JEE Surface Tension Questions

The shape of the liquid surface in a capillary tube depends upon the magnitude of the contact angle $$\theta$$ with the glass wall.

• If $$\theta \lt 90^{\circ}$$, the liquid wets glass. The meniscus is concave and $$\cos\theta$$ is positive.
• If $$\theta = 90^{\circ}$$, the liquid does not rise or fall. The meniscus is flat and $$\cos\theta = 0$$.
• If $$\theta \gt 90^{\circ}$$, the liquid does not wet glass. The meniscus is convex and $$\cos\theta$$ is negative.

For the two liquids A and B we are given $$K = \dfrac{\cos\theta_A}{\cos\theta_B}$$.

Case 1: $$K \gt 0$$
Both $$\cos\theta_A$$ and $$\cos\theta_B$$ have the same sign, so both menisci are of the same type (either both concave or both convex).

Case 2: $$K = 0$$
This is possible only when $$\cos\theta_A = 0$$. Hence $$\theta_A = 90^{\circ}$$ (flat meniscus for liquid A) while liquid B can have any meniscus depending on $$\theta_B$$.

Case 3: $$K \lt 0$$
The ratio is negative, so $$\cos\theta_A$$ and $$\cos\theta_B$$ have opposite signs. Exactly one liquid has $$\theta \lt 90^{\circ}$$ (concave) and the other has $$\theta \gt 90^{\circ}$$ (convex).
Let $$\cos\theta_A$$ be positive and $$\cos\theta_B$$ be negative (the reverse choice still makes $$K$$ negative, but only one option in the list matches a concave A and convex B). Hence

 • liquid A: $$\theta_A \lt 90^{\circ}$$ ⇒ concave meniscus
 • liquid B: $$\theta_B \gt 90^{\circ}$$ ⇒ convex meniscus

This situation is described by Option C (statement 3): “K is negative, then liquid A has concave meniscus and liquid B has convex meniscus.”

Therefore, the correct statement is Option C (3).

When a liquid drop is formed, its surface possesses energy equal to (surface tension) × (surface area).
Hence the change in surface energy equals the change in surface area multiplied by the surface tension $$T$$.

Step 1: Initial surface area
Two identical drops, each of radius $$r$$, have total surface area
$$A_i = 2 \times 4\pi r^{2} = 8\pi r^{2}$$

Step 2: Radius of the bigger drop after coalescence
Volume is conserved during coalescence.
Initial volume: $$V_i = 2 \times \frac{4}{3}\pi r^{3} = \frac{8}{3}\pi r^{3}$$
Let $$R$$ be the radius of the bigger drop.
Final volume: $$V_f = \frac{4}{3}\pi R^{3}$$

Equating volumes:
$$\frac{4}{3}\pi R^{3} = \frac{8}{3}\pi r^{3}$$
$$\Rightarrow R^{3} = 2\,r^{3}$$
$$\Rightarrow R = 2^{\frac{1}{3}}\,r$$

Step 3: Final surface area
$$A_f = 4\pi R^{2} = 4\pi \left(2^{\frac{1}{3}}r\right)^{2} = 4\pi r^{2}\,2^{\frac{2}{3}}$$

Step 4: Decrease in surface area
$$\Delta A = A_i - A_f = 8\pi r^{2} - 4\pi r^{2}\,2^{\frac{2}{3}}$$
Factorising:
$$\Delta A = 4\pi r^{2}\left[2 - 2^{\frac{2}{3}}\right]$$

Step 5: Surface energy released
Released energy $$E = T \times \Delta A$$
$$E = 4\pi r^{2}\,T \left[2 - 2^{\frac{2}{3}}\right]$$

Thus, the surface energy released is
$$\boxed{4\pi r^{2}T\left[2 - 2^{\frac{2}{3}}\right]}$$

Option A is correct.

We need to evaluate two statements about fluid properties.

Statement I: "Hot water flows faster than cold water."

This is TRUE. The viscosity of a liquid decreases with increasing temperature because higher kinetic energy of the molecules helps them overcome intermolecular attractive forces more easily. Since viscosity resists flow, lower viscosity means hot water flows faster than cold water.

Statement II: "Soap water has higher surface tension as compared to fresh water."

This is FALSE. Soap is a surfactant that accumulates at the water surface and disrupts the cohesive hydrogen bonds between water molecules, thereby reducing the surface tension. Fresh water has a surface tension of about $$0.073$$ N/m at room temperature, and adding soap lowers this value significantly. This is precisely why soap helps in cleaning — the reduced surface tension allows water to spread and wet surfaces more effectively.

The correct answer is Option (1): Statement I is true but Statement II is false.

The capillary rise formula for a tube kept vertical is  $$h = \dfrac{2 T \cos \phi}{\rho g r}$$  where

$$T$$ = surface tension,  $$\phi$$ = contact angle,  $$\rho$$ = density of liquid,  $$g$$ = acceleration due to gravity,  $$r$$ = radius of the tube.

Given data (in SI units):
Radius, $$r = 0.1 \text{ mm} = 0.1 \times 10^{-3}\text{ m} = 1.0 \times 10^{-4}\text{ m}$$
Surface tension, $$T = 70 \text{ dyn/cm}$$. Since $$1 \text{ dyn} = 10^{-5}\text{ N}$$ and $$1 \text{ cm} = 10^{-2}\text{ m}$$,
$$T = 70 \times 10^{-5}\text{ N} \big/ 10^{-2}\text{ m} = 70 \times 10^{-3}\text{ N/m} = 0.07\text{ N/m}$$
Contact angle with glass, $$\phi \simeq 0^\circ \;\Rightarrow\; \cos\phi = 1$$
Density of water, $$\rho = 1000\text{ kg/m}^3$$
Acceleration, $$g = 9.8\text{ m/s}^2$$

Substituting into the formula, the vertical rise is

$$h = \dfrac{2 \times 0.07 \times 1}{1000 \times 9.8 \times 1.0 \times 10^{-4}} = \dfrac{0.14}{0.98} = 0.142857\text{ m} = 14.2857\text{ cm}$$

The tube is inclined at $$30^\circ$$ to the vertical. If $$L$$ is the length of water column along the axis of the tube, its vertical component equals $$h$$, so

$$L \cos 30^\circ = h$$ $$\Rightarrow\; L = \dfrac{h}{\cos 30^\circ} = \dfrac{14.2857}{\dfrac{\sqrt3}{2}} = \dfrac{200}{7\sqrt3}\text{ cm} \approx 16.5\text{ cm}$$

Among the given options, $$\dfrac{82}{5}\text{ cm} = 16.4\text{ cm}$$ is the closest and matches the calculated value.

Hence the length of water risen in the inclined capillary is approximately $$\mathbf{16.4\;cm}$$, so the correct choice is Option A.

Let us analyze each statement about fluid properties.

Statement A: Surface tension arises due to extra energy of molecules at the interior compared to molecules at the surface.

This is INCORRECT. Surface tension arises because molecules at the surface have higher potential energy than those in the interior (not the other way around). Interior molecules are surrounded by neighbors on all sides and are in a lower energy state.

Statement B: As temperature of liquid rises, the coefficient of viscosity increases.

This is INCORRECT. For liquids, viscosity decreases with increasing temperature because higher thermal energy helps molecules overcome intermolecular forces more easily.

Statement C: As temperature of gas increases, the coefficient of viscosity increases.

This is CORRECT. For gases, viscosity increases with temperature. This is because viscosity in gases arises from momentum transfer between molecular layers, and higher temperature means greater molecular speeds and more momentum transfer. The relationship is $$\eta \propto \sqrt{T}$$.

Statement D: The onset of turbulence is determined by Reynolds number.

This is CORRECT. The Reynolds number $$Re = \frac{\rho v d}{\eta}$$ determines whether flow is laminar or turbulent. Turbulence sets in when $$Re$$ exceeds a critical value.

Statement E: In steady flow, two streamlines never intersect.

This is CORRECT. If streamlines intersected, a fluid element at the intersection would have two velocity directions simultaneously, which is impossible in steady flow.

The correct statements are C, D, and E. The answer is Option A) C, D, E Only.

For a soap bubble in air, the excess pressure inside it is given by $$\Delta P = \frac{4T}{r}$$ where $$T$$ is the surface tension of the soap film and $$r$$ is the radius of the bubble.

Let the radii of bubbles A and B be $$r_A$$ and $$r_B$$ respectively.
Then $$\Delta P_A = \frac{4T}{r_A}$$ $$\Delta P_B = \frac{4T}{r_B}$$

The statement “excess pressure inside bubble A is half the excess pressure inside bubble B” translates to $$\Delta P_A = \frac{1}{2}\,\Delta P_B$$ Substituting from above, $$\frac{4T}{r_A} = \frac{1}{2}\left(\frac{4T}{r_B}\right)$$

Simplifying (the factor $$4T$$ cancels out): $$\frac{1}{r_A} = \frac{1}{2}\,\frac{1}{r_B}$$ which gives $$r_B = \frac{1}{2}\,r_A$$ or equivalently $$r_A = 2\,r_B$$.

The volume of a spherical bubble is $$V = \frac{4}{3}\,\pi r^{3}$$. Therefore, the ratio of the volumes of bubbles A and B is $$\frac{V_A}{V_B} = \left(\frac{r_A}{r_B}\right)^3 = \left(2\right)^3 = 8$$.

Given that $$V_A = n\,V_B$$, we have $$n = 8$$.

Hence the required value of $$n$$ is $$8$$.

We need to evaluate both statements about capillarity and contact angle.

Statement I: When a capillary tube is dipped into a liquid, the liquid neither rises nor falls. The contact angle may be $$0°$$.

If the liquid neither rises nor falls, the contact angle must be $$90°$$ (not $$0°$$). When the contact angle is $$0°$$, the liquid completely wets the surface and rises in the capillary. Statement I is false.

Statement II: The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well.

This is correct. The contact angle depends on the nature of both the solid surface and the liquid (as well as the surrounding medium). It depends on the relative strengths of adhesive forces (solid-liquid) and cohesive forces (liquid-liquid). Statement II is true.

The correct answer is Option (3): Statement I is false but Statement II is true.

1000 small drops coalesce to 1 big drop. Volume conserved: $$1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow R = 10r$$.

Surface energy: $$E = \sigma \times 4\pi r^2 \times n$$ for droplets, $$E_{big} = \sigma \times 4\pi R^2$$.

