Capillary Tube and Capillarity Questions in English

(A) When two glass plates are placed very close to each other in water,the water rises between them due to capillary action. The pressure of the water between the plates becomes less than the atmospheric pressure outside. This pressure difference creates a net force that pulls the two plates together. Therefore,there is a force of attraction between the two glass plates.

(D) For a concave meniscus (as formed by water in a glass capillary),the pressure on the concave side is greater than the pressure on the convex side.
Since the water surface is concave,the pressure just below the surface (inside the liquid) is less than the pressure just above the surface (atmospheric pressure).
Therefore,the pressure below the curved surface is lesser than the pressure on the upper side.

(D) The height of the liquid rise in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{rdg}$.
Here,$T$ is the surface tension,$\theta$ is the angle of contact,$r$ is the radius of the capillary tube,$d$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Since $T$,$\theta$,$d$,and $g$ are constants for a given liquid and tube material,we have $h \propto \frac{1}{r}$.
This implies that the height of the liquid rise is inversely proportional to the radius (or diameter) of the capillary tube.
Therefore,the liquid will rise more in the tube with a smaller diameter.

(D) The height of the liquid column in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
In the state of weightlessness or in a gravity-free space,the effective acceleration due to gravity $g$ becomes $0$.
As $g \to 0$,the height $h$ tends to infinity $(h \to \infty)$.
This means that the water will continue to rise until it reaches the upper end of the capillary tube,regardless of the length of the tube.

(B) Let the width of each plate be $b$. Due to surface tension, the liquid will rise to a height $h$.
The upward force due to surface tension acts along the two contact lines of length $b$ each:
$F_{\text{up}} = 2 \times (T \cos \theta \times b) = 2Tb \cos \theta$ ... $(i)$
The weight of the liquid column raised between the plates is given by:
$W = \text{Volume} \times \text{density} \times g = (b \times x \times h) \times d \times g$ ... (ii)
At equilibrium, the upward force balances the weight of the liquid column:
$2Tb \cos \theta = bxhdg$
Dividing both sides by $b$, we get:
$2T \cos \theta = xhdg$
Therefore, the height of the liquid rise is:
$h = \frac{2T \cos \theta}{xdg}$

(D) The upward force due to surface tension $(F_s)$ is given by the product of surface tension $(T)$ and the inner circumference $(C)$ of the capillary tube.
$F_s = T \times C$
Given that the upward force is balanced by the weight of the liquid $(W = 75 \times 10^{-4} \, N)$,
$T \times C = 75 \times 10^{-4} \, N$
Given $T = 6 \times 10^{-2} \, N m^{-1}$,we have:
$6 \times 10^{-2} \times C = 75 \times 10^{-4}$
$C = \frac{75 \times 10^{-4}}{6 \times 10^{-2}}$
$C = 12.5 \times 10^{-2} \, m$
Thus,the inner circumference is $12.5 \times 10^{-2} \, m$.

(D) The correct answer is $D$. Blotting paper and newspaper have very fine pores that act as capillary tubes. When an ink pen touches the surface,the ink is immediately absorbed into the paper due to the phenomenon of capillarity,causing the ink to spread and blur,making it impossible to write clearly.

(B) 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 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.
Since $T, \theta, \rho$,and $g$ are constant for the same liquid and same material of the tube,we have $h \propto \frac{1}{r}$.
Given that the height in capillary $P$ is $h_P = \frac{2}{3} h_Q$,we can write the ratio of the radii as $\frac{r_P}{r_Q} = \frac{h_Q}{h_P}$.
Substituting the given values,we get $\frac{r_P}{r_Q} = \frac{h_Q}{(2/3)h_Q} = \frac{3}{2}$.
Since the diameter $D = 2r$,the ratio of the diameters is $\frac{D_P}{D_Q} = \frac{2r_P}{2r_Q} = \frac{r_P}{r_Q} = \frac{3}{2}$.

(C) The height of liquid rise in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
Since the liquid,material,and contact angle are the same,we have $h \propto \frac{1}{r}$,which implies $r \propto \frac{1}{h}$.
Given $h_1 = 2.2 \ cm$ and $h_2 = 6.6 \ cm$.
The ratio of the radii is $\frac{r_1}{r_2} = \frac{h_2}{h_1}$.
Substituting the values,$\frac{r_1}{r_2} = \frac{6.6}{2.2} = \frac{3}{1}$.
Therefore,the ratio of their radii is $3:1$.

(C) The height of liquid rise in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
Since the material of the capillary and the liquid are the same,the surface tension $T$,contact angle $\theta$,density $\rho$,and acceleration due to gravity $g$ are constant.
Therefore,$h \propto \frac{1}{r}$,which implies $h_1 r_1 = h_2 r_2$.
Given $r_1 = 1 \, mm$,$r_2 = 2 \, mm$,and $h_1 = 30 \, cm$.
Substituting the values: $30 \times 1 = h_2 \times 2$.
$h_2 = \frac{30}{2} = 15 \, cm$.

(B) When a capillary tube of length $l < h$ is dipped in water,the water rises to the top of the tube.
At the top,the radius of curvature of the meniscus adjusts itself such that the pressure balance is maintained.
The water does not overflow because the radius of curvature $R$ of the meniscus increases until the pressure difference $2T/R$ equals the hydrostatic pressure $\rho gh$ corresponding to the length of the tube.
Thus,the water stays at the top of the tube and forms a meniscus with a larger radius of curvature.

(B) The height to which water rises in a capillary tube is given by $h = \frac{2T \cos \theta}{\rho r g}$. Given that the capillary rise is $10 \, cm$,the water column will naturally try to reach this height. If the tube is cut or immersed such that the length above the water surface is less than the equilibrium height $(8 \, cm < 10 \, cm)$,the water will rise to the top of the tube. To maintain equilibrium,the radius of curvature of the meniscus will adjust such that $h' = \frac{2T \cos \theta'}{\rho r' g} = 8 \, cm$. Thus,the water will rise to the upper end and form a spherical surface with a different radius of curvature to satisfy the pressure balance.

(C) The formula for capillary rise is given by $h = \frac{2T \cos \theta}{r \rho g}$.
Assuming the contact angle $\theta = 0^\circ$ for water in a glass capillary,$\cos 0^\circ = 1$.
Given: $h = 3\, cm = 3 \times 10^{-2}\, m$,$T = 75 \times 10^{-3}\, N/m$,$\rho = 10^3\, kg/m^3$,and $g = 10\, m/s^2$.
Substituting the values: $3 \times 10^{-2} = \frac{2 \times 75 \times 10^{-3}}{r \times 10^3 \times 10}$.
$3 \times 10^{-2} = \frac{150 \times 10^{-3}}{r \times 10^4} = \frac{15 \times 10^{-3}}{r \times 10^3} = \frac{15 \times 10^{-6}}{r}$.
$r = \frac{15 \times 10^{-6}}{3 \times 10^{-2}} = 5 \times 10^{-4}\, m = 0.5\, mm$.
The diameter $D = 2r = 2 \times 0.5\, mm = 1.0\, mm$.

(C) The angle of contact $(\theta)$ between mercury and glass is obtuse $(\theta > 90^{\circ})$.
Due to this, the cohesive forces between mercury molecules are stronger than the adhesive forces between mercury and glass.
As a result, the meniscus of mercury in the capillary tube is convex upwards.
This leads to a depression of the mercury level inside the capillary tube relative to the level outside.

(D) The height to which water rises in a capillary tube is determined by the surface tension and the radius of the tube,given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
When the capillary tube is inserted into water to a depth $l$,the water rises to a height $h$ above the free surface of the water.
If the lower end is closed while inside the water and the tube is removed,the column of water trapped inside the tube remains at height $h$ because the capillary action is a property of the tube-liquid interface.
When the closed end is opened,the water level will adjust to the equilibrium height $h$ determined by the capillary forces,provided the tube length is sufficient.
Therefore,the water will remain at height $h$.

(B) The height $h$ to which a liquid rises in a capillary tube of radius $r$ is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
From this relation,we can see that the height of the liquid column is inversely proportional to the radius of the tube,i.e.,$h \propto \frac{1}{r}$.
Since the diameter $d = 2r$,the radius is also directly proportional to the diameter $(r \propto d)$.
Therefore,$h \propto \frac{1}{d}$.
If the diameter is doubled $(d' = 2d)$,the new height $h'$ becomes $h' = \frac{h}{2}$.
Thus,the height of the liquid that will rise is half of the original height.

(B) The formula for capillary rise is given by $h = \frac{2T \cos \theta}{r \rho g}$.
Given:
Surface tension $T = 0.06 \, N/m$.
Diameter $d = 1 \, mm = 10^{-3} \, m$,so radius $r = 0.5 \times 10^{-3} \, m$.
Angle of contact $\theta = 0^\circ$,so $\cos 0^\circ = 1$.
Density $\rho = 1000 \, kg/m^3$.
Acceleration due to gravity $g = 10 \, m/s^2$.
Substituting the values:
$h = \frac{2 \times 0.06 \times 1}{0.5 \times 10^{-3} \times 1000 \times 10} = \frac{0.12}{5} = 0.024 \, m$.
Converting to centimeters: $0.024 \, m = 2.4 \, cm$.
Rounding to the nearest provided option,the correct answer is $2.44 \, cm$.

(B) The height $h$ to which a liquid rises in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$,where $T$ is surface tension,$\theta$ is the angle of contact,$r$ is the radius of the tube,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Since the same liquid is used in both tubes,$T$,$\theta$,$\rho$,and $g$ are constant.
Therefore,$h \propto \frac{1}{r}$,which implies $h_1 r_1 = h_2 r_2$.
The ratio of the heights is $\frac{h_1}{h_2} = \frac{r_2}{r_1}$.
Given $r_1 = 0.2 \ cm$ and $r_2 = 0.4 \ cm$,we have $\frac{h_1}{h_2} = \frac{0.4}{0.2} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(B) The vertical height $(h)$ of the liquid column in a capillary tube depends on the surface tension,density of the liquid,and the radius of the tube. It is independent of the inclination of the tube.
When the tube is tilted at an angle $\theta$ with the vertical,the vertical height $h$ remains the same.
If $l$ is the length of the liquid column along the tube,then from the geometry of the setup,we have:
$h = l \cos \theta$
Given:
$h = 3 \, cm$
$\theta = {60^\circ}$
Substituting the values:
$3 = l \cos {60^\circ}$
$3 = l \times \frac{1}{2}$
$l = 3 \times 2 = 6 \, cm$
Therefore,the length of the liquid column along the tube is $6 \, cm$.

(C) The tip of the nib of a writing pen is split to provide the capillary action.
Due to capillarity,the ink rises in the split of the nib,which allows the pen to write continuously by maintaining a steady flow of ink to the paper.

(A) The capillary rise $h$ in a capillary tube of radius $r$ is given by the formula:
$h = \frac{2T \cos \theta}{rdg}$
where $T$ is the surface tension,$\theta$ is the angle of contact,$d$ is the density of the liquid,and $g$ is the acceleration due to gravity.
To find the radius $r$,we rearrange the formula:
$r = \frac{2T \cos \theta}{hdg}$
Therefore,the correct relation is $r = \frac{2T \cos \theta}{hdg}$.

(D) The height $h$ to which a liquid rises in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{rdg}$,where $T$ is 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.
From this relation,we can see 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 $g_{eff} < g$,the value of $h$ will increase.
Therefore,the correct option is $D$.

(C) When a liquid rises in a capillary tube,the meniscus is formed at the interface of the liquid and the air.
As the liquid rises,the contact angle between the liquid and the glass remains constant for a given pair of materials.
However,the surface of contact between the air and the liquid (the meniscus) maintains its shape and curvature to satisfy the pressure difference across the interface (Young-Laplace equation).
In the context of capillary action,the interface that defines the geometry of the rise is the air-liquid interface.

(B) The height of liquid rise 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 for the same liquid,we have $h \propto \frac{1}{r}$.
This implies $h_1 r_1 = h_2 r_2$.
Given $h_1 = 1.2 \ mm$ and $r_2 = \frac{r_1}{2}$.
Substituting these values,we get $1.2 \times r_1 = h_2 \times \frac{r_1}{2}$.
Solving for $h_2$,we get $h_2 = 1.2 \times 2 = 2.4 \ mm$.

(A) The rise of a liquid in a capillary tube is governed 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 tube,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Performing the experiment in a vacuum does not affect the surface tension or the contact angle of the liquid.
Since the atmospheric pressure acts equally on the liquid surface in the container and the liquid inside the capillary,its absence in a vacuum does not change the pressure difference across the meniscus.
Therefore,the liquid will still rise in the capillary tube as it would under normal atmospheric conditions.

(A) The height of the liquid column $h$ in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
From this relation,we can see that $h \propto \frac{1}{r}$,which implies $rh = \text{constant}$.
If the liquid level $h$ falls (decreases),then for the product $rh$ to remain constant,the radius $r$ of the capillary must increase.
Therefore,the correct option is $A$.

(D) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{rdg}$.
Since the diameter $(r)$,contact angle $(\theta)$,and acceleration due to gravity $(g)$ are constant for both tubes,the height is proportional to the surface tension $(T)$ and inversely proportional to the density $(d)$: $h \propto \frac{T}{d}$.
Therefore,the ratio of the heights is given by: $\frac{h_1}{h_2} = \frac{T_1}{T_2} \times \frac{d_2}{d_1}$.
Given values are $T_1 = 60 \text{ dyne/cm}$,$T_2 = 50 \text{ dyne/cm}$,$d_1 = 0.8$,and $d_2 = 0.6$.
Substituting these values: $\frac{h_1}{h_2} = \frac{60}{50} \times \frac{0.6}{0.8} = \frac{6}{5} \times \frac{6}{8} = \frac{36}{40} = \frac{9}{10}$.

(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 tube is inclined at an angle $\alpha$ with the vertical,the vertical height $h$ remains the same because the pressure difference at the meniscus depends only on the vertical height.
If $l$ is the length of the water column along the inclined tube,then $h = l \cos \alpha$.
Given $h = 2.0 \, cm$ and $\alpha = 60^{\circ}$.
Therefore,$l = \frac{h}{\cos \alpha} = \frac{2.0}{\cos 60^{\circ}} = \frac{2.0}{0.5} = 4.0 \, cm$.

(D) The formula for the rise of a liquid in a capillary tube is given by $T = \frac{rh\rho g}{2 \cos \theta}$,where $T$ is the surface tension,$r$ is the radius of the capillary tube,$h$ is the height of the liquid column,$\rho$ is the density of the liquid,$g$ is the acceleration due to gravity,and $\theta$ is the angle of contact.
For pure water in a glass capillary tube,the angle of contact $\theta$ is approximately $0^\circ$.
Since $\cos(0^\circ) = 1$,the formula simplifies to $T = \frac{rh\rho g}{2}$.

(C) The height of the water column in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
In a freely falling elevator,the effective acceleration due to gravity $g_{eff}$ becomes $0$ because the elevator is in a state of weightlessness.
As $g_{eff} \to 0$,the height of the water column $h$ 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 until it fills the entire length of the capillary tube,which is $20 \,cm$.

