Mechanical Properties of Fluids

(a) Blood pressure greater at feet than at brain:

The pressure in a fluid at rest increases with depth according to $P = P_0 + \rho g h$. The feet are at a much greater vertical distance (height $h$) below the heart compared to the brain. Therefore, the blood pressure at the feet is greater than at the brain by an amount $\rho g h$, where $h$ is the height difference between the brain and the feet.

(b) Atmospheric pressure halves at ~6 km:

The atmosphere is not a liquid of uniform density. The density of air decreases rapidly with altitude. Most of the mass of the atmosphere is concentrated in the lower layers (near the Earth's surface). Since pressure at any point equals the weight of the air column above it per unit area, and since most of the atmospheric mass lies below 6 km, the pressure at 6 km is nearly half the sea-level pressure. The upper layers (above 6 km) are very thin and contribute little to the total pressure.

(c) Hydrostatic pressure is a scalar:

Although pressure is defined as force per unit area ($P = F/A$), the force here is the normal force exerted by the fluid on a surface. When we say pressure at a point in a fluid, it acts equally in all directions (Pascal's law). There is no unique direction associated with pressure at a point — it has the same magnitude regardless of the orientation of the surface element chosen. Hence, hydrostatic pressure is a scalar quantity.

(a) Angle of contact — mercury (obtuse) vs water (acute):

The angle of contact depends on the relative magnitudes of cohesive forces (between liquid molecules) and adhesive forces (between liquid and solid molecules).
- For mercury–glass: Cohesive forces between mercury molecules are much stronger than the adhesive forces between mercury and glass. The mercury surface curves upward (convex meniscus), giving an obtuse angle of contact (> 90°).
- For water–glass: Adhesive forces between water and glass are stronger than the cohesive forces between water molecules. The water surface curves upward at the edges (concave meniscus), giving an acute angle of contact (< 90°).

(b) Water spreads on glass; mercury forms drops:

- Water on glass: Adhesive force (water–glass) > Cohesive force (water–water). Water molecules are attracted more to glass than to each other, so water spreads out to maximise contact with glass (wets glass).
- Mercury on glass: Cohesive force (mercury–mercury) > Adhesive force (mercury–glass). Mercury molecules prefer to stay together, so mercury pulls itself into drops to minimise contact with glass (does not wet glass).

(c) Surface tension is independent of area of the surface:

Surface tension is a property of the liquid–interface system at a given temperature. It is defined as the force per unit length (or surface energy per unit area) and arises from the intermolecular forces at the surface. These intermolecular forces are intrinsic to the liquid and do not depend on how large the surface is. Hence, surface tension is independent of the area of the surface.

(d) Detergent solution has small angle of contact:

Detergent molecules are surfactants — they have one hydrophilic (water-loving) end and one hydrophobic end. When dissolved in water, they reduce the surface tension of water significantly. A lower surface tension means the adhesive forces between the solution and the solid surface become relatively more dominant, reducing the angle of contact. A smaller angle of contact means better wetting, which is why detergents help in cleaning.

(e) A liquid drop is always spherical under no external forces:

In the absence of external forces, a liquid drop tends to minimise its surface energy (since surface energy = surface tension × area). For a given volume, the shape with the minimum surface area is a sphere. Therefore, surface tension causes the drop to contract to a spherical shape to minimise surface energy.

(a) Surface tension of liquids generally decreases with temperature.

*Reason*: As temperature increases, the intermolecular forces weaken, reducing surface tension.

(b) Viscosity of gases increases with temperature, whereas viscosity of liquids decreases with temperature.

*Reason*: In gases, viscosity arises from molecular collisions; more collisions at higher temperature → higher viscosity. In liquids, viscosity arises from intermolecular cohesion; higher temperature weakens cohesion → lower viscosity.

(c) For solids with elastic modulus of rigidity, the shearing force is proportional to shear strain, while for fluids it is proportional to rate of shear strain.

*Reason*: Solids resist deformation (Hooke's law: stress ∝ strain). Fluids continuously deform under shear stress, so stress ∝ rate of strain (Newton's law of viscosity).

(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows conservation of mass (equation of continuity: $Av = \text{constant}$).

*Reason*: The continuity equation $A_1v_1 = A_2v_2$ directly gives the increase in speed at a constriction. Bernoulli's principle then explains the pressure change.