Ratio = $$\frac{1000 \times r^2}{R^2} = \frac{1000r^2}{100r^2} = 10 = \frac{10}{1}$$.

So $$x = 1$$.

The answer is 1.

Volume of 1000 small drops = volume of big drop.

$$1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow R = 10r$$.

$$E_1 = 1000 \times 4\pi r^2 \times T$$

$$E_2 = 4\pi R^2 \times T = 4\pi (10r)^2 T = 400\pi r^2 T$$

$$\frac{E_1}{E_2} = \frac{1000 \times 4\pi r^2 T}{400\pi r^2 T} = \frac{4000}{400} = 10$$

So $$E_1 : E_2 = 10 : 1$$, meaning $$x = 10$$.

Therefore, the answer is $$\boxed{10}$$.

A liquid column of height 0.04 cm balances the excess pressure of a soap bubble. Find the diameter of the bubble.

$$\Delta P = \frac{4S}{R}$$

where $$S$$ is the surface tension and $$R$$ is the radius. (Factor of 4 because a soap bubble has two surfaces.)

$$\Delta P = \rho g h = 8000 \times 10 \times 0.04 \times 10^{-2} = 8000 \times 10 \times 4 \times 10^{-4} = 32 \text{ Pa}$$

$$\frac{4S}{R} = \rho g h \implies R = \frac{4S}{\rho g h} = \frac{4 \times 0.28}{32} = \frac{1.12}{32} = 0.035 \text{ m}$$

$$D = 2R = 0.07 \text{ m} = 7 \text{ cm}$$

The correct answer is 7.

Work done in blowing a soap bubble from radius $$r_1$$ to $$r_2$$:

$$W = 8\pi T(r_2^2 - r_1^2)$$ (factor 8 because soap bubble has two surfaces)

Initial diameter = 7 cm, so $$r_1 = 3.5$$ cm. $$T = 40$$ dyne/cm. $$W = 36960$$ erg.

$$36960 = 8 \times \frac{22}{7} \times 40 \times (r_2^2 - 12.25)$$

$$36960 = \frac{7040}{7}(r_2^2 - 12.25)$$

$$36960 \times 7 = 7040(r_2^2 - 12.25)$$

$$r_2^2 - 12.25 = \frac{258720}{7040} = 36.75$$

$$r_2^2 = 49$$, $$r_2 = 7$$ cm

The answer is 7.

Let the outside (chamber) pressure initially be $$P_0 = 10^{5}\,\text{Pa}$$.
Inside the soap bubble the pressure is higher by the excess pressure $$\Delta P$$ produced by surface tension, so

$$P_{\text{in,1}} = P_0 + \Delta P.$$

For a soap bubble (two surfaces) the excess pressure is related to the surface tension $$T$$ and the bubble radius $$R$$ by

$$\Delta P = \dfrac{4T}{R} \qquad -(1)$$

The chamber pressure is now reduced to $$P_2 = \dfrac{8P_0}{27}$$ while temperature is kept constant. Because no air escapes from the bubble, the amount of gas inside remains the same, so the ideal-gas relation at constant temperature applies:

$$P_{\text{in,1}}V_1 = P_{\text{in,2}}V_2,$$

where $$V = \dfrac{4}{3}\pi R^{3}$$ is the bubble volume. Using $$P_{\text{in,1}} = P_0 + \Delta P$$ and $$P_{\text{in,2}} = P_2 + \Delta P_2$$, we get

$$(P_0 + \Delta P)R^{3} = (P_2 + \Delta P_2)R_2^{3}. \qquad -(2)$$

At the new equilibrium the excess pressure is

$$\Delta P_2 = \dfrac{4T}{R_2}. \qquad -(3)$$

Divide $$(1)$$ by $$(3)$$ to relate the two radii:

$$\dfrac{\Delta P}{\Delta P_2} = \dfrac{R_2}{R} \;\;\Longrightarrow\;\; R_2 = R\dfrac{\Delta P}{\Delta P_2}. \qquad -(4)$$

The statement “$$\Delta P$$ is much smaller than the chamber pressure” lets us neglect the terms $$\Delta P$$ and $$\Delta P_2$$ compared with $$P_0$$ and $$P_2$$. Thus $$P_{\text{in,1}}\approx P_0$$ and $$P_{\text{in,2}}\approx P_2$$, and equation $$(2)$$ simplifies to

$$P_0 R^{3} = P_2 R_2^{3}. \qquad -(5)$$

Substituting $$R_2$$ from $$(4)$$ into $$(5)$$:

$$P_0 R^{3} = P_2\!\left(R\dfrac{\Delta P}{\Delta P_2}\right)^{3} \;\;\Longrightarrow\;\; \dfrac{\Delta P}{\Delta P_2} = \left(\dfrac{P_2}{P_0}\right)^{1/3}. \qquad -(6)$$

Rearranging $$(6)$$ gives

$$\Delta P_2 = \Delta P\left(\dfrac{P_2}{P_0}\right)^{1/3}. \qquad -(7)$$

Insert the numerical values: $$\Delta P = 144\,\text{Pa}$$ and $$P_2/P_0 = \dfrac{8}{27}.$$ Because $$\left(\dfrac{8}{27}\right)^{1/3} = \dfrac{2}{3},$$

$$\Delta P_2 = 144 \times \dfrac{2}{3} = 96\,\text{Pa}.$$

Therefore, the new excess pressure in the bubble is 96 Pa.

We need to identify which of the given materials is not a semiconductor.

Analysing each material:

Option 1: Silicon (Si)

Silicon is one of the most widely used semiconductor materials. It has a band gap of approximately 1.1 eV at room temperature. It is the basis of modern electronics and integrated circuits. Silicon is a semiconductor.

Option 2: Copper oxide (Cu$$_2$$O / CuO)

Copper oxide is a p-type semiconductor. Historically, copper oxide rectifiers (based on Cu$$_2$$O) were among the first semiconductor devices used commercially. It has a band gap of about 1.2-2.0 eV depending on the form. Copper oxide is a semiconductor.

Option 3: Germanium (Ge)

Germanium is a classic semiconductor with a band gap of approximately 0.67 eV at room temperature. It was the first material used in transistors before being largely replaced by silicon. Germanium is a semiconductor.

Option 4: Graphite

Graphite is an allotrope of carbon with a layered structure. In each layer, carbon atoms are sp$$^2$$ hybridised, and the unhybridised p-orbitals form a delocalised $$\pi$$-electron system extending over the entire layer. These delocalised electrons are free to move within the layers, making graphite a good electrical conductor (resistivity $$\sim 3-60 \times 10^{-6}$$ ohm-m). Graphite is classified as a semimetal or conductor, not a semiconductor.

The correct answer is Option 4: Graphite.

Excess pressure inside a soap bubble: $$P = \frac{4T}{r}$$, where $$T$$ is surface tension and $$r$$ is radius.

Given $$P_1 = 3P_2$$: $$\frac{4T}{r_1} = 3 \cdot \frac{4T}{r_2} \implies r_2 = 3r_1$$.

Volume ratio: $$\frac{V_1}{V_2} = \frac{r_1^3}{r_2^3} = \frac{r_1^3}{27r_1^3} = \frac{1}{27}$$.

The correct answer is Option 3: $$1:27$$.

Find how surface energy changes when 1000 small droplets coalesce into one big drop.

We relate the radii by conserving the total volume: $$1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$ which gives $$R^3 = 1000r^3$$ and hence $$R = 10r$$.

The total surface area of the 1000 small drops is $$A_{\text{small}} = 1000 \times 4\pi r^2$$, while the surface area of the single large drop is $$A_{\text{big}} = 4\pi R^2 = 4\pi(10r)^2 = 400\pi r^2$$.

Since the surface energy is given by $$E = T \times A$$ with constant surface tension $$T$$, the ratio of the energies is $$\frac{E_{\text{big}}}{E_{\text{small}}} = \frac{A_{\text{big}}}{A_{\text{small}}} = \frac{400\pi r^2}{4000\pi r^2} = \frac{1}{10}$$.

Therefore, the surface energy becomes $$\frac{1}{10}$$th of its original value.

The correct answer is Option D: $$\frac{1}{10}$$th.

We need to evaluate two statements about capillary rise in cold and hot water.

The height of liquid rise in a capillary tube is given by Jurin's law:

$$h = \frac{2T\cos\theta}{\rho g r}$$

where $$T$$ is the surface tension of the liquid, $$\theta$$ is the contact angle, $$\rho$$ is the liquid density, $$g$$ is acceleration due to gravity, and $$r$$ is the radius of the capillary tube.

Surface tension of a liquid decreases with increasing temperature. This is because as temperature increases, the kinetic energy of molecules increases, weakening the intermolecular attractive forces at the surface. Therefore:

$$T_{hot} < T_{cold}$$

Since $$h \propto T$$ (surface tension), and the surface tension decreases with temperature:

$$h_{hot} < h_{cold}$$

This means the height of capillary rise is smaller in hot water than in cold water.

Statement I: "The height of capillary rise will be smaller in hot water." This is TRUE, as shown above.

Statement II: "The height of capillary rise will be smaller in cold water." This is FALSE; it is the opposite of what actually happens.

The correct answer is Option (3): Statement I is true but Statement II is false.

Evaluate the two statements about contact angle and capillary rise.

Statement I: "The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well." This is TRUE. The contact angle depends on the interfacial tensions between three phases: solid-liquid ($$\gamma_{SL}$$), solid-vapor ($$\gamma_{SV}$$), and liquid-vapor ($$\gamma_{LV}$$). These tensions are determined by the molecular properties of both the solid and the liquid. For example, water on glass has a different contact angle than water on wax, because the solid-liquid interaction differs. Similarly, mercury on glass has a different contact angle than water on glass, because the liquid is different.

Statement II: "The rise of a liquid in a capillary tube does not depend on the inner radius of the tube." This is FALSE. The capillary rise formula is:

$$h = \frac{2T\cos\theta}{\rho g r}$$

where $$T$$ is the surface tension, $$\theta$$ is the contact angle, $$\rho$$ is the liquid density, $$g$$ is gravitational acceleration, and $$r$$ is the inner radius of the tube. Clearly, $$h \propto \frac{1}{r}$$, so the capillary rise does depend on the inner radius. A narrower tube gives a greater rise.

The correct answer is Option A: Statement I is true but Statement II is false.