(B) The weight of the liquid column supported in the capillary is balanced by the upward force due to surface tension along the circumference of the tube.
The force due to surface tension is given by $F = T \times (2\pi r)$,where $T$ is the surface tension and $r$ is the radius of the capillary.
Given: Weight $W = F = 6.28 \times 10^{-4} \ N$,radius $r = 2 \times 10^{-3} \ m$,and $\pi \approx 3.14$.
Rearranging the formula for surface tension: $T = \frac{F}{2\pi r}$.
Substituting the values: $T = \frac{6.28 \times 10^{-4}}{2 \times 3.14 \times 2 \times 10^{-3}}$.
$T = \frac{6.28 \times 10^{-4}}{12.56 \times 10^{-3}} = \frac{6.28}{12.56} \times 10^{-1} = 0.5 \times 10^{-1} = 5 \times 10^{-2} \ N/m$.
Thus,the correct option is $B$.

(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 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 for the same liquid and the same material of the tube,we have $h \propto \frac{1}{r}$.
Given that $R_B > R_A$,it follows that $\frac{1}{R_A} > \frac{1}{R_B}$.
Therefore,the height of the liquid rise $h_A$ in tube $A$ will be greater than the height $h_B$ in tube $B$ $(h_A > h_B)$.

(A) The height of liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
For water,the contact angle $\theta = 0^\circ$,so $\cos 0^\circ = 1$.
Given: $h = 3 \, cm = 3 \times 10^{-2} \, m$,$T = 7.2 \times 10^{-2} \, N/m$,$\rho = 10^3 \, kg/m^3$,and $g = 10 \, m/s^2$.
Rearranging the formula for radius $r$: $r = \frac{2T}{h \rho g}$.
Substituting the values: $r = \frac{2 \times 7.2 \times 10^{-2}}{3 \times 10^{-2} \times 10^3 \times 10} = \frac{14.4 \times 10^{-2}}{3 \times 10^2} = 4.8 \times 10^{-4} \, m$.
The diameter $d$ is $2r$: $d = 2 \times 4.8 \times 10^{-4} \, m = 9.6 \times 10^{-4} \, m$.

(C) The height of the liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{rdg}$.
Assuming the angle of contact $\theta = 0^\circ$ for water and glass,$\cos 0^\circ = 1$.
Given: $h = 0.015 \ m$,$T = 75 \times 10^{-3} \ N/m$,$g = 10 \ m/s^2$,and density of water $\rho = 10^3 \ kg/m^3$.
Rearranging the formula for radius $r$: $r = \frac{2T}{h \rho g}$.
Substituting the values: $r = \frac{2 \times 75 \times 10^{-3}}{0.015 \times 10^3 \times 10}$.
$r = \frac{150 \times 10^{-3}}{150} = 10^{-3} \ m$.
Since $10^{-3} \ m = 1 \ mm$,the radius of the capillary is $1 \ mm$.

(D) The height of water rise 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 for the given liquid and tube material,we have $h \propto \frac{1}{r}$.
Let $h_1 = 3\, mm$ and $r_1 = r$.
For the second tube,$r_2 = \frac{r}{3}$.
Using the relation $\frac{h_2}{h_1} = \frac{r_1}{r_2}$,we get $\frac{h_2}{3} = \frac{r}{r/3} = 3$.
Therefore,$h_2 = 3 \times 3 = 9\, mm$.

(A) The phenomenon of a liquid rising in a narrow tube or porous material is known as capillarity.
In a lantern,the wick acts as a bundle of fine capillary tubes.
Due to the surface tension of the kerosene oil,the adhesive forces between the oil and the wick material are stronger than the cohesive forces within the oil,causing the oil to rise against gravity through the wick.

(A) When a capillary tube is dipped in water,the meniscus formed is concave.
According to the Young-Laplace equation,the pressure on the concave side is greater than the pressure on the convex side.
Since the water surface inside the capillary is concave,the pressure just below the meniscus is less than the atmospheric pressure acting on the free surface of the water outside the tube.
This pressure difference creates an upward force that pushes the water column up into the capillary tube until the hydrostatic pressure of the column balances this pressure difference.
Therefore,the correct option is $A$.

(B) The mass of the liquid in the capillary tube is given by $M = V \times \rho = (\pi R^2 H) \rho$.
From the theory of capillarity,the height of the liquid column $H$ is inversely proportional to the radius $R$,i.e.,$H \propto 1/R$.
Substituting this into the mass equation: $M \propto R^2 \times (1/R) = R$.
Therefore,$M \propto R$.
If the radius is halved $(R' = R/2)$,the new mass $M'$ will be $M' = M \times (R'/R) = M \times (1/2) = M/2$.

(C) 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,we have $h \propto \frac{1}{r}$,where $r$ is the radius of the capillary tube.
Given that the diameter is halved,the radius $r$ is also halved $(r_2 = \frac{r_1}{2})$.
Therefore,the new height $h_2$ is given by: $\frac{h_2}{h_1} = \frac{r_1}{r_2} = \frac{r_1}{r_1/2} = 2$.
Thus,$h_2 = 2h_1 = 2h$.

(D) 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 tube,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
In an artificial satellite revolving around the Earth,the effective acceleration due to gravity $g_{eff}$ is zero because the satellite is in a state of weightlessness.
Substituting $g = 0$ into the formula,we get $h = \frac{2T \cos \theta}{r \rho (0)} = \infty$.
Since the height cannot be infinite,the liquid will rise until it reaches the top end of the capillary tube. Therefore,the water will rise to the full length of the capillary tube.

(B) The height or depression $h$ of a liquid 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,$d$ is the density of the liquid,$g$ is the acceleration due to gravity,and $r$ is the radius of the capillary tube.
From this formula,we can see that $h \propto \frac{1}{r}$.
This relationship represents a rectangular hyperbola. Therefore,the graph between $h$ and $r$ is a rectangular hyperbola,which corresponds to the curve shown in option $B$.

(A) The fluid resistance $R_f$ of a capillary tube is given by $R_f = \frac{8 \eta l}{\pi r^4}$.
For the first capillary,$R_1 = \frac{8 \eta L}{\pi R^4}$.
For the second capillary,$R_2 = \frac{8 \eta (2L)}{\pi (2R)^4} = \frac{16 \eta L}{\pi (16 R^4)} = \frac{\eta L}{\pi R^4} = \frac{1}{8} R_1$.
When connected in series,the equivalent resistance is $R_{eq} = R_1 + R_2 = R_1 + \frac{1}{8} R_1 = \frac{9}{8} R_1$.
The rate of flow $Q$ is given by $Q = \frac{P}{R_{eq}}$.
Since $X = \frac{P}{R_1}$,the new rate of flow $Q = \frac{P}{\frac{9}{8} R_1} = \frac{8}{9} \left( \frac{P}{R_1} \right) = \frac{8}{9} X$.

(B) The height of the water column in a vertical capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g} = 2.0 \, cm$.
When the capillary tube is tilted at an angle $\alpha = 60^\circ$ with the vertical,the vertical height $h$ remains constant because it depends on the pressure balance at the meniscus.
If $l$ is the length of the water column along the tube,then $h = l \cos \alpha$.
Therefore,$l = \frac{h}{\cos \alpha} = \frac{2.0 \, cm}{\cos 60^\circ}$.
Since $\cos 60^\circ = 0.5$,we get $l = \frac{2.0}{0.5} = 4.0 \, cm$.