(e) For the model of a plane in a wind tunnel, turbulence occurs at a greater speed for turbulence for an actual plane.

*Reason*: The Reynolds number $Re = \rho v L/\eta$ determines onset of turbulence. The model is smaller (smaller $L$), so to achieve the same $Re$ (and hence same flow conditions), the speed $v$ must be greater for the model.

(a) Blowing over the paper keeps it horizontal:

By Bernoulli's principle, $P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$. When you blow over the paper, the air speed above the paper increases, so the pressure above decreases. The pressure below (atmospheric) is now greater than the pressure above. This net upward pressure difference provides an upward force that keeps the paper horizontal. If you blow under, the pressure below decreases and the paper droops.

(b) Fast jets through finger gaps:

When we partially close the tap with fingers, we reduce the area of cross-section of the opening. By the equation of continuity ($Av = \text{constant}$), as the area decreases, the velocity of water increases. Hence fast jets gush through the narrow openings between the fingers.

(c) Needle size controls flow rate better than thumb pressure:

By Poiseuille's law, the volume flow rate through a tube is:
$$Q = \frac{\pi r^4 \Delta P}{8 \eta l}$$
The flow rate depends on the fourth power of the radius $r$ of the needle. A small change in the needle's radius produces a very large change in flow rate ($Q \propto r^4$), whereas the flow rate depends only linearly on the pressure difference $\Delta P$. Therefore, the needle size (radius) is a far more sensitive control of flow rate than the thumb pressure.

(d) Backward thrust on the vessel:

This is a consequence of conservation of momentum (Newton's third law). When fluid flows out through a hole with momentum in the forward direction, by Newton's third law, an equal and opposite reaction force (backward thrust) acts on the vessel. This is the same principle as rocket propulsion.

(e) Spinning cricket ball does not follow parabolic trajectory:

A spinning cricket ball experiences the Magnus effect. Due to spin, the ball drags air around it. On one side, the spin-induced air velocity adds to the wind velocity (higher speed, lower pressure by Bernoulli's principle), and on the other side it opposes (lower speed, higher pressure). This pressure difference creates a net sideways (or upward/downward) force called the Magnus force, which deflects the ball from its parabolic trajectory. Hence the ball swings or curves in the air.

Given:
- Mass of girl, $m = 50\ \text{kg}$
- Diameter of heel, $d = 1.0\ \text{cm} = 1.0 \times 10^{-2}\ \text{m}$
- Radius of heel, $r = 0.5\ \text{cm} = 0.5 \times 10^{-2}\ \text{m}$
- $g = 9.8\ \text{m s}^{-2}$

Force exerted by heel on floor:
$$F = mg = 50 \times 9.8 = 490\ \text{N}$$

Area of heel (circular):
$$A = \pi r^2 = \pi \times (0.5 \times 10^{-2})^2 = \pi \times 0.25 \times 10^{-4}\ \text{m}^2$$
$$A = 7.854 \times 10^{-5}\ \text{m}^2$$

Pressure:
$$P = \frac{F}{A} = \frac{490}{7.854 \times 10^{-5}}$$
$$\boxed{P \approx 6.24 \times 10^{6}\ \text{Pa}}$$

This is about 60 times the atmospheric pressure — very high pressure due to the small area of the heel.

Given:
- Density of wine, $\rho_{\text{wine}} = 984\ \text{kg m}^{-3}$
- Normal atmospheric pressure, $P_0 = 1.01 \times 10^5\ \text{Pa}$
- $g = 9.8\ \text{m s}^{-2}$

Concept: At the base of the barometer column, the pressure equals atmospheric pressure:
$$P_0 = \rho g h$$

Solving for height $h$:
$$h = \frac{P_0}{\rho_{\text{wine}} \cdot g} = \frac{1.01 \times 10^5}{984 \times 9.8}$$
$$h = \frac{1.01 \times 10^5}{9643.2}$$
$$\boxed{h \approx 10.5\ \text{m}}$$

This is much taller than the mercury barometer (~0.76 m) because wine is much less dense than mercury ($\rho_{\text{Hg}} = 13,600\ \text{kg m}^{-3}$).

Given:
- Maximum stress the structure can withstand: $P_{\text{max}} = 10^9\ \text{Pa}$
- Depth of ocean: $h = 3\ \text{km} = 3 \times 10^3\ \text{m}$
- Density of sea water: $\rho \approx 10^3\ \text{kg m}^{-3}$
- $g = 9.8\ \text{m s}^{-2}$
- Atmospheric pressure: $P_0 = 1.01 \times 10^5\ \text{Pa}$

Pressure at depth $h$:
$$P = P_0 + \rho g h$$
$$P = 1.01 \times 10^5 + (10^3)(9.8)(3 \times 10^3)$$
$$P = 1.01 \times 10^5 + 2.94 \times 10^7$$
$$P \approx 2.94 \times 10^7\ \text{Pa} \approx 3 \times 10^7\ \text{Pa}$$

Comparison:
$$P \approx 3 \times 10^7\ \text{Pa} \ll 10^9\ \text{Pa} = P_{\text{max}}$$

Conclusion: The pressure at the ocean depth of 3 km is approximately $3 \times 10^7$ Pa, which is well below the maximum stress of $10^9$ Pa that the structure can withstand. Therefore, yes, the structure is suitable for the oil well.

Given:
- Maximum mass of car, $m = 3000\ \text{kg}$
- Area of larger piston, $A = 425\ \text{cm}^2 = 425 \times 10^{-4}\ \text{m}^2$
- $g = 9.8\ \text{m s}^{-2}$

Concept: By Pascal's law, the pressure transmitted through the hydraulic fluid is the same everywhere. The pressure on the larger piston equals the weight of the car divided by its area.

Force on larger piston:
$$F = mg = 3000 \times 9.8 = 2.94 \times 10^4\ \text{N}$$

Pressure on larger piston (= pressure on smaller piston):
$$P = \frac{F}{A} = \frac{2.94 \times 10^4}{425 \times 10^{-4}}$$
$$P = \frac{2.94 \times 10^4}{4.25 \times 10^{-2}}$$
$$\boxed{P \approx 6.92 \times 10^5\ \text{Pa}}$$

The smaller piston must bear a maximum pressure of approximately $6.92 \times 10^5$ Pa.

Given:
- Height of water column, $h_w = 10.0\ \text{cm}$
- Height of spirit column, $h_s = 12.5\ \text{cm}$
- Mercury levels are equal in both arms.
- Density of water, $\rho_w = 1.0\ \text{g cm}^{-3}$
- Let density of spirit = $\rho_s$

Concept: Since the mercury levels are the same in both arms, the pressure exerted by the water column on one side must equal the pressure exerted by the spirit column on the other side at the mercury surface level.

Equating pressures at the mercury level:
$$\rho_w \cdot g \cdot h_w = \rho_s \cdot g \cdot h_s$$

$$\rho_s = \rho_w \times \frac{h_w}{h_s} = 1.0 \times \frac{10.0}{12.5}$$

$$\rho_s = 0.8\ \text{g cm}^{-3}$$

Specific gravity of spirit:
$$\text{Specific gravity} = \frac{\rho_s}{\rho_w} = \frac{0.8}{1.0} = \boxed{0.8}$$

Given (from previous problem):
- Specific gravity of spirit, $\rho_s = 0.8\ \text{g cm}^{-3}$
- Specific gravity of mercury, $\rho_{Hg} = 13.6\ \text{g cm}^{-3}$
- Additional water poured = 15.0 cm
- Additional spirit poured = 15.0 cm

Setup: Let the mercury level in the water arm fall by $x$ cm. Then the mercury level in the spirit arm rises by $x$ cm (mercury is incompressible, so the total mercury volume is conserved).

Total height of water column (original 10 cm + 15 cm added + mercury dropped by $x$):
$$h_w^{\text{total}} = 10 + 15 + x = (25 + x)\ \text{cm}$$

Total height of spirit column (original 12.5 cm + 15 cm added + mercury rose by $x$, so spirit column effectively loses $x$ from below... wait — let me reconsider):

Actually, when mercury in the water arm falls by $x$, the water column above it increases by $x$. When mercury in the spirit arm rises by $x$, the spirit column above it decreases by $x$ (spirit is pushed up).

Wait — the spirit column height above mercury = $12.5 + 15 - x = (27.5 - x)$ cm.
The water column height above mercury = $10 + 15 + x = (25 + x)$ cm.

Equating pressures at the lower mercury level (in the water arm):

Pressure at the bottom of the water arm mercury surface = