Capillary Tube and Capillarity Questions in English

(A) The behavior of a liquid in a capillary tube depends on the angle of contact between the liquid and the glass surface.
For water and glass, the angle of contact is acute $( < 90^{\circ})$, which results in an upward force due to surface tension, causing the water to rise in the capillary.
For mercury and glass, the angle of contact is obtuse $( > 90^{\circ})$, which results in a downward force due to surface tension, causing the mercury level to be depressed in the capillary.
Therefore, water rises up and mercury is depressed.

(N/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 of the liquid.
$\theta$ is the angle of contact between the liquid and the capillary wall.
$r$ is the radius of the capillary tube.
$\rho$ is the density of the liquid.
$g$ is the acceleration due to gravity.

(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 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.
$(i)$ Since $h \propto \frac{1}{r}$,as the radius $r$ of the capillary tube decreases,the height $h$ of the column increases. Therefore,the answer is 'more'.
$(ii)$ For a convex meniscus,the angle of contact $\theta > 90^\circ$,making $\cos \theta$ negative,which results in a depression of the liquid level. For a concave meniscus,the angle of contact $\theta < 90^\circ$,making $\cos \theta$ positive,which results in the liquid rising up. Therefore,the answers are 'depressed' and 'rises up' respectively.

(N/A) The lighting of a lamp is due to the phenomenon of $Capillarity$.
In a lamp,the wick consists of fine pores or capillaries.
When the lamp is filled with oil,the oil rises up through these fine pores of the wick due to the surface tension of the liquid.
This process is known as capillary action,which allows the oil to reach the flame at the top of the wick,enabling the lamp to burn continuously.

(A) The height of capillary rise $h$ 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,$\rho$ is the density,and $g$ is the acceleration due to gravity.
From this formula,we see that $h \propto \frac{1}{r}$.
Given that the initial height $h = 20 \ mm$ and the new radius $r' = \frac{r}{3}$.
Using the proportionality $h' r' = h r$,we get:
$h' = h \left( \frac{r}{r'} \right) = 20 \times \left( \frac{r}{r/3} \right) = 20 \times 3 = 60 \ mm$.

(N/A) The wick of an oil lamp is made of thin cotton fibers. These fibers act as tiny capillary tubes. Due to the phenomenon of capillarity,the oil rises through the wick against gravity. This continuous supply of oil to the top of the wick sustains the flame,allowing the lamp to produce light.

(N/A) Soil contains fine capillary tubes formed by the arrangement of soil particles. Through these capillaries,water from the deeper layers rises to the surface due to capillary action and evaporates,causing the soil to dry out. When a farm is ploughed,these capillary tubes are broken. This disruption prevents the water from rising to the surface,thereby retaining moisture in the soil.

(N/A) Cotton clothes contain thin fibers that act like capillaries. Due to the phenomenon of capillarity,these fibers absorb sweat from the body. When this absorbed sweat is exposed to the atmosphere,it undergoes evaporation. Since evaporation is an endothermic process,it absorbs latent heat from the body,providing a cooling effect and keeping the body dry.

(N/A) The ink must move against the force of gravity while writing with such a pen on a whiteboard. In a fiber refill,the ink moves due to the capillary phenomenon within the fiber-made capillaries. This allows the ink to flow continuously,enabling smooth writing on the board.

(N/A) The rise of a liquid in a capillary tube is governed by the formula $h = \frac{2S \cos \theta}{r \rho g}$,where $S$ 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.
For water in a glass capillary,the angle of contact $\theta$ is acute (i.e.,$\theta < 90^{\circ}$),which makes $\cos \theta$ positive.
Since $S, r, \rho,$ and $g$ are all positive constants,the height $h$ becomes positive,indicating that the liquid level rises inside the capillary tube to balance the pressure difference caused by the curved meniscus.

(N/A) The height of a liquid column in a capillary tube is given by the formula $h = \frac{2S \cos \theta}{r \rho g}$,where $S$ 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.
For mercury and glass,the angle of contact $\theta$ is obtuse (approximately $135^{\circ}$),which is greater than $90^{\circ}$.
Since $\cos \theta$ is negative for angles greater than $90^{\circ}$,the value of $h$ becomes negative.
$A$ negative height indicates that the level of mercury in the capillary tube is lower than the level of the mercury in the surrounding container. This phenomenon is known as capillary depression.

(C) The height of the water column in a capillary tube is given by the formula $h = \frac{2S \cos \theta}{r \rho g}$,where $S$ 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.
In a freely falling elevator,the effective acceleration due to gravity is $g_{eff} = 0$.
As $g_{eff} \to 0$,the height $h$ of the water column tends to infinity $(h \propto \frac{1}{g})$.
However,the water column is constrained by the length of the capillary tube.
Therefore,the water will rise to fill the entire length of the tube.
Thus,the length of the water column in the tube will be $20 \ cm$.

(N/A) The height $h$ to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Given values are: $T = 7.28 \times 10^{-2} \ N/m$,$r = 2.5 \times 10^{-5} \ m$,$\theta = 0^{\circ}$,$\rho = 10^3 \ kg/m^3$ (density of water),and $g = 9.8 \ m/s^2$.
Substituting these values into the formula:
$h = \frac{2 \times (7.28 \times 10^{-2}) \times \cos 0^{\circ}}{(2.5 \times 10^{-5}) \times 10^3 \times 9.8}$
$h = \frac{14.56 \times 10^{-2}}{2.5 \times 10^{-2} \times 9.8} = \frac{14.56}{24.5} \approx 0.594 \ m \approx 0.6 \ m$.
Since the height reached by capillary action is only about $0.6 \ m$,it is insufficient to supply water to the tops of tall trees (which can be $10 \ m$ to $100 \ m$ high). Therefore,surface tension alone cannot account for the supply of water to the top of all trees.

(C) Let $r$ be the radius of the capillary tube and $R$ be the radius of the meniscus.
From the geometry of the meniscus,the angle of contact $\theta$ is related to the angle between the tangents. Since the tangents make an angle of $60^{\circ}$ with each other,the angle between the tangent and the vertical wall is $30^{\circ}$. Thus,the contact angle $\theta = 30^{\circ}$.
From the geometry,$\cos \theta = \frac{r}{R}$,so $R = \frac{r}{\cos 30^{\circ}} = \frac{r}{\sqrt{3}/2} = \frac{2r}{\sqrt{3}}$.
Given $r = 0.15 \times 10^{-3} m$,we have $R = \frac{2 \times 0.15 \times 10^{-3}}{\sqrt{3}} = \frac{0.3 \times 10^{-3}}{\sqrt{3}} m$.
The height $h$ of the liquid column is given by $h = \frac{2T \cos \theta}{\rho g r}$.
Substituting the values: $h = \frac{2 \times 0.05 \times \cos 30^{\circ}}{667 \times 10 \times 0.15 \times 10^{-3}}$.
Alternatively,using $h = \frac{2T}{\rho g R}$:
$h = \frac{2 \times 0.05}{667 \times 10 \times (\frac{0.3 \times 10^{-3}}{\sqrt{3}})} = \frac{0.1 \times \sqrt{3}}{6670 \times 0.3 \times 10^{-3}} = \frac{0.1732}{2.001} \approx 0.0865\, m$.
Rounding to the nearest value,$h \approx 0.087\, m$.

(C) The formula for capillary rise is given by $h = \frac{2S \cos \theta}{\rho gr}$.
Rearranging for surface tension $S$,we get $S = \frac{\rho grh}{2 \cos \theta}$.
Given values:
Radius $r = 0.015 \; cm = 1.5 \times 10^{-4} \; m$.
Height $h = 15 \; cm = 0.15 \; m$.
Density $\rho = 900 \; kg \; m^{-3}$.
Acceleration due to gravity $g = 10 \; ms^{-2}$.
Contact angle $\theta = 0^{\circ}$,so $\cos \theta = 1$.
Substituting the values:
$S = \frac{900 \times 10 \times 0.15 \times 1.5 \times 10^{-4}}{2 \times 1}$
$S = \frac{9000 \times 0.225 \times 10^{-4}}{2}$
$S = \frac{2025 \times 10^{-4}}{2} = 1012.5 \times 10^{-4} \; N/m$.
Converting to $mN/m$ $(1 \; N/m = 1000 \; mN/m)$:
$S = 1012.5 \times 10^{-4} \times 10^3 \; mN/m = 101.25 \; mN/m$.
The closest integer is $101$.

(D) The height of water rise in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
The mass of water in the capillary tube is $m = V \rho = (\pi r^2 h) \rho$.
Substituting the expression for $h$:
$m = \pi r^2 \left( \frac{2T \cos \theta}{r \rho g} \right) \rho = \frac{2 \pi r T \cos \theta}{g}$.
Since $T$,$\theta$,and $g$ are constants,we find that $m \propto r$.
For the first tube,$m_1 = 5 \, g$ and $r_1 = r$.
For the second tube,$r_2 = 2r$.
Using the proportionality $m \propto r$:
$\frac{m_2}{m_1} = \frac{r_2}{r_1} = \frac{2r}{r} = 2$.
Therefore,$m_2 = 2 \times m_1 = 2 \times 5 \, g = 10 \, g$.

(B) The pressure at points $A$ and $B$ at the same horizontal level must be equal,so $P_A = P_B$.
The pressure just below the meniscus in the two limbs is given by $P_{atm} - \frac{2T}{r_1}$ and $P_{atm} - \frac{2T}{r_2}$ respectively.
Equating the pressures at the same horizontal level:
$P_{atm} - \frac{2T}{r_1} + \rho g(x + \Delta h) = P_{atm} - \frac{2T}{r_2} + \rho g x$
$\rho g \Delta h = 2T \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$
Given $r_1 = 2.5 \, mm = 2.5 \times 10^{-3} \, m$ and $r_2 = 4.0 \, mm = 4.0 \times 10^{-3} \, m$.
$\Delta h = \frac{2T}{\rho g} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$
$\Delta h = \frac{2 \times 7.3 \times 10^{-2}}{1.0 \times 10^3 \times 10} \left( \frac{1}{2.5 \times 10^{-3}} - \frac{1}{4.0 \times 10^{-3}} \right)$
$\Delta h = \frac{14.6 \times 10^{-2}}{10^4} \times 10^3 \left( \frac{1}{2.5} - \frac{1}{4.0} \right)$
$\Delta h = 14.6 \times 10^{-3} \times (0.4 - 0.25) = 14.6 \times 10^{-3} \times 0.15 = 2.19 \times 10^{-3} \, m = 2.19 \, mm$.

(C) The net force acting on the liquid column of height $h$ is given by the sum of the surface tension force,the weight of the liquid,and the viscous drag force.
The upward force due to surface tension is $F_s = T(2 \pi a)$.
The downward force due to gravity is $F_g = m g = (\pi a^2 h \rho) g$.
According to Poiseuille's law,the pressure drop per unit length is $\frac{\Delta P}{h} = \frac{8 \eta v}{a^2}$. The viscous force acting downwards is $F_v = \Delta P \cdot A = (\frac{8 \eta v}{a^2} h) (\pi a^2) = 8 \pi \eta h v$,where $v = \frac{dh}{dt}$.
By Newton's second law,the rate of change of momentum is equal to the net force:
$\frac{dp}{dt} = F_s - F_g - F_v$
Substituting the expressions:
$\frac{dp}{dt} = T(2 \pi a) - (\pi a^2 \rho g h) - 8 \pi \eta h \frac{dh}{dt}$
Comparing this with the given expression $\frac{dp}{dt} = -\pi a^2 \rho gh + F$,we get:
$F = T(2 \pi a) - 8 \pi \eta h \frac{dh}{dt}$
Thus,the correct option is $C$.

(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 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.
Since $T, \theta, \rho,$ and $g$ are constant for the given liquid and tube material,we have:
$h \propto \frac{1}{r}$
Since the diameter $d = 2r$,it follows that $h \propto \frac{1}{d}$.
Given $h_1 = 4 \,cm$ and $d_2 = \frac{d_1}{2}$,we can write:
$\frac{h_1}{h_2} = \frac{d_2}{d_1}$
$\frac{4}{h_2} = \frac{d_1 / 2}{d_1} = \frac{1}{2}$
$h_2 = 4 \times 2 = 8 \,cm$.
Therefore,the water will rise to a height of $8 \,cm$.

(D) The correct answer is $D$.
The phenomenon responsible for the rise of kerosene in the wick of a lamp is capillary action.
$A$ wick consists of a bundle of fibers that create numerous fine pores or narrow spaces,which act as capillaries.
Due to the surface tension of the liquid and the adhesive forces between the liquid and the wick material,the kerosene rises through these capillaries against gravity,allowing it to reach the flame.

(B) Ploughing helps in retaining moisture in the soil by breaking the capillary tubes formed in the soil.
Capillaries are fine pores or channels formed in the soil through which water can rise to the surface due to surface tension and evaporate.
By breaking these capillaries,the upward movement of water is stopped,thereby preventing evaporation and helping the soil retain its moisture.

(A) The height $h$ of the liquid column in a capillary tube is given by the formula: $h = \frac{2S \cos \theta}{r \rho g}$,where $S$ is the surface tension,$\theta$ is the contact angle,$\rho$ is the density of the liquid,$r$ is the radius of the tube,and $g$ is the acceleration due to gravity.
The mass $M$ of the liquid in the capillary tube is given by: $M = \text{Volume} \times \text{Density} = (\pi r^2 h) \rho$.
Substituting the expression for $h$ into the mass equation:
$M = \pi r^2 \left( \frac{2S \cos \theta}{r \rho g} \right) \rho$
$M = \frac{2 \pi S \cos \theta}{g} \times r$.
From this expression,we can see that the mass $M$ is directly proportional to the radius $r$ of the capillary tube $(M \propto r)$.
If the radius of the tube is doubled $(r' = 2r)$,the new mass $M'$ will be:
$M' \propto r' = 2r$
$M' = 2M$.
Therefore,the mass of the liquid that rises in the capillary tube will be $2M$.

(D) The capillary rise $h$ is given by $h = \frac{2T}{r \rho g}$,where $T$ is surface tension,$r$ is the radius,$\rho$ is density,and $g$ is acceleration due to gravity.
To keep the water level in the capillary at the same height as the vessel,we must apply an external pressure $P$ to counteract the capillary pressure.
The pressure difference across the meniscus is $\Delta P = \frac{2T}{r} = \frac{4T}{d}$,where $d$ is the diameter.
Given $T = 0.07 \,N/m$ and $d = 0.28 \times 10^{-3} \,m$,the pressure difference is $\Delta P = \frac{4 \times 0.07}{0.28 \times 10^{-3}} = \frac{0.28}{0.28 \times 10^{-3}} = 10^3 \,N/m^2$.
The total pressure $P$ required is $P = P_0 + \Delta P$,where $P_0 = 10^5 \,N/m^2$.
$P = 10^5 + 10^3 = 100 \times 10^3 + 1 \times 10^3 = 101 \times 10^3 \,N/m^2$.
Thus,the value is $101$.

(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 the liquid,and $g$ is the acceleration due to gravity.
In a freely falling elevator,the effective acceleration due to gravity $g_{eff} = g - a$. Since the elevator is in free fall,$a = g$,therefore $g_{eff} = 0$.
As $g_{eff} \to 0$,the height of the water column $h \to \infty$.
However,the water will rise until it reaches the top of the capillary tube. Since the tube is $20 \, cm$ long,the water will fill the entire length of the tube.

(A) The maximum height $h$ to which water can 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 and glass,the maximum height is $h_{max} = \frac{2T}{r \rho g}$.
Substituting the values: $h_{max} = \frac{2 \times 0.07}{0.25 \times 10^{-3} \times 1000 \times 9.8} = \frac{0.14}{2.45} = 0.0571 \, m = 5.71 \, cm$.
Since the tube projects only $2 \, cm$ above the water surface,the water will rise to the top of the tube and adjust its contact angle $\theta$ to satisfy the equilibrium condition $h' = \frac{2T \cos \theta}{r \rho g}$,where $h' = 2 \, cm = 0.02 \, m$.
Thus,$\cos \theta = \frac{h' r \rho g}{2T} = \frac{0.02 \times 0.25 \times 10^{-3} \times 1000 \times 9.8}{2 \times 0.07} = \frac{0.049}{0.14} = 0.35$.
Calculating the angle,$\theta = \cos^{-1}(0.35) \approx 69.5^{\circ} \approx 70^{\circ}$.

(A) The water column in the capillary tube forms two concave menisci at both ends.
Let $P_0$ be the atmospheric pressure.
The pressure just below the upper meniscus is $P_{upper} = P_0 - \frac{2T}{r}$.
The pressure just below the lower meniscus is $P_{lower} = P_0 - \frac{2T}{r}$.
Moving from the top meniscus to the bottom meniscus through the water column of length $L$,the pressure changes as:
$P_{lower} = P_{upper} + \rho g L$
Substituting the values:
$P_0 - \frac{2T}{r} = (P_0 - \frac{2T}{r}) + \rho g L$
Wait,this implies the column is in equilibrium under the pressure difference. The pressure at the top is $P_0 - \frac{2T}{r}$ and at the bottom is $P_0 - \frac{2T}{r}$.
Actually,for a water column of length $L$ to be in equilibrium,the pressure difference between the two ends must balance the hydrostatic pressure $\rho g L$.
Since both ends have concave menisci,the pressure at the top is $P_0 - \frac{2T}{r}$ and the pressure at the bottom is $P_0 + \frac{2T}{r}$ (if the meniscus is convex relative to the water). However,in a capillary tube,both ends form concave menisci. The pressure at the top is $P_0 - \frac{2T}{r}$ and the pressure at the bottom is $P_0 - \frac{2T}{r}$. This would mean the column cannot be in equilibrium unless the pressure difference is balanced by gravity.
Correct approach: The pressure at the top is $P_0 - \frac{2T}{r}$. The pressure at the bottom is $P_0 - \frac{2T}{r}$. The pressure difference across the column is $\rho g L$. Thus,$\rho g L = \frac{2T}{r} - (-\frac{2T}{r}) = \frac{4T}{r}$.
Therefore,$L = \frac{4T}{r \rho g}$.

(C) The capillary rise formula is given by $h = \frac{2S \cos \theta}{r \rho g}$.
Assuming the angle of contact $\theta$ and radius $r$ remain constant,we have $h \propto \frac{S}{\rho}$.
Given for liquid $A$: $h_1 = 5 \ cm$,surface tension $= S_1$,density $= \rho_1$.
Given for liquid $B$: surface tension $S_2 = 2S_1$,density $\rho_2 = 2\rho_1$.
Using the ratio: $\frac{h_1}{h_2} = \frac{S_1}{S_2} \times \frac{\rho_2}{\rho_1}$.
Substituting the values: $\frac{5}{h_2} = \frac{S_1}{2S_1} \times \frac{2\rho_1}{\rho_1} = \frac{1}{2} \times 2 = 1$.
Therefore,$h_2 = 5 \ cm = 0.05 \ m$.

(C) The height of capillary rise $h$ is given by the formula $h = \frac{2T \cos \theta}{\rho gr}$,where $T$ is the surface tension,$\theta$ is the contact angle,$\rho$ is the density,$g$ is the acceleration due to gravity,and $r$ is the radius of the capillary tube.
As the temperature of water increases,the surface tension $T$ of water decreases.
Since $h \propto T$,a decrease in surface tension leads to a decrease in the height of the capillary rise.
Therefore,the height of the capillary rise is smaller in hot water compared to cold water.
Thus,$Statement$ $I$ is true and $Statement$ $II$ is false.

(C) Statement $I$ is correct because the contact angle depends on the nature of the solid and liquid surfaces,as well as the cohesive and adhesive forces between the molecules.
Statement $II$ is incorrect because 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 contact angle,$\rho$ is the density of the liquid,$g$ is the acceleration due to gravity,and $r$ is the inner radius of the capillary tube. Since $h \propto \frac{1}{r}$,the rise of the liquid depends on the radius of the tube.
Therefore,Statement $I$ is true and Statement $II$ is false.

(B) The pressure balance at the meniscus is given by $\frac{2 \sigma}{R} = \rho g h$,where $R$ is the radius of the meniscus.
From the geometry of the meniscus,$R = \frac{r}{\cos \theta}$,where $r$ is the radius of the capillary and $\theta$ is the angle of contact.
Substituting $R$ into the height equation: $h = \frac{2 \sigma \cos \theta}{\rho g r}$.
$(A)$ For a given material,$\theta$ is constant,so $h \propto \frac{1}{r}$. Thus,$h$ decreases as $r$ increases. This is true.
$(B)$ From the formula,$h \propto \sigma$,so $h$ depends on $\sigma$. This is false.
$(C)$ If the lift accelerates upward with acceleration $a$,the effective gravity becomes $g_{\text{eff}} = g + a$. The new height is $h' = \frac{2 \sigma \cos \theta}{\rho (g+a) r}$. Since $g+a > g$,$h'$ decreases. This is true.
$(D)$ From the formula,$h \propto \cos \theta$,not $\theta$. This is false.
Therefore,statements $(A)$ and $(C)$ are true.

(D) Let $R_c$ be the radius of curvature of the meniscus. From the geometry of the capillary tube,the angle between the vertical axis and the wall of the tube is $\alpha/2$. The contact angle $\theta$ is measured between the tangent to the liquid surface and the wall of the tube.
From the geometry shown in the figure,we have $\cos(\theta + \alpha/2) = \frac{b}{R_c}$,which implies $R_c = \frac{b}{\cos(\theta + \alpha/2)}$.
The pressure difference across the curved meniscus is given by $\Delta P = \frac{2S}{R_c}$.
Equating the pressure at the level of the meniscus inside the tube to the atmospheric pressure $P_0$,we have $P_0 - \frac{2S}{R_c} + h\rho g = P_0$,which simplifies to $h\rho g = \frac{2S}{R_c}$.
Substituting the expression for $R_c$,we get $h\rho g = \frac{2S \cos(\theta + \alpha/2)}{b}$.
Therefore,$h = \frac{2S}{b\rho g} \cos(\theta + \alpha/2)$.

(A) The vertical height $h$ of the water column in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{\rho gr}$.
Given values are: $T = 70 \ dyn/cm$,$\theta = 0^{\circ}$,$\rho = 1 \ g/cm^3$,$g = 980 \ cm/s^2$,and $r = 0.1 \ mm = 0.01 \ cm$.
Substituting these values into the formula:
$h = \frac{2 \times 70 \times \cos 0^{\circ}}{1 \times 980 \times 0.01} = \frac{140}{9.8} = \frac{1400}{98} = \frac{100}{7} \ cm$.
The tube is inclined at $30^{\circ}$ with the vertical,which means the angle $\alpha$ with the horizontal is $90^{\circ} - 30^{\circ} = 60^{\circ}$.
The length $\ell$ of the water column along the tube is related to the vertical height $h$ by the relation $\ell = \frac{h}{\sin \alpha}$.
Substituting $\alpha = 60^{\circ}$:
$\ell = \frac{100/7}{\sin 60^{\circ}} = \frac{100/7}{\sqrt{3}/2} = \frac{200}{7\sqrt{3}} \approx 16.49 \ cm$.
Thus,the correct option is $A$.

(C) The height of the liquid column in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
When the tube is long enough,the liquid rises to height $h = 2 \ cm$ with an angle of contact $\theta = 0^{\circ}$.
When the tube is cut or lowered such that its length above the liquid surface is $h' = 1 \ cm$,the liquid will rise to the top of the tube and adjust its angle of contact to $\theta'$ to maintain equilibrium.
Since $T$,$r$,$\rho$,and $g$ are constant,we have $h \cos \theta = h' \cos \theta'$.
Substituting the values: $2 \times \cos(0^{\circ}) = 1 \times \cos(\theta')$.
Since $\cos(0^{\circ}) = 1$,we get $2 \times 1 = \cos(\theta')$,which implies $\cos(\theta') = 0.5$.
Therefore,$\theta' = \cos^{-1}(0.5) = 60^{\circ}$.

(D) The formula for the height of a liquid column in a capillary tube is $h = \frac{2T \cos \theta}{r \rho g}$.
Rearranging for surface tension $T$,we get $T = \frac{h r \rho g}{2 \cos \theta}$.
Given values: $h = 10 \ cm = 0.1 \ m$,$r = 2 \ mm = 2 \times 10^{-3} \ m$,$\rho = 10^3 \ kg/m^3$,$\theta = 0^{\circ}$,and $g = 10 \ m/s^2$ (assuming standard gravity).
Substituting these values:
$T = \frac{0.1 \times (2 \times 10^{-3}) \times 10^3 \times 10}{2 \times \cos(0^{\circ})}$
$T = \frac{0.1 \times 2 \times 10}{2 \times 1} = \frac{2}{2} = 1 \ N/m$.

(B) The height of the water column in a capillary tube is given by $h = \frac{2S \cos \theta}{\rho rg}$.
Initially,the water rises to $h_1 = 8 \ cm$ with an angle of contact $\theta_1 = 0^{\circ}$ (for water and glass).
When the tube is cut at $h_2 = 6 \ cm$,the water will fill the entire tube up to the top because $6 \ cm < 8 \ cm$.
Thus,the new height of the water column is $6 \ cm$.
Since $h \propto \cos \theta$,we have $\frac{h_1}{\cos \theta_1} = \frac{h_2}{\cos \theta_2}$.
Substituting the values: $\frac{8}{\cos 0^{\circ}} = \frac{6}{\cos \theta_2}$.
Since $\cos 0^{\circ} = 1$,we get $\cos \theta_2 = \frac{6}{8} = \frac{3}{4}$.
Therefore,the new angle of contact is $\theta_2 = \cos^{-1} \left( \frac{3}{4} \right)$.

(B) The upward force due to surface tension acting on the liquid column is given by $F = (2 \pi R) T \cos \theta_C$.
In equilibrium,the upward force due to surface tension balances the weight of the liquid column: $(2 \pi R) T \cos \theta_C = W$.
Assuming the contact angle $\theta_C = 0^{\circ}$ for a liquid that rises in a capillary,we have $\cos 0^{\circ} = 1$.
Thus,$T = \frac{W}{2 \pi R}$.
Substituting the given values: $T = \frac{6.2 \times 10^{-4}}{2 \times 3.14 \times 2 \times 10^{-3}}$.
$T = \frac{6.2 \times 10^{-4}}{12.56 \times 10^{-3}} \approx 0.04936 \,N/m \approx 5 \times 10^{-2} \,N/m$.

(C) The capillary rise or depression is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Given values: Surface tension $T = 0.465 \ N/m$,radius $r = 1.00 \ mm = 10^{-3} \ m$,density $\rho = 13.6 \times 10^3 \ kg/m^3$,angle of contact $\theta = 135^{\circ}$,and acceleration due to gravity $g = 9.8 \ m/s^2$.
Substituting the values: $h = \frac{2 \times 0.465 \times \cos(135^{\circ})}{10^{-3} \times 13.6 \times 10^3 \times 9.8}$.
Since $\cos(135^{\circ}) = -\frac{1}{\sqrt{2}} \approx -0.707$,the negative sign indicates a depression.
$h = \frac{2 \times 0.465 \times (-0.707)}{13.6 \times 9.8} \times 10^3 \approx -4.93 \times 10^{-3} \ m$.
$h \approx -5 \times 10^{-3} \ m = -5 \ mm$.
The magnitude of the depression is approximately $5 \ mm$.