(D) The height of a liquid in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{rdg}$,where $T$ is surface tension,$r$ is the radius of the tube,$d$ is the density,and $g$ is the acceleration due to gravity.
Assuming the contact angle $\theta$ and the radius $r$ are the same for both liquids,the ratio of heights is given by:
$\frac{h_1}{h_2} = \left( \frac{T_1}{T_2} \right) \times \left( \frac{d_2}{d_1} \right)$
Given $T_1 = 60 \, dyne/cm$,$T_2 = 50 \, dyne/cm$,$d_1 = 0.8 \, g/cm^3$,and $d_2 = 0.6 \, g/cm^3$:
$\frac{h_1}{h_2} = \left( \frac{60}{50} \right) \times \left( \frac{0.6}{0.8} \right)$
$\frac{h_1}{h_2} = \left( \frac{6}{5} \right) \times \left( \frac{6}{8} \right) = \frac{36}{40} = \frac{9}{10}$.

(B) The mass of water in a capillary tube is given by $M = V \rho = (\pi R^2 H) \rho$.
We know that the height of the water column in a capillary tube is inversely proportional to the radius,i.e.,$H \propto \frac{1}{R}$.
Substituting this into the mass equation: $M \propto R^2 \times (\frac{1}{R}) = R$.
Therefore,$M \propto R$.
If the radius is halved $(R' = R/2)$,the new mass $M'$ will be $M' = M \times (R'/R) = M \times (1/2) = M/2$.

(C) The height of a liquid column in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
Since the material of the tubes and the liquid are the same,the surface tension $T$,contact angle $\theta$,density $\rho$,and acceleration due to gravity $g$ are constant.
Therefore,$h \propto \frac{1}{r}$,which implies $h_1 r_1 = h_2 r_2$.
The ratio of the radii is given by $\frac{r_1}{r_2} = \frac{h_2}{h_1}$.
Given $h_1 = 2.2 \ cm$ and $h_2 = 6.6 \ cm$,we have $\frac{r_1}{r_2} = \frac{6.6}{2.2} = \frac{3}{1}$.
Thus,the ratio of the radii is $3:1$.

(A) The vertical height $h$ of the liquid column in a capillary tube remains constant regardless of the inclination of the tube.
If the tube is inclined at an angle $\alpha$ with the vertical,the length $l$ of the liquid column along the tube is given by $l = \frac{h}{\cos \alpha}$.
For the two given angles $\alpha_1 = 30^o$ and $\alpha_2 = 60^o$,the lengths are $l_1 = \frac{h}{\cos 30^o}$ and $l_2 = \frac{h}{\cos 60^o}$.
The ratio of the lengths is $\frac{l_1}{l_2} = \frac{\cos 60^o}{\cos 30^o}$.
Substituting the values: $\frac{l_1}{l_2} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.
Therefore,the ratio is $1:\sqrt{3}$.

(C) The height to which water rises in a capillary tube is given by $h = \frac{2T \cos \theta}{r \rho g}$.
If the actual length of the tube $L$ is less than the equilibrium height $h$,the water will rise to the top of the tube.
At the top,the radius of curvature $R$ of the meniscus will adjust such that $h' = \frac{2T \cos \theta'}{R \rho g} = L$,where $h' < h$.
Since the pressure at the top remains balanced by the surface tension,the water will not overflow but will stay at the top of the tube with a modified radius of curvature.

(D) The height of capillary rise is given by $h = \frac{2T \cos \theta}{r \rho g}$.
For a given value of $T$ and $r$,we have $h \propto \frac{\cos \theta}{\rho}$.
Since the liquids rise to the same height $(h_1 = h_2 = h_3)$,we have $\frac{\cos \theta_1}{\rho_1} = \frac{\cos \theta_2}{\rho_2} = \frac{\cos \theta_3}{\rho_3}$.
Given $\rho_1 > \rho_2 > \rho_3$,it follows that $\cos \theta_1 > \cos \theta_2 > \cos \theta_3$.
Since the cosine function is a decreasing function in the interval $[0, \frac{\pi}{2}]$,we have $\theta_1 < \theta_2 < \theta_3$.
Therefore,$0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$.