We need to find the excess pressure inside a soap bubble over the outside pressure.

Key Concept: Excess pressure in a bubble with surface tension.

For a liquid drop (with one surface), the excess pressure inside due to surface tension is given by the Young-Laplace equation:

$$\Delta P_{drop} = \frac{2S}{R}$$

where $$S$$ is the surface tension and $$R$$ is the radius.

For a soap bubble: A soap bubble has two surfaces -- an inner surface and an outer surface. Both surfaces contribute to the excess pressure. Each surface contributes $$\frac{2S}{R}$$.

Therefore, the total excess pressure inside a soap bubble is:

$$\Delta P_{bubble} = \frac{2S}{R} + \frac{2S}{R} = \frac{4S}{R}$$

This is double the excess pressure of a single-surface liquid drop because the soap film has two air-liquid interfaces.

The correct answer is Option 4: $$\frac{4S}{R}$$.

A liquid drop of radius $$R$$ is divided into 27 identical smaller drops. By conserving volume, we have $$\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3$$, which simplifies to $$R^3 = 27r^3$$ and hence $$r = \frac{R}{3}$$.

The initial surface area of the drop is $$A_i = 4\pi R^2$$. After division into 27 drops of radius $$r$$, the total surface area becomes $$A_f = 27 \times 4\pi r^2 = 27 \times 4\pi \times \frac{R^2}{9} = 12\pi R^2$$. Therefore, the change in surface area is $$\Delta A = A_f - A_i = 12\pi R^2 - 4\pi R^2 = 8\pi R^2$$.

The work done in increasing the surface area against surface tension is given by $$W = T \times \Delta A = T \times 8\pi R^2 = 8\pi R^2 T$$.

The work done is $$8\pi R^2 T$$, which corresponds to Option 1.

The excess pressure inside an air bubble in a liquid:

$$\Delta P = \frac{2T}{r} + \rho g h$$

where $$T$$ = surface tension, $$r$$ = radius, $$\rho$$ = density, $$h$$ = depth.

Note: An air bubble in liquid has only one surface, so excess pressure due to surface tension is $$\frac{2T}{r}$$ (not $$4T/r$$).

$$\frac{2T}{r} = \frac{2 \times 0.075}{1 \times 10^{-3}} = 150 \text{ Pa}$$

$$\rho g h = 1000 \times 10 \times 0.1 = 1000 \text{ Pa}$$

Total excess pressure above atmospheric: $$150 + 1000 = 1150 \text{ Pa}$$

This matches the answer key value of $$\mathbf{1150}$$.

A soap bubble has two surfaces, so the work done to change its radius is:

$$W = 2T \times \Delta A = 2T \times 4\pi(R_2^2 - R_1^2)$$

We are given: $$T = 3.5 \times 10^{-2}$$ N/m, $$R_1 = 10$$ cm = 0.1 m, $$R_2 = 20$$ cm = 0.2 m

$$W = 2 \times 3.5 \times 10^{-2} \times 4 \times \frac{22}{7} \times (0.04 - 0.01)$$

$$= 2 \times 3.5 \times 10^{-2} \times \frac{88}{7} \times 0.03$$

$$= 7 \times 10^{-2} \times \frac{88}{7} \times 0.03$$

$$= 88 \times 10^{-2} \times 0.03$$

$$= 2.64 \times 10^{-2} = 264 \times 10^{-4} \text{ J}$$

We need to find the work done to increase the radius of a soap bubble from 3.5 cm to 7 cm.

A soap bubble has two surfaces (inner and outer), so:

$$W = 2T \times \Delta A = 2T \times 4\pi(R_2^2 - R_1^2) = 8\pi T(R_2^2 - R_1^2)$$

$$T = 2.0 \times 10^{-2}$$ N/m, $$R_1 = 3.5$$ cm $$= 0.035$$ m, $$R_2 = 7$$ cm $$= 0.07$$ m, $$\pi = \dfrac{22}{7}$$.

$$R_2^2 - R_1^2 = (0.07)^2 - (0.035)^2 = 0.0049 - 0.001225 = 0.003675 \text{ m}^2$$

$$W = 8 \times \frac{22}{7} \times 2.0 \times 10^{-2} \times 0.003675$$

$$W = 8 \times \frac{22}{7} \times 7.35 \times 10^{-5}$$

$$W = 8 \times 22 \times \frac{7.35}{7} \times 10^{-5} = 8 \times 22 \times 1.05 \times 10^{-5}$$

$$W = 184.8 \times 10^{-5} = 18.48 \times 10^{-4} \text{ J}$$

The correct answer is Option C: $$18.48 \times 10^{-4}$$ J.

We have a mercury drop of radius $$R = 10^{-3}$$ m broken into $$n = 125$$ equal droplets, with surface tension $$T = 0.45$$ N/m.

By conservation of volume:

$$\frac{4}{3}\pi R^3 = 125 \times \frac{4}{3}\pi r^3$$

so $$R^3 = 125r^3$$, giving $$r = \frac{R}{5} = \frac{10^{-3}}{5} = 2 \times 10^{-4}$$ m.

The initial surface area is $$A_i = 4\pi R^2 = 4\pi \times 10^{-6}$$ m$$^2$$. The final total surface area (for all 125 droplets) is

$$A_f = 125 \times 4\pi r^2 = 125 \times 4\pi \times (2 \times 10^{-4})^2 = 125 \times 4\pi \times 4 \times 10^{-8} = 20\pi \times 10^{-6} \text{ m}^2$$

Now the change in surface area is

$$\Delta A = A_f - A_i = 20\pi \times 10^{-6} - 4\pi \times 10^{-6} = 16\pi \times 10^{-6} \text{ m}^2$$

So the gain in surface energy is (since surface energy equals surface tension times area)

$$\Delta E = T \times \Delta A = 0.45 \times 16\pi \times 10^{-6} = 7.2\pi \times 10^{-6} = 22.6 \times 10^{-6} \text{ J} = 2.26 \times 10^{-5} \text{ J}$$

Hence, the correct answer is $$2.26 \times 10^{-5}$$ J.

Given: 1000 droplets each of radius $$r = 1$$ mm $$= 10^{-3}$$ m, surface tension $$T = 0.07$$ N/m.

To begin, finding the radius of the combined drop,

By conservation of volume:

$$ 1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 $$

$$ R = 10r = 10 \times 10^{-3} = 10^{-2} \text{ m} $$

Next, change in surface area,

Initial total surface area = $$1000 \times 4\pi r^2 = 4000\pi \times 10^{-6}$$ m$$^2$$

Final surface area = $$4\pi R^2 = 4\pi \times 10^{-4}$$ m$$^2$$

$$ \Delta A = 4000\pi \times 10^{-6} - 4\pi \times 10^{-4} = 4\pi(1000 \times 10^{-6} - 10^{-4}) $$

$$ = 4\pi(10^{-3} - 10^{-4}) = 4\pi \times 9 \times 10^{-4} = 36\pi \times 10^{-4} $$

From this, released surface energy,

$$ E = T \times \Delta A = 0.07 \times 36 \times \frac{22}{7} \times 10^{-4} $$

$$ = 0.01 \times 36 \times 22 \times 10^{-4} = 0.01 \times 792 \times 10^{-4} = 7.92 \times 10^{-4} \text{ J} $$

We need to find the capillary rise in liquid B, given that the rise in liquid A is 5 cm.

$$h = \frac{2T\cos\theta}{\rho g r}$$

For liquid B: $$T_B = 2T_A$$ and $$\rho_B = 2\rho_A$$

Assuming same contact angle and same capillary tube.

$$\frac{h_B}{h_A} = \frac{T_B}{T_A} \times \frac{\rho_A}{\rho_B} = \frac{2T_A}{T_A} \times \frac{\rho_A}{2\rho_A} = 2 \times \frac{1}{2} = 1$$

$$h_B = h_A = 5$$ cm $$= 0.05$$ m

The height of liquid column raised in liquid B is $$0.05$$ m.

The correct answer is Option 3: $$0.05$$ m.

Consider a spherical soap bubble of radius 3 cm inside another spherical soap bubble of radius 6 cm. We need to find the radius $$r$$ of a single soap bubble whose internal pressure equals the internal pressure of the smaller bubble (radius 3 cm) in this system.

For any soap bubble, there are two surfaces (inner and outer), and the excess pressure inside is given by $$\Delta P = \frac{4S}{R}$$, where $$S$$ is the surface tension and $$R$$ is the radius. Let $$P_0$$ be the atmospheric pressure outside the outer bubble.

Define the pressures:

• $$P_1$$: Pressure inside the outer bubble but outside the inner bubble (in the space between them).
• $$P_2$$: Pressure inside the inner bubble (radius 3 cm).

For the outer bubble (radius $$R_1 = 6$$ cm), the pressure difference is:

For the inner bubble (radius $$R_2 = 3$$ cm), the pressure difference is:

Substitute $$P_1$$ from the outer bubble into this equation:

Simplify the expression inside the parentheses:

Now, for a single soap bubble of radius $$r$$, the internal pressure $$P_r$$ is:

The problem states that $$P_2 = P_r$$, so:

Subtract $$P_0$$ from both sides:

Divide both sides by $$4S$$ (assuming $$S \neq 0$$):

Therefore, $$r = 2$$ cm.

Hence, the correct answer is 2.

A water drop of diameter 2 cm is broken into 64 equal droplets. The radius of the original drop is $$R = 1$$ cm $$= 0.01$$ m. By conservation of volume:

$$\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3$$

It follows that $$R^3 = 64r^3$$ and hence $$r = \frac{R}{4} = \frac{0.01}{4} = 0.0025$$ m.

The initial surface area is $$4\pi R^2 = 4\pi (0.01)^2 = 4\pi \times 10^{-4}$$ m$$^2$$, while the final surface area is $$64 \times 4\pi r^2 = 64 \times 4\pi (0.0025)^2 = 64 \times 4\pi \times 6.25 \times 10^{-6}$$

$$= 64 \times 25\pi \times 10^{-6} = 1600\pi \times 10^{-6} = 16\pi \times 10^{-4}$$ m$$^2$$.

The increase in surface area is $$\Delta A = 16\pi \times 10^{-4} - 4\pi \times 10^{-4} = 12\pi \times 10^{-4}$$ m$$^2$$, so the gain in surface energy is given by

$$\Delta E = \text{Surface tension} \times \Delta A = 0.075 \times 12\pi \times 10^{-4}$$

$$= 0.075 \times 12 \times 3.1416 \times 10^{-4}$$

$$= 0.075 \times 37.699 \times 10^{-4}$$

$$= 2.827 \times 10^{-4}$$ J

$$\approx 2.8 \times 10^{-4}$$ J. The correct answer is Option A.

A water drop of radius $$R = 1 \text{ cm}$$ is broken into $$729$$ equal droplets. The surface tension of water is $$S = 75 \text{ dyne/cm}$$. We need to find the gain in surface energy.

By conservation of volume:

$$\frac{4}{3}\pi R^3 = 729 \times \frac{4}{3}\pi r^3$$

$$R^3 = 729 \cdot r^3$$

$$r = \frac{R}{9} = \frac{1}{9} \text{ cm}$$

Initial surface area (one big drop):

$$A_i = 4\pi R^2 = 4\pi (1)^2 = 4\pi \text{ cm}^2$$

Final surface area (729 small drops):

$$A_f = 729 \times 4\pi r^2 = 729 \times 4\pi \times \frac{1}{81} = 36\pi \text{ cm}^2$$

$$\Delta A = A_f - A_i = 36\pi - 4\pi = 32\pi \text{ cm}^2$$

$$\Delta E = S \times \Delta A = 75 \times 32\pi \text{ dyne/cm} \times \text{cm}^2 = 2400\pi \text{ erg}$$

Using $$\pi = 3.14$$:

$$\Delta E = 2400 \times 3.14 = 7536 \text{ erg}$$

Since $$1 \text{ erg} = 10^{-7} \text{ J}$$:

$$\Delta E = 7536 \times 10^{-7} \text{ J} = 7.536 \times 10^{-4} \text{ J}$$

Rounding to the first decimal place: $$\Delta E \approx 7.5 \times 10^{-4} \text{ J}$$.

The correct answer is Option C: $$7.5 \times 10^{-4} \text{ J}$$.

A liquid drop of density $$\rho$$ floats half-immersed in a liquid of density $$\sigma$$ with surface tension $$T = 7.5 \times 10^{-4} \text{ N cm}^{-1} = 7.5 \times 10^{-2} \text{ N m}^{-1}$$.

Set up the equilibrium condition.

For the floating drop, three forces act: weight (downward), buoyancy (upward), and surface tension (upward, acting along the contact circle of radius $$r$$).

The buoyancy equals the weight of liquid displaced by the immersed half-sphere (volume $$\frac{2}{3}\pi r^3$$), and the surface tension acts along the circular contact line of circumference $$2\pi r$$.

Simplify the equation.

Substitute values and find r in cm.

Working in SI units:

The correct answer is Option A: $$\dfrac{15}{\sqrt{2\rho - \sigma}}$$.

We consider the Assertion (A) which states that clothes with oil or grease stains cannot be cleaned by water wash alone. This is a well-known fact — water alone cannot remove oil or grease stains because water does not wet oily surfaces effectively; a detergent or soap is needed. So the Assertion is true.

Now we consider the Reason (R), which states that the angle of contact between oil/grease and water is obtuse. When water comes in contact with a greasy surface, the adhesive forces between water and grease are much weaker than the cohesive forces within water. This results in an obtuse (greater than $$90^\circ$$) contact angle, meaning water does not spread over the greasy surface. Since the contact angle is obtuse, water cannot penetrate and dissolve the grease, which is precisely why water alone fails to clean such stains. So the Reason is true, and it correctly explains the Assertion.

Hence, the correct answer is Option A.

Let each small drop have radius $$R$$. The surface energy of each drop is $$4\pi R^2 T$$, so the total surface energy before coalescence is $$E_i = 2 \times 4\pi R^2 T = 8\pi R^2 T$$.

After coalescence, volume is conserved: $$2 \times \frac{4}{3}\pi R^3 = \frac{4}{3}\pi r^3$$, giving $$r^3 = 2R^3$$, so $$r = 2^{1/3}R$$.

The surface energy of the large drop is $$E_f = 4\pi r^2 T = 4\pi (2^{1/3}R)^2 T = 4\pi \cdot 2^{2/3} R^2 T$$.

The ratio of initial to final surface energy is $$\frac{E_i}{E_f} = \frac{8\pi R^2 T}{4\pi \cdot 2^{2/3} R^2 T} = \frac{8}{4 \cdot 2^{2/3}} = \frac{2}{2^{2/3}} = 2^{1-2/3} = 2^{1/3}$$.

Therefore the ratio of total surface energy before to after is $$2^{1/3} : 1$$.

The excess (gauge) pressure inside a soap bubble over the pressure of the medium that surrounds the bubble is given by the formula $$\Delta P = \dfrac{4T}{r}$$, where $$T$$ is the surface tension of the soap film and $$r$$ is the radius of the bubble. This factor 4 appears because a soap bubble has two free surfaces, an inner and an outer one.

We have two concentric bubbles. The larger bubble, whose radius is $$r_2 = 6 \, \text{cm}$$, is in direct contact with the atmosphere. The smaller bubble, of radius $$r_1 = 3 \, \text{cm}$$, is lodged inside the larger one and is therefore surrounded not by the atmosphere but by the air that already exists inside the larger bubble.

First, we calculate the excess pressure inside the larger bubble over atmospheric pressure. Using the stated formula,

$$\Delta P_{\text{large}} = \dfrac{4T}{r_2} = \dfrac{4T}{6} = \dfrac{2T}{3}.$$ Thus the pressure just inside the wall of the larger bubble is higher than atmospheric pressure by $$\dfrac{2T}{3}$$.

Next, we determine the excess pressure inside the smaller bubble relative to the air that surrounds it (which is the air inside the larger bubble). Again applying the formula,

$$\Delta P_{\text{small w.r.t. large}} = \dfrac{4T}{r_1} = \dfrac{4T}{3}.$$

Therefore, the total excess pressure of the air at the centre of the smaller bubble over the atmospheric pressure is obtained by simple addition of these two pressure rises, because pressures add algebraically along the same line of action:

$$\Delta P_{\text{small w.r.t. atmosphere}} = \Delta P_{\text{large}} + \Delta P_{\text{small w.r.t. large}} = \dfrac{2T}{3} + \dfrac{4T}{3} = \dfrac{6T}{3} = 2T.$$

We are asked for the radius $$r_{\text{eq}}$$ of a single isolated soap bubble that would have exactly this same excess pressure $$2T$$ over the atmosphere. Setting the general expression equal to this value, we write

$$\dfrac{4T}{r_{\text{eq}}} = 2T.$$

Dividing both sides by $$T$$ and then by 2, we carry out the straightforward algebraic steps:

$$\dfrac{4}{r_{\text{eq}}} = 2 \;\;\Longrightarrow\;\; r_{\text{eq}} = \dfrac{4}{2} = 2 \, \text{cm}.$$

So, the answer is $$2 \text{ cm}$$.

We have a U-shaped tube whose two limbs are capillaries of different radii, both in contact with water. Because water wets glass (angle of contact $$\theta = 0^{\circ} \Rightarrow \cos\theta = 1$$), it rises inside each limb due to surface tension. At any horizontal level inside the continuous water column, the pressures in the two limbs must be the same. That requirement allows us to relate the capillary rises in the two limbs.

First, recall the standard capillary rise formula for a single capillary tube:

$$h = \frac{2T\cos\theta}{\rho g r}$$

where

$$T = \text{surface tension}, \quad \theta = \text{angle of contact}, \quad \rho = \text{density of the liquid}, \quad g = \text{acceleration due to gravity}, \quad r = \text{radius of the tube}.$$

Because both limbs are filled with the same water, we apply this formula separately to each limb, obtaining

$$h_1 = \frac{2T}{\rho g r_1}, \qquad h_2 = \frac{2T}{\rho g r_2}$$

with $$r_1$$ and $$r_2$$ being the radii of the narrower and the wider limbs, respectively.

Inside the continuous water column, the hydrostatic pressures at the same horizontal level must balance. That means the column of water in the narrower limb must rise higher than that in the wider limb by exactly the amount that equalises the pressures. Hence the difference in heights of the two free surfaces is simply

$$\Delta h = h_1 - h_2 = \frac{2T}{\rho g}\left(\frac{1}{r_1} - \frac{1}{r_2}\right).$$

Now we substitute the numerical data. The diameters are 5.0 mm and 8.0 mm, so the radii are

$$r_1 = \frac{5.0\ \text{mm}}{2} = 2.5\ \text{mm} = 2.5 \times 10^{-3}\ \text{m},$$

$$r_2 = \frac{8.0\ \text{mm}}{2} = 4.0\ \text{mm} = 4.0 \times 10^{-3}\ \text{m}.$$

Compute the reciprocal radii:

$$\frac{1}{r_1} = \frac{1}{2.5 \times 10^{-3}} = 0.4 \times 10^{3}\ \text{m}^{-1} = 400\ \text{m}^{-1},$$

$$\frac{1}{r_2} = \frac{1}{4.0 \times 10^{-3}} = 0.25 \times 10^{3}\ \text{m}^{-1} = 250\ \text{m}^{-1}.$$

Thus

$$\frac{1}{r_1} - \frac{1}{r_2} = 400 - 250 = 150\ \text{m}^{-1}.$$

Next, evaluate the coefficient $$\dfrac{2T}{\rho g}$$:

$$2T = 2 \times 7.3 \times 10^{-2}\ \text{N\,m}^{-1} = 1.46 \times 10^{-1}\ \text{N\,m}^{-1},$$

$$\rho g = 1.0 \times 10^{3}\ \text{kg\,m}^{-3} \times 10\ \text{m\,s}^{-2} = 1.0 \times 10^{4}\ \text{N\,m}^{-3}.$$

Therefore

$$\frac{2T}{\rho g} = \frac{1.46 \times 10^{-1}}{1.0 \times 10^{4}} = 1.46 \times 10^{-5}\ \text{m}.$$

Finally, multiply to find the height difference:

$$\Delta h = 1.46 \times 10^{-5}\ \text{m} \times 150 = 2.19 \times 10^{-3}\ \text{m}.$$

Convert metres to millimetres:

$$2.19 \times 10^{-3}\ \text{m} = 2.19\ \text{mm}.$$

Hence, the correct answer is Option C.