(D) The formula for capillary rise is given by $h = \frac{2T \cos \theta}{r \rho g_{eff}}$,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 $g_{eff}$ is the effective acceleration due to gravity.
From this relation,we see that $h \propto \frac{1}{g_{eff}}$.
$A$: On Jupiter,$g$ is much larger than on Earth,so $h$ will be less than $h_{earth}$. This is true.
$B$: In a lift moving up with constant acceleration $a$,$g_{eff} = g + a > g$,so $h$ will be less than $h_{earth}$. This is true.
$C$: On the Moon,$g$ is less than on Earth,so $h$ will be more than $h_{earth}$. This is true.
$D$: In a lift moving down with constant acceleration $a$,$g_{eff} = g - a < g$,so $h$ will be greater than $h_{earth}$. The statement that it is less than $h$ is false.

(C) The capillary rise $h$ is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Since the cross-sectional area $A = \pi r^2$, we have $r = \sqrt{\frac{A}{\pi}}$, which implies $r \propto \sqrt{A}$.
Substituting this into the formula, we get $h \propto \frac{1}{\sqrt{A}}$.
Given the initial height $h_1 = 20 \,mm$ for area $A_1 = A$.
For the new area $A_2 = \frac{A}{4}$, the new height $h_2$ is calculated as:
$\frac{h_2}{h_1} = \sqrt{\frac{A_1}{A_2}} = \sqrt{\frac{A}{A/4}} = \sqrt{4} = 2$.
Therefore, $h_2 = 2 \times h_1 = 2 \times 20 \,mm = 40 \,mm$.
Converting to centimeters, $h_2 = 4 \,cm$.

(A) The upward force due to surface tension $(F)$ is given by the formula $F = T \times L$,where $T$ is the surface tension and $L$ is the inner circumference of the capillary tube.
Given: $F = 98 \text{ dyne} = 98 \times 10^{-5} \text{ N}$ (since $1 \text{ dyne} = 10^{-5} \text{ N}$).
Surface tension $T = 7 \times 10^{-2} \text{ Nm}^{-1}$.
We need to find the circumference $L$.
Using the formula $F = T \times L$,we get $L = F / T$.
$L = (98 \times 10^{-5} \text{ N}) / (7 \times 10^{-2} \text{ Nm}^{-1})$.
$L = 14 \times 10^{-3} \text{ m}$.
$L = 0.014 \text{ m} = 1.4 \text{ cm}$.

(C) The height of a 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 capillary,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Since $T, \theta, \rho,$ and $g$ are constant for the same liquid and glass,we have $h \propto \frac{1}{r}$ or $h \propto \frac{1}{d}$,where $d$ is the diameter.
Let $h_P = 2.4 \ cm$ and $d_P$ be the diameter of capillary $P$.
For capillary $Q$,the diameter $d_Q = 0.80 \times d_P$.
Using the relation $h_P d_P = h_Q d_Q$,we get:
$h_Q = h_P \times \frac{d_P}{d_Q} = 2.4 \times \frac{d_P}{0.80 \times d_P} = \frac{2.4}{0.80} = 3.0 \ cm$.
Therefore,the rise of liquid in capillary $Q$ is $3.0 \ cm$.

(B) The height of water 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 constant,we have $h \propto \frac{1}{g}$.
At the surface of the earth,the acceleration due to gravity is $g_0$. Thus,$x = \frac{k}{g_0}$.
At a depth $d$ inside a mine,the acceleration due to gravity is given by $g_d = g_0(1 - \frac{d}{R}) = g_0(\frac{R-d}{R})$.
Thus,the new height is $Y = \frac{k}{g_d} = \frac{k}{g_0(\frac{R-d}{R})} = \frac{k}{g_0} \cdot \frac{R}{R-d}$.
Substituting $x = \frac{k}{g_0}$,we get $Y = x \cdot \frac{R}{R-d}$.
Therefore,the ratio $Y:x = \frac{R}{R-d}$,which is $R:(R-d)$.

(B) The height of a 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 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.
Since the diameter (and thus radius $r$) is the same for both tubes,and the angle of contact $\theta$ is the same,the height $h$ is proportional to $\frac{T}{\rho}$.
Therefore,the ratio of heights is $\frac{h_1}{h_2} = \left(\frac{T_1}{T_2}\right) \times \left(\frac{\rho_2}{\rho_1}\right)$.
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,we get $\frac{h_1}{h_2} = \left(\frac{6}{5}\right) \times \left(\frac{3}{4}\right) = \frac{18}{20} = \frac{9}{10}$.

(B) 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,we see that $h \propto \frac{1}{g}$.
Let $g_e$ be the acceleration due to gravity on Earth and $g_m$ be the acceleration due to gravity on the Moon.
Given that $g_e = 6g_m$,or $g_m = \frac{g_e}{6}$.
Let $h_e$ be the height on Earth and $h_m$ be the height on the Moon.
Then,$\frac{h_m}{h_e} = \frac{g_e}{g_m} = \frac{g_e}{g_e / 6} = 6$.
Therefore,$h_m = 6h_e$.
The rise of the liquid column on the Moon's surface is six times that on the Earth's surface.

(A) The height $h$ to which a liquid rises in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$,where $T$ is surface tension,$\rho$ is density,and $r$ is the radius of the tube.
Since $h \propto \frac{1}{r}$,if the radius becomes $\frac{r}{5}$,the new height $h' = 5h$.
The mass of water in the capillary is given by $m = \text{Volume} \times \text{Density} = (\pi r^2 h) \rho$.
For the new capillary,the new mass $m'$ is $m' = \pi (r')^2 h' \rho$.
Substituting $r' = \frac{r}{5}$ and $h' = 5h$:
$m' = \pi (\frac{r}{5})^2 (5h) \rho = \pi (\frac{r^2}{25}) (5h) \rho = \frac{1}{5} (\pi r^2 h \rho) = \frac{m}{5}$.

(C) 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 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.
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 $r \propto \sqrt{a}$.
Therefore,$h \propto \frac{1}{\sqrt{a}}$.
Let $h_1 = h$ for area $a_1 = a$,and $h_2$ be the height for area $a_2 = 4a$.
Then,$\frac{h_2}{h_1} = \sqrt{\frac{a_1}{a_2}} = \sqrt{\frac{a}{4a}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,$h_2 = \frac{h}{2}$.

(A) The upward force due to surface tension $(F)$ is given by the formula: $F = T \cdot L$,where $T$ is the surface tension and $L$ is the inner circumference of the capillary tube.
Given: $F = 105 \text{ dyne} = 105 \times 10^{-5} \text{ N} = 1.05 \times 10^{-3} \text{ N}$.
Surface tension $T = 7 \times 10^{-2} \text{ N/m}$.
We need to find the circumference $L$.
Using the formula $L = F / T$:
$L = (1.05 \times 10^{-3} \text{ N}) / (7 \times 10^{-2} \text{ N/m})$
$L = 0.15 \times 10^{-1} \text{ m} = 0.015 \text{ m}$.
Converting to centimeters: $L = 0.015 \times 100 \text{ cm} = 1.5 \text{ cm}$.
Thus,the inner circumference of the capillary tube is $1.5 \text{ cm}$.

(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 angle of contact,$r$ is the radius of the capillary,$\rho$ is the density of water,and $g$ is the acceleration due to gravity.
Since $T, \theta, r,$ and $\rho$ remain constant,we have $h \propto \frac{1}{g}$.
Therefore,$\frac{h_2}{h_1} = \frac{g_1}{g_2}$,where $g_1$ is the acceleration due to gravity on the Earth's surface and $g_2$ is the acceleration due to gravity at depth $d$.
The acceleration due to gravity at depth $d$ is given by $g_2 = g_1 \left(1 - \frac{d}{R}\right) = g_1 \left(\frac{R-d}{R}\right)$.
Substituting this into the ratio,we get $\frac{h_2}{h_1} = \frac{g_1}{g_1 \left(\frac{R-d}{R}\right)} = \frac{R}{R-d}$.
Thus,the correct option is $D$.

(D) The upward force due to surface tension $(F)$ acting on the circumference of the capillary is given by the formula: $F = T \times L$,where $T$ is the surface tension and $L$ is the inner circumference of the capillary.
Given: $F = 108 \ dyne = 108 \times 10^{-5} \ N$ (since $1 \ dyne = 10^{-5} \ N$).
Given: $T = 7.2 \times 10^{-2} \ N/m$.
We need to find the circumference $L$.
Using the formula $F = T \times L$,we get:
$L = F / T$
$L = (108 \times 10^{-5}) / (7.2 \times 10^{-2})$
$L = (108 / 7.2) \times 10^{-3}$
$L = 15 \times 10^{-3} \ m$
$L = 1.5 \times 10^{-2} \ m = 1.5 \ cm$.
Therefore,the inner circumference of the capillary is $1.5 \ cm$.