-3} \;m$ and $r_2 = 3.0 \times 10^{-3} \;m$.
The difference in levels is $\Delta h = h_1 - h_2 = \frac{2S \cos \theta}{\rho g} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.
Substituting the values: $\Delta h = \frac{2 \times 7.3 \times 10^{-2} \times 1}{10^3 \times 9.8} \left( \frac{1}{1.5 \times 10^{-3}} - \frac{1}{3.0 \times 10^{-3}} \right)$.
$\Delta h = \frac{14.6 \times 10^{-2}}{9.8 \times 10^3} \left( \frac{2 - 1}{3.0 \times 10^{-3}} \right) = \frac{14.6 \times 10^{-2}}{9.8 \times 10^3} \times \frac{1}{3.0 \times 10^{-3}} = \frac{14.6}{9.8 \times 3} \times 10^{-2} \;m$.
$\Delta h \approx 0.4966 \times 10^{-2} \;m = 4.966 \;mm \approx 5 \;mm$.

(N/A) The phenomenon of the rise or fall of a liquid in a capillary tube (held vertically in a liquid) is called capillarity. In this phenomenon,the surface tension of the liquid plays an important role.
The word 'capilla' means hair in Latin. If the tube is as thin as a hair,the rise would be very large for a liquid with an acute angle of contact. This type of tube is known as a capillary tube.
Consider a vertical capillary tube of circular cross-section (radius $a$) inserted into an open vessel of water. The contact angle $\theta$ between water and glass is acute $(\theta < 90^{\circ})$. Thus,the surface of the water in the capillary is concave. This implies a pressure difference between the two sides of the top surface.
$P_{i} - P_{0} = \frac{2S}{r}$,where $r$ is the radius of the meniscus ...$(1)$
From the geometry,$\cos \theta = \frac{a}{r}$,therefore $r = \frac{a}{\cos \theta}$ ...$(2)$
Substituting equation $(2)$ into $(1)$:
$P_{i} - P_{0} = \frac{2S \cos \theta}{a}$ ...$(3)$
At the same horizontal level as the free surface of the liquid in the vessel,the pressure inside the tube must be equal to the atmospheric pressure $P_{0}$.
Thus,$P_{0} = P_{i} + h \rho g$,where $h$ is the height of the liquid column,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
$P_{0} - P_{i} = h \rho g$
Comparing this with equation $(3)$,we get:
$h \rho g = \frac{2S \cos \theta}{a}$
Therefore,the height of the liquid rise is:
$h = \frac{2S \cos \theta}{a \rho g}$

(N/A) Capillarity is the phenomenon of the rise or fall of a liquid in a capillary tube (a tube with a very fine bore) when it is dipped in the liquid. This effect is caused by the surface tension of the liquid and the adhesive and cohesive forces acting at the liquid-solid interface.
Two practical illustrations of capillarity are:
$1$. The rise of oil in the wick of a lamp: The oil rises through the fine spaces between the threads of the wick due to capillary action,allowing it to reach the flame.
$2$. Absorption of ink by a blotting paper: The blotting paper contains a large number of fine pores which act as capillary tubes,drawing the ink into the paper through capillary action.