Let the surface tension of the soap film be $$T$$. Because a soap bubble has two liquid surfaces, the excess pressure inside a bubble of radius $$r$$ in vacuum is given by the formula

$$P = \dfrac{4T}{r}$$

Here the outside pressure is zero (perfect vacuum), so the whole pressure inside the bubble is only this excess pressure.

For the first bubble of radius $$r_1$$ we therefore have

$$P_1 = \dfrac{4T}{r_1}$$

For the second bubble of radius $$r_2$$ we have in the same way

$$P_2 = \dfrac{4T}{r_2}$$

The volumes of the two initial bubbles are, respectively,

$$V_1 = \dfrac{4}{3}\pi r_1^{3}, \qquad V_2 = \dfrac{4}{3}\pi r_2^{3}$$

When the two bubbles coalesce isothermally, the gas contained in them is conserved. At constant temperature the ideal-gas relation $$PV = nRT$$ tells us that the product $$PV$$ is directly proportional to the number of moles of gas. Hence the total $$PV$$ before coalescence equals the final $$PV$$ after coalescence.

So we write

$$P_1 V_1 + P_2 V_2 = P_f V_f$$

where $$P_f$$ and $$V_f$$ are the pressure and volume of the single final bubble. Let the radius of this final bubble be $$R$$. Then

$$P_f = \dfrac{4T}{R}, \qquad V_f = \dfrac{4}{3}\pi R^{3}$$

Substituting all these expressions, we get

$$\dfrac{4T}{r_1}\left(\dfrac{4}{3}\pi r_1^{3}\right)\;+\;\dfrac{4T}{r_2}\left(\dfrac{4}{3}\pi r_2^{3}\right) \;=\;\dfrac{4T}{R}\left(\dfrac{4}{3}\pi R^{3}\right)$$

Every term on both sides contains the common factor $$\dfrac{4T}{\phantom{2}}\times\dfrac{4}{3}\pi$$, so we cancel it out completely. After cancelling we are left with

$$\dfrac{r_1^{3}}{r_1}\;+\;\dfrac{r_2^{3}}{r_2}\;=\;\dfrac{R^{3}}{R}$$

which simplifies step by step as follows:

$$r_1^{2} + r_2^{2} = R^{2}$$

Now taking the square root of both sides (and keeping only the positive root because radius is positive) we find

$$R = \sqrt{\,r_1^{2} + r_2^{2}\,}$$

This expression exactly matches Option C.

Hence, the correct answer is Option C.

We first collect the data given in the question. The radius of the capillary is $$r = 0.015 \text{ cm}$$ and the height to which the liquid rises is $$h = 15 \text{ cm}$$. We convert these centimetre values into metre because SI units must be used throughout. Since $$1 \text{ cm} = 10^{-2} \text{ m}$$, we have $$r = 0.015 \times 10^{-2} \text{ m} = 0.00015 \text{ m}$$ and $$h = 15 \times 10^{-2} \text{ m} = 0.15 \text{ m}$$. The density of the liquid is $$\rho = 900 \text{ kg m}^{-3}$$ and the acceleration due to gravity is $$g = 10 \text{ m s}^{-2}$$. The contact angle is nearly $$0^{\circ}$$, so $$\cos \theta \approx 1$$.

Now we recall the capillary rise formula. For a liquid that wets the walls of a capillary (contact angle $$\theta$$) the equilibrium height $$h$$ to which the liquid rises is related to the surface tension $$T$$ by the expression $$h = \frac{2T \cos\theta}{\rho g r}.$$

Because $$\theta \approx 0^{\circ}$$, we can put $$\cos\theta = 1$$, and we rearrange the formula to make $$T$$ the subject: $$T = \frac{h \, \rho \, g \, r}{2}.$$

Substituting each numerical value step by step, we write $$T = \frac{0.15 \times 900 \times 10 \times 0.00015}{2} \; \text{N m}^{-1}.$$

We now carry out the multiplication in the numerator: $$0.15 \times 900 = 135,$$ so the expression becomes $$T = \frac{135 \times 10 \times 0.00015}{2}.$$ Next, $$135 \times 10 = 1350,$$ giving $$T = \frac{1350 \times 0.00015}{2}.$$ Multiplying further, $$1350 \times 0.00015 = 0.2025,$$ and therefore $$T = \frac{0.2025}{2} \; \text{N m}^{-1}.$$ Dividing by $$2$$ we obtain $$T = 0.10125 \; \text{N m}^{-1}.$$

We are asked for the answer in milliNewton per metre. Remembering that $$1 \text{ N} = 1000 \text{ mN}$$, we multiply by $$1000$$: $$T = 0.10125 \times 1000 = 101.25 \; \text{mN m}^{-1}.$$

Finally, rounding to the nearest integer as demanded in the closed-integer answer format, we get $$T \approx 101 \text{ mN m}^{-1}.$$

So, the answer is $$101$$.

For a liquid that wets the tube, the height to which it rises in a capillary of radius $$r$$ is obtained from the balance between the upward vertical component of surface-tension force and the downward weight of the liquid column.

First we recall 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 radius of the capillary tube.

We are told that the two tangents drawn to the meniscus from opposite sides make an angle of $$60^{\circ}$$ with each other. Since these two tangents are symmetric about the vertical, the angle each tangent makes with the wall is the contact angle. Hence

$$2\theta = 60^{\circ}\;\;\Longrightarrow\;\;\theta = 30^{\circ}.$$

The numerical values given are

$$T = 0.05 \;\text{N\,m}^{-1}, \qquad \rho = 667 \;\text{kg\,m}^{-3}, \qquad g = 10 \;\text{m\,s}^{-2}, \qquad r = 0.15 \;\text{mm}.$$

Converting the radius to metres,

$$r = 0.15 \times 10^{-3} \;\text{m} = 1.5 \times 10^{-4} \;\text{m}.$$

Now we substitute every quantity into the capillary-rise formula.

First evaluate the numerator:

$$2T\cos\theta = 2 \times 0.05 \times \cos 30^{\circ}.$$

Since $$\cos 30^{\circ} = \frac{\sqrt3}{2} \approx 0.8660,$$ we have

$$2T\cos\theta = 0.10 \times 0.8660 = 0.0866 \;\text{N}.$$

Next evaluate the denominator:

$$\rho g r = 667 \times 10 \times 1.5 \times 10^{-4}.$$

Multiplying step by step,

$$667 \times 10 = 6670,$$

and

$$6670 \times 1.5 = 10005,$$

so

$$\rho g r = 10005 \times 10^{-4} \;\text{N\,m}^{-1} = 1.0005 \;\text{N\,m}^{-1}.$$

Finally, divide the numerator by the denominator to obtain the height:

$$h = \frac{0.0866}{1.0005} \;\text{m} \approx 0.0866 \;\text{m}.$$

Rounding to three significant figures gives

$$h \approx 0.087\;\text{m}.$$

Hence, the correct answer is Option B.

Let us denote the atmospheric (external) pressure by $$P_{0}$$. For a thin soap bubble the excess pressure inside the bubble is given by the well-known formula

$$\Delta P \;=\; P_{\text{inside}} - P_{0} \;=\; \dfrac{4T}{r}$$

Here $$T$$ is the surface tension of the liquid film and $$r$$ is the radius of the spherical bubble. This expression contains the factor $$4T$$ (and not $$2T$$) because a soap bubble has two liquid surfaces in contact with air, one on the inside and one on the outside.

We are told that the absolute internal pressures of the two bubbles are

$$P_{1}=1.01\;\text{atm},\qquad P_{2}=1.02\;\text{atm}$$

and, naturally, both bubbles are in the same atmosphere so

$$P_{0}=1.00\;\text{atm}$$

Using the formula for each bubble, we first calculate their respective excess pressures:

For the first bubble $$\Delta P_{1}=P_{1}-P_{0}=1.01-1.00=0.01\;\text{atm}$$

For the second bubble $$\Delta P_{2}=P_{2}-P_{0}=1.02-1.00=0.02\;\text{atm}$$

Now we substitute these excess pressures into the relation $$\Delta P=\dfrac{4T}{r}$$ to connect each pressure to the corresponding radius.

For the first bubble $$\dfrac{4T}{r_{1}}=0.01\;\text{atm}\qquad\Longrightarrow\qquad r_{1}=\dfrac{4T}{0.01\;\text{atm}}$$

For the second bubble $$\dfrac{4T}{r_{2}}=0.02\;\text{atm}\qquad\Longrightarrow\qquad r_{2}=\dfrac{4T}{0.02\;\text{atm}}$$

To compare the radii, we take the ratio of the two expressions:

$$\dfrac{r_{2}}{r_{1}} =\dfrac{\dfrac{4T}{0.02}}{\dfrac{4T}{0.01}} =\dfrac{0.01}{0.02} =\dfrac{1}{2}$$

So we find $$r_{2}=\dfrac{r_{1}}{2}$$ meaning the second bubble (with the higher internal pressure) has half the radius of the first bubble.

Each bubble is spherical, so its volume is given by the standard geometric formula

$$V=\dfrac{4}{3}\pi r^{3}$$

Therefore, the ratio of their volumes becomes

$$\dfrac{V_{1}}{V_{2}} =\dfrac{\dfrac{4}{3}\pi r_{1}^{3}}{\dfrac{4}{3}\pi r_{2}^{3}} =\left(\dfrac{r_{1}}{r_{2}}\right)^{3} =\left(\dfrac{r_{1}}{\,r_{1}/2}\right)^{3} =(2)^{3}=8$$

Thus

$$V_{1}:V_{2}=8:1$$

The option that matches this ratio is Option C.

Hence, the correct answer is Option C.

Let the radius of the spherical droplet be $$r$$. We start by writing every force that acts on the droplet while it floats exactly half-immersed in the liquid.

First, the weight of the droplet. The volume of a sphere is $$\frac{4}{3}\pi r^{3}$$, so using the definition $$\text{weight} = \text{(mass)}\,g = (\text{density}\times\text{volume})\,g$$ we have

$$W = \frac{4}{3}\pi r^{3}\,d\,g.$$

Next, the buoyant force. Archimedes’ principle states that the buoyant force equals the weight of the liquid displaced. Because the droplet is half-immersed, only half its volume is displaced:

$$V_{\text{submerged}} = \frac{1}{2}\left(\frac{4}{3}\pi r^{3}\right)=\frac{2}{3}\pi r^{3}.$$

Hence the upward buoyant force is

$$F_{b} = V_{\text{submerged}}\;\rho\,g = \frac{2}{3}\pi r^{3}\,\rho\,g.$$

Now we consider the force due to surface tension. Surface tension $$T$$ pulls all along the circular line of contact, whose circumference is $$2\pi r$$. For a droplet that meets the liquid surface at a right angle (which is the case when it is exactly half immersed) the entire tension around the rim contributes vertically upward. Therefore, the upward force supplied by surface tension is

$$F_{T} = T\;(2\pi r)=2\pi r\,T.$$

Because the droplet is in equilibrium, the sum of the upward forces equals the downward weight:

$$W = F_{b} + F_{T}.$$

Substituting the three expressions just obtained, we get

$$\frac{4}{3}\pi r^{3}d g = \frac{2}{3}\pi r^{3}\rho g + 2\pi r T.$$

We now perform careful algebra step by step. First divide the entire equation by $$\pi r$$ (this removes one power of $$r$$ from every term):

$$\frac{4}{3}r^{2} d g = \frac{2}{3}r^{2} \rho g + 2 T.$$

Shift the buoyant term to the left side:

$$\frac{4}{3}r^{2} d g - \frac{2}{3}r^{2} \rho g = 2 T.$$

Take $$\frac{2}{3}r^{2}g$$ common on the left:

$$\left(\frac{2}{3}r^{2}g\right)(2d - \rho)=2T.$$

Divide both sides by 2 to simplify:

$$\left(\frac{1}{3}r^{2}g\right)(2d - \rho)=T.$$

Finally solve for $$r^{2}$$:

$$r^{2}= \frac{3T}{(2d-\rho)g}.$$

Taking the positive square root (radius is always positive), we obtain

$$r=\sqrt{\frac{3T}{(2d-\rho)g}}.$$

This matches Option D.

Hence, the correct answer is Option D.

For a liquid contained in a narrow capillary tube, the height of rise or depression is governed by the capillary formula

$$h \;=\; \frac{2T\cos\theta}{\rho g\,r}\;,$$

where $$T$$ is the surface tension of the liquid, $$\theta$$ is the contact angle with the tube material, $$\rho$$ is the density of the liquid, $$g$$ is the acceleration due to gravity, and $$r$$ is the inner radius of the capillary tube. This expression gives a positive height for rise (when $$\cos\theta>0$$) and a negative height for depression (when $$\cos\theta<0$$). However, the magnitude of the height is always

$$|h| \;=\; \frac{2T\,|\cos\theta|}{\rho g\,r}.$$

According to the statement of the problem, the magnitude of the depression of mercury equals the magnitude of the rise of water and is denoted by the same symbol $$h$$. Hence we write, for mercury (subscript $$m$$) and water (subscript $$w$$)

$$h \;=\; \frac{2T_m\,|\cos\theta_m|}{\rho_m g\,r_1} \quad\text{and}\quad h \;=\; \frac{2T_w\,|\cos\theta_w|}{\rho_w g\,r_2}.$$

Since both right-hand sides are equal to the same $$h$$, we equate them directly:

$$\frac{2T_m\,|\cos\theta_m|}{\rho_m g\,r_1} \;=\; \frac{2T_w\,|\cos\theta_w|}{\rho_w g\,r_2}.$$

The common factors $$2$$ and $$g$$ cancel from both numerators and denominators, leaving

$$\frac{T_m\,|\cos\theta_m|}{\rho_m\,r_1} \;=\; \frac{T_w\,|\cos\theta_w|}{\rho_w\,r_2}.$$

Rearranging to isolate the ratio of the radii, we obtain

$$\frac{r_1}{r_2} \;=\;\frac{T_m}{T_w}\; \frac{\rho_w}{\rho_m}\; \frac{|\cos\theta_m|}{|\cos\theta_w|}.$$

Now we substitute the numerical data supplied in the question:

Surface-tension ratio: $$\dfrac{T_m}{T_w}=7.5.$$

Density ratio: $$\dfrac{\rho_m}{\rho_w}=13.6 \;\Longrightarrow\; \dfrac{\rho_w}{\rho_m}=\dfrac{1}{13.6}.$$

Contact angles:
For mercury $$\theta_m=135^\circ\,,\quad |\cos\theta_m|=|\cos135^\circ|=\dfrac{\sqrt2}{2}\approx0.707,$$
For water $$\theta_w=0^\circ\,,\quad |\cos\theta_w|=|\cos0^\circ|=1.$$

Substituting these three pieces of information into the radius ratio gives

$$\frac{r_1}{r_2} \;=\; 7.5\; \left(\frac1{13.6}\right)\; \left(\frac{0.707}{1}\right) \;=\; \frac{7.5\times0.707}{13.6}.$$

First, multiplying $$7.5\times0.707$$ yields

$$7.5\times0.707\;\approx\;5.30.$$

Next, dividing by $$13.6$$ produces

$$\frac{5.30}{13.6}\;\approx\;0.39.$$

Converting to an exact fraction close to this decimal, we note that

$$0.39\;\text{is almost equal to}\;\frac{2}{5}=0.40.$$

Among the options offered, $$\dfrac25$$ is clearly the closest value.

Hence, the correct answer is Option D.

We begin with the well-known expression for the height to which a liquid of density $$\rho$$ rises in a capillary tube of radius $$r$$ due to surface tension. The formula is stated as

$$h=\frac{2T\cos\theta}{\rho g\,r},$$

where $$T$$ is the surface-tension coefficient of the liquid, $$\theta$$ is the angle of contact, and $$g$$ is the acceleration due to gravity.

Now, the volume of the liquid column inside the tube is obtained from the usual cylinder relation

$$V=\pi r^{2}h.$$

To convert this volume into mass, we multiply by the density $$\rho$$ because mass and volume are connected through the formula $$m=\rho V$$. Substituting the above volume, we have

$$m=\rho\bigl(\pi r^{2}h\bigr).$$

At this stage we substitute the value of $$h$$ from Jurin’s law. Thus

$$m=\rho\bigl(\pi r^{2}\bigr)\left(\frac{2T\cos\theta}{\rho g\,r}\right).$$

Notice that the factor $$\rho$$ appears in both the numerator and the denominator, so it cancels out completely. Carrying out this cancellation and also cancelling one factor of $$r$$ between the numerator and denominator, we obtain

$$m=\frac{2\pi T\cos\theta}{g}\,r.$$

This final algebraic form shows clearly that the mass $$m$$ of the liquid column is directly proportional to the radius $$r$$ of the capillary tube. In simpler words, we can write

$$m\propto r.$$

We are told that when the radius is $$r$$, the mass of the water column is $$M$$. Symbolically,

$$M=k\,r,$$

where $$k=\dfrac{2\pi T\cos\theta}{g}$$ is the proportionality constant comprising only fixed physical quantities.

Next, we increase the radius to $$2r$$. According to the direct proportionality, the new mass $$M'$$ must satisfy

$$M'=k\,(2r)=2\,k\,r.$$

But $$k\,r$$ is nothing but our original mass $$M$$. Therefore, substituting, we find

$$M'=2M.$$

So doubling the radius doubles the mass of water that rises in the capillary.

Hence, the correct answer is Option B.

For a thin soap bubble the excess pressure inside over the outside is given by the Laplace formula:

$$\Delta P=\dfrac{4T}{r},$$

where $$T$$ is the surface tension and $$r$$ is the radius of the bubble. A soap bubble has two liquid surfaces, hence the factor 4 in the numerator.

Let us denote

$$P_0=\text{outside atmospheric pressure},$$

$$P_1$$ = pressure in the air layer between the two bubbles,

$$P_2=\text{pressure inside the smaller (inner) bubble}.$$

We have an outer bubble of radius $$6\ \text{cm}$$. Applying the formula to this outer bubble, the pressure difference between its inside and the atmosphere is

$$P_1-P_0=\dfrac{4T}{6\ \text{cm}}=\dfrac{2T}{3}.$$

Next, there is an inner bubble of radius $$4\ \text{cm}$$. The pressure difference between the inside of this inner bubble and the air immediately outside it (which is at pressure $$P_1$$) is

$$P_2-P_1=\dfrac{4T}{4\ \text{cm}}=T.$$

Now we add these two differences to obtain the total excess pressure inside the inner bubble relative to the atmosphere:

$$P_2-P_0=(P_2-P_1)+(P_1-P_0)=T+\dfrac{2T}{3}=\dfrac{5T}{3}.$$

Suppose we have a single soap bubble of unknown radius $$R$$ whose inside pressure exceeds the outside pressure by this same amount $$P_2-P_0$$. Using the Laplace formula once again, we write

$$P_2-P_0=\dfrac{4T}{R}.$$

Substituting the value already found for $$P_2-P_0$$, we get

$$\dfrac{4T}{R}=\dfrac{5T}{3}.$$

Dividing both sides by $$T$$ and cross-multiplying,

$$4\cdot 3 = 5R \;\;\Longrightarrow\;\; R=\dfrac{12}{5}\ \text{cm}=2.4\ \text{cm}.$$

Hence, the correct answer is Option A.

The pressure difference ($$\Delta P$$) across a curved liquid surface is determined by the surface tension ($$T$$) and the two principal radii of curvature ($$R_1$$ and $$R_2$$) is $$\Delta P = T \left[ \frac{1}{R_1} + \frac{1}{R_2} \right]$$

This excess pressure is the same as the excess pressure inside a drop or air bubble.

So,  $$R_1 = R_2 = R$$

$$\Delta P = T \left[ \frac{1}{R} + \frac{1}{R} \right]$$

$$\Delta P = \frac{2T}{R}$$

We are given:

• Radius of air bubble, $$ r = 0.1 \, \text{cm} $$. Convert to meters: $$ 0.1 \, \text{cm} = 0.1 \times 10^{-2} \, \text{m} = 0.001 \, \text{m} $$.
• Surface tension, $$ S = 0.06 \, \text{N/m} $$.
• Density of liquid, $$ \rho = 10^3 \, \text{kg/m}^3 = 1000 \, \text{kg/m}^3 $$.
• Excess pressure inside bubble over atmospheric pressure, $$ \Delta P = 1100 \, \text{N/m}^2 $$.
• Acceleration due to gravity, $$ g = 9.8 \, \text{m/s}^2 $$.

We need to find the depth $$ h $$ (in meters) of the bubble below the surface of the liquid.

For an air bubble inside a liquid, the excess pressure inside the bubble over the atmospheric pressure is the sum of two components:

1. The pressure due to the depth of the bubble: $$ \rho g h $$.
2. The pressure due to surface tension: $$ \frac{2S}{r} $$.

Therefore, the equation for excess pressure is:

$$ \Delta P = \rho g h + \frac{2S}{r} $$

Substitute the given values:

$$ 1100 = (1000) \times (9.8) \times h + \frac{2 \times 0.06}{0.001} $$

First, calculate the surface tension term:

$$ \frac{2 \times 0.06}{0.001} = \frac{0.12}{0.001} = 120 \, \text{N/m}^2 $$

So the equation becomes:

$$ 1100 = 9800h + 120 $$

Now, solve for $$ h $$. Subtract 120 from both sides:

$$ 1100 - 120 = 9800h $$

$$ 980 = 9800h $$

Divide both sides by 9800:

$$ h = \frac{980}{9800} $$

Simplify the fraction:

$$ h = \frac{98}{980} = \frac{98 \div 98}{980 \div 98} = \frac{1}{10} = 0.1 \, \text{m} $$

Therefore, the depth of the bubble is 0.1 m.

Comparing with the options:

• A. 0.1 m
• B. 0.15 m
• C. 0.20 m
• D. 0.25 m

Hence, the correct answer is Option A.

To solve this problem, we need to find the correct relation between the change in volume (V), change in surface area (S), surface tension (T), and atmospheric pressure (P) when two soap bubbles coalesce into one. We start by recalling the physics of soap bubbles.

A soap bubble has an excess pressure inside due to surface tension. Since a soap bubble has two surfaces (inner and outer), the excess pressure is given by $$\Delta P = \frac{4T}{r}$$, where $$r$$ is the radius. Therefore, the pressure inside a bubble is $$P_{\text{inside}} = P + \frac{4T}{r}$$, with $$P$$ being atmospheric pressure.

Let the two initial bubbles have radii $$r_1$$ and $$r_2$$. After coalescing, they form a single bubble of radius $$r$$. The air inside is conserved, and assuming constant temperature, we use the ideal gas law. The number of moles of air is conserved.

For the first bubble, the number of moles is:

$$n_1 = \frac{\left(P + \frac{4T}{r_1}\right) \cdot \frac{4}{3}\pi r_1^3}{RT}$$

For the second bubble:

$$n_2 = \frac{\left(P + \frac{4T}{r_2}\right) \cdot \frac{4}{3}\pi r_2^3}{RT}$$

The total initial moles are:

$$n_1 + n_2 = \frac{\frac{4}{3}\pi}{RT} \left[ \left(P + \frac{4T}{r_1}\right) r_1^3 + \left(P + \frac{4T}{r_2}\right) r_2^3 \right]$$

Simplifying the terms inside the brackets:

$$\left(P + \frac{4T}{r_1}\right) r_1^3 = P r_1^3 + 4T r_1^2$$

$$\left(P + \frac{4T}{r_2}\right) r_2^3 = P r_2^3 + 4T r_2^2$$

So:

$$n_1 + n_2 = \frac{\frac{4}{3}\pi}{RT} \left[ P r_1^3 + 4T r_1^2 + P r_2^3 + 4T r_2^2 \right]$$

For the final single bubble:

$$n = \frac{\left(P + \frac{4T}{r}\right) \cdot \frac{4}{3}\pi r^3}{RT} = \frac{\frac{4}{3}\pi}{RT} \left[ P r^3 + 4T r^2 \right]$$

Since moles are conserved, $$n_1 + n_2 = n$$:

$$P r_1^3 + 4T r_1^2 + P r_2^3 + 4T r_2^2 = P r^3 + 4T r^2$$

Rearranging:

$$P r_1^3 + P r_2^3 - P r^3 + 4T r_1^2 + 4T r_2^2 - 4T r^2 = 0$$

$$P (r_1^3 + r_2^3 - r^3) + 4T (r_1^2 + r_2^2 - r^2) = 0 \quad ...(1)$$

Now, define the changes. The change in volume of contained air, $$V$$, is the final volume minus the initial total volume. The volume of a bubble is $$\frac{4}{3}\pi r^3$$, so:

$$V = \frac{4}{3}\pi r^3 - \left( \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 \right) = \frac{4}{3}\pi (r^3 - r_1^3 - r_2^3)$$

Solving for $$r^3 - r_1^3 - r_2^3$$:

$$r^3 - r_1^3 - r_2^3 = \frac{3V}{4\pi} \quad ...(2)$$

The change in total surface area, $$S$$, is defined as the final surface area minus the initial total surface area. For a soap bubble, the surface area is often considered as the area of one surface (e.g., the outer surface), which is $$4\pi r^2$$. Thus:

Initial total surface area = $$4\pi r_1^2 + 4\pi r_2^2 = 4\pi (r_1^2 + r_2^2)$$

Final surface area = $$4\pi r^2$$

So:

$$S = 4\pi r^2 - 4\pi (r_1^2 + r_2^2) = 4\pi (r^2 - r_1^2 - r_2^2)$$

Solving for $$r^2 - r_1^2 - r_2^2$$:

$$r^2 - r_1^2 - r_2^2 = \frac{S}{4\pi} \quad ...(3)$$

Note that equation (1) contains $$r_1^3 + r_2^3 - r^3$$ and $$r_1^2 + r_2^2 - r^2$$. Using equations (2) and (3):

$$r_1^3 + r_2^3 - r^3 = - (r^3 - r_1^3 - r_2^3) = - \frac{3V}{4\pi}$$

$$r_1^2 + r_2^2 - r^2 = - (r^2 - r_1^2 - r_2^2) = - \frac{S}{4\pi}$$

Substitute these into equation (1):

$$P \left( - \frac{3V}{4\pi} \right) + 4T \left( - \frac{S}{4\pi} \right) = 0$$

$$- \frac{3PV}{4\pi} - \frac{4T S}{4\pi} = 0$$

Simplify the second term:

$$- \frac{3PV}{4\pi} - \frac{T S}{\pi} = 0$$

Multiply both sides by $$-1$$:

$$\frac{3PV}{4\pi} + \frac{T S}{\pi} = 0$$

To eliminate denominators, multiply both sides by $$4\pi$$:

$$4\pi \cdot \frac{3PV}{4\pi} + 4\pi \cdot \frac{T S}{\pi} = 0$$

$$3PV + 4 T S = 0$$

Since $$T S = ST$$, we write:

$$3PV + 4ST = 0$$

This matches option B.

Hence, the correct answer is Option B.

The height of the water column in a capillary tube is given by the formula:

where $$ S $$ is the surface tension, $$ \theta $$ is the contact angle, $$ \rho $$ is the density of water, $$ g $$ is the acceleration due to gravity, and $$ r $$ is the radius of the capillary tube.

Since the capillary tube and the liquid (water) are the same in both scenarios, $$ S $$, $$ \theta $$, $$ \rho $$, and $$ r $$ are constant. Therefore, the height $$ h $$ is inversely proportional to $$ g $$:

At the Earth's surface, the height is $$ x $$, so:

In the mine at depth $$ d $$, the height is $$ y $$, so:

The ratio $$ \frac{x}{y} $$ is:

Now, we need to find how $$ g $$ changes with depth. The acceleration due to gravity at a depth $$ d $$ below the Earth's surface is given by:

where $$ R $$ is the radius of the Earth. This is because, at depth $$ d $$, only the mass of the Earth within the sphere of radius $$ R - d $$ contributes to gravity, and under the assumption of uniform density, $$ g $$ decreases linearly with depth.

Therefore, substituting this into the ratio:

So, the ratio $$ \frac{x}{y} $$ is:

Comparing with the options:

A. $$ \left(\frac{R-d}{R+d}\right) $$

B. $$ \left(\frac{R+d}{R-d}\right) $$

C. $$ \left(1 - \frac{2d}{R}\right) $$

D. $$ \left(1 - \frac{d}{R}\right) $$

Option D matches the derived expression.

Hence, the correct answer is Option D.

We are given that a large number of liquid drops, each of radius $$r$$, coalesce to form a single big drop of radius $$R$$. The energy released during coalescence is converted into the kinetic energy of the big drop. We need to find the speed of the big drop, given the surface tension $$T$$ and density $$\rho$$.

First, recall that the surface energy of a liquid drop is given by the surface tension multiplied by the surface area. For a single small drop of radius $$r$$, the surface area is $$4\pi r^2$$, so its surface energy is $$T \times 4\pi r^2$$.

Let the number of small drops be $$n$$. The volume of one small drop is $$\frac{4}{3}\pi r^3$$, and the volume of the big drop is $$\frac{4}{3}\pi R^3$$. Since volume is conserved during coalescence:

$$n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$

Canceling $$\frac{4}{3}\pi$$ from both sides:

$$n r^3 = R^3$$

Solving for $$n$$:

$$n = \frac{R^3}{r^3}$$

The total surface energy before coalescence is the sum of the surface energies of all small drops:

$$\text{Initial surface energy} = n \times T \times 4\pi r^2 = \frac{R^3}{r^3} \times T \times 4\pi r^2 = T \times 4\pi \frac{R^3}{r}$$

The surface energy after coalescence is that of the big drop:

$$\text{Final surface energy} = T \times 4\pi R^2$$

The energy released is the difference between the initial and final surface energies:

$$E_{\text{released}} = \text{Initial surface energy} - \text{Final surface energy} = T \times 4\pi \frac{R^3}{r} - T \times 4\pi R^2$$

Factor out $$T \times 4\pi$$:

$$E_{\text{released}} = T \times 4\pi \left( \frac{R^3}{r} - R^2 \right)$$

Factor $$R^2$$ from the expression inside the parentheses:

$$E_{\text{released}} = T \times 4\pi R^2 \left( \frac{R}{r} - 1 \right)$$

Note that $$\frac{R}{r} - 1 = \frac{R - r}{r}$$. Also, observe that $$\frac{1}{r} - \frac{1}{R} = \frac{R - r}{R r}$$. Therefore:

$$R - r = R r \left( \frac{1}{r} - \frac{1}{R} \right)$$

Substitute this into the expression:

$$E_{\text{released}} = T \times 4\pi R^2 \times \frac{R - r}{r} = T \times 4\pi R^2 \times \frac{R r \left( \frac{1}{r} - \frac{1}{R} \right)}{r}$$

Simplify by canceling $$r$$ in the numerator and denominator:

$$E_{\text{released}} = T \times 4\pi R^2 \times R \left( \frac{1}{r} - \frac{1}{R} \right) = T \times 4\pi R^3 \left( \frac{1}{r} - \frac{1}{R} \right)$$

This energy is converted into the kinetic energy of the big drop. The kinetic energy is given by $$\frac{1}{2} M v^2$$, where $$M$$ is the mass of the big drop and $$v$$ is its speed.

The mass $$M$$ is the density times volume:

$$M = \rho \times \frac{4}{3}\pi R^3$$

Set the energy released equal to the kinetic energy:

$$T \times 4\pi R^3 \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{1}{2} M v^2$$

Substitute $$M$$:

$$T \times 4\pi R^3 \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{1}{2} \times \rho \times \frac{4}{3}\pi R^3 \times v^2$$

Divide both sides by $$\pi R^3$$ (assuming $$R \neq 0$$):

$$T \times 4 \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{1}{2} \times \rho \times \frac{4}{3} \times v^2$$

Simplify the right side:

$$\frac{1}{2} \times \frac{4}{3} = \frac{4}{6} = \frac{2}{3}$$

So:

$$4T \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{2}{3} \rho v^2$$

Solve for $$v^2$$. Multiply both sides by $$\frac{3}{2}$$:

$$v^2 = 4T \left( \frac{1}{r} - \frac{1}{R} \right) \times \frac{3}{2} \times \frac{1}{\rho}$$

Simplify:

$$v^2 = \frac{4 \times 3}{2} \times \frac{T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{12}{2} \times \frac{T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right) = 6 \times \frac{T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)$$

Therefore:

$$v^2 = \frac{6T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)$$

Take the square root of both sides:

$$v = \sqrt{ \frac{6T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right) }$$

Comparing with the options, this matches option B.

Hence, the correct answer is Option B.

$$F_B = \text{Volume} \times \rho_w \times g = \frac{4}{3}\pi R^3 \rho_w g$$ (Buoyancy Force)

The surface tension of water pulls downward along the circular contact perimeter of radius $$r$$. From the geometry of the contact region, let $$\theta$$ be the angle that the bubble's surface makes with the vertical at the contact rim. For a spherical shape where $$r \ll R$$: $$\sin\theta \approx \frac{r}{R}$$

The downward component of the surface tension force acting along the circular perimeter of radius $$r$$ ($$2\pi r$$) is $$F_{ST} = T \cdot (2\pi r) \cdot \sin\theta$$

$$F_{ST} = 2\pi r T \left(\frac{r}{R}\right) = \frac{2\pi r^2 T}{R}$$

Just before detachment, the upward buoyancy force is exactly counterbalanced by the downward surface tension force holding the bubble to the bottom:

$$F_{ST} = F_B$$

$$\frac{2\pi r^2 T}{R} = \frac{4}{3}\pi R^3 \rho_w g$$

$$r^2 = \frac{4 R^4 \rho_w g}{6 T} = \frac{2 \rho_w g}{3T} R^4$$

$$r = R^2 \sqrt{\frac{2\rho_w g}{3T}}$$

We consider a spherical drop of liquid of radius $$r$$. Its surface energy is equal to the product of its surface tension and its surface area, while the energy required to vaporise a small mass of the liquid is obtained from the definition of latent heat. We wish to see whether the surface energy released when the drop shrinks by an infinitesimal amount can supply the latent heat needed for the corresponding infinitesimal mass to evaporate, all at the same temperature.

The surface area of a sphere is $$A = 4\pi r^{2}$$. If the radius decreases by an infinitesimal amount $$dr$$ (with $$dr<0$$ as the drop is getting smaller), the change in surface area is obtained by differentiating:

$$dA = \frac{d}{dr}(4\pi r^{2})\,dr = 8\pi r\,dr.$$

Surface energy equals $$T\times A$$. Therefore the change in surface energy is

$$dE_{\text{s}} = T\,dA = T\,(8\pi r\,dr)=8\pi rT\,dr.$$

(Because $$dr<0$$, $$dE_{\text{s}}$$ is negative, signifying that energy is released. We will shortly compare magnitudes, so the sign will not matter.)

Next, the volume of the drop is $$V = \dfrac{4}{3}\pi r^{3}$$. Its infinitesimal change is

$$dV = \frac{d}{dr}\!\left(\frac{4}{3}\pi r^{3}\right)\!dr = 4\pi r^{2}\,dr.$$

The mass corresponding to this change of volume is, using density $$\rho$$,

$$dm = \rho\,dV = \rho\,(4\pi r^{2}\,dr)=4\pi r^{2}\rho\,dr.$$

Latent heat of vaporisation $$L$$ is defined as the heat required per unit mass to convert liquid to vapour at constant temperature. Hence the heat needed to vaporise the mass $$dm$$ is

$$dQ = L\,dm = L\,(4\pi r^{2}\rho\,dr)=4\pi r^{2}\rho L\,dr.$$

For the evaporation to proceed solely at the expense of the drop’s surface energy, the magnitude of the surface energy released must at least equal the latent heat required, so we set

$$|dE_{\text{s}}| = dQ.$$

Using the expressions obtained, this becomes

$$8\pi rT\,|dr| = 4\pi r^{2}\rho L\,|dr|.$$

The common factors $$4\pi|dr|$$ cancel immediately, leaving

$$2T = r\rho L.$$

Solving for the radius, we get

$$r = \frac{2T}{\rho L}.$$

This value represents the minimum radius. For a drop of exactly this radius, the surface energy released during an infinitesimal contraction is just sufficient to supply the latent heat of the infinitesimal mass that evaporates. If the drop were any smaller, the released surface energy would be insufficient.

Hence, the correct answer is Option B.

Let us understand both statements step by step. We start by recalling the formula for the height $$h$$ to which a liquid rises in a capillary tube. The formula is:

$$ h = \frac{2S \cos \theta}{\rho g r} $$

where $$S$$ is the surface tension of the liquid, $$\theta$$ is the angle of contact, $$\rho$$ is the density of the liquid, $$g$$ is the acceleration due to gravity, and $$r$$ is the radius of the capillary tube.

Statement-1 claims that when the temperature of the liquid is raised, the height $$h$$ increases, assuming the density $$\rho$$ and the angle of contact $$\theta$$ remain constant.

Statement-2 states that the surface tension $$S$$ of a liquid decreases with a rise in temperature.

We know that Statement-2 is a well-established fact. The surface tension of liquids generally decreases as temperature increases because the increased thermal energy reduces the cohesive forces between molecules at the surface.

Now, let us analyze Statement-1 using the formula. Given that $$\rho$$, $$\theta$$, $$g$$, and $$r$$ are constant (as per the problem and since the capillary tube doesn't change), the height $$h$$ is directly proportional to the surface tension $$S$$. Therefore:

$$ h \propto S $$

If the temperature rises, Statement-2 tells us that $$S$$ decreases. Since $$h$$ is proportional to $$S$$, a decrease in $$S$$ would lead to a decrease in $$h$$. However, Statement-1 claims that $$h$$ increases with temperature. This is a contradiction.

Therefore, Statement-1 is false because the height $$h$$ should decrease, not increase, when temperature rises (given that $$\rho$$ and $$\theta$$ are constant).

In summary:

• Statement-1 is false.
• Statement-2 is true.

Hence, the correct answer is Option B.

First, recall the formula for the height $$ h $$ to which a liquid rises in a capillary tube:

Here, $$ S $$ is the surface tension of the liquid (water), $$ \theta $$ is the angle of contact, $$ \rho $$ is the density of water, $$ g $$ is the acceleration due to gravity, and $$ r $$ is the radius of the capillary tube. The radius $$ r $$, density $$ \rho $$, gravity $$ g $$, and surface tension $$ S $$ remain constant. Only $$ \theta $$ changes when the tube is waxed.

For an unwaxed glass capillary tube, water wets the surface, and the angle of contact $$ \theta $$ is acute (less than 90°). Typically, $$ \theta $$ is close to 0° for pure water and clean glass. Thus, $$ \cos \theta $$ is positive, and $$ h $$ is positive, meaning water rises in the tube.

When the inner wall is coated with wax, the surface becomes hydrophobic (water-repelling). Wax increases the angle of contact $$ \theta $$ because water does not wet the surface as well. For wax, $$ \theta $$ becomes obtuse (greater than 90°), often around 106° for paraffin wax. Therefore, $$ \theta $$ increases compared to the unwaxed tube.

Now, examine the effect on $$ h $$. Since $$ \theta $$ increases from less than 90° to more than 90°, $$ \cos \theta $$ changes:

• When $$ \theta < 90^\circ $$, $$ \cos \theta > 0 $$ (positive).
• When $$ \theta > 90^\circ $$, $$ \cos \theta < 0 $$ (negative).

Substituting into the formula:

For unwaxed tube: $$ h_{\text{unwaxed}} = \frac{2S \cos \theta_{\text{unwaxed}}}{\rho g r} > 0 $$ (since $$ \cos \theta_{\text{unwaxed}} > 0 $$).

For waxed tube: $$ h_{\text{waxed}} = \frac{2S \cos \theta_{\text{waxed}}}{\rho g r} < 0 $$ (since $$ \cos \theta_{\text{waxed}} < 0 $$).

A negative $$ h $$ means capillary depression occurs—water falls below the surrounding level instead of rising. The question asks for the height "up to which water rises." In the waxed tube, water does not rise; it depresses. Therefore, compared to the positive rise in the unwaxed tube, the height of rise $$ h $$ decreases (it becomes negative, so it is less than the positive value).

Thus:

• The angle of contact $$ \theta $$ increases (from acute to obtuse).
• The height $$ h $$ to which water rises decreases (from positive to negative).

Comparing with the options:

A. $$ \theta $$ increases and $$ h $$ increases → Incorrect, because $$ h $$ decreases.

B. $$ \theta $$ decreases and $$ h $$ decreases → Incorrect, because $$ \theta $$ increases.

C. $$ \theta $$ increases and $$ h $$ decreases → Correct.

D. $$ \theta $$ decreases and $$ h $$ increases → Incorrect, because $$ \theta $$ increases and $$ h $$ decreases.

Hence, the correct answer is Option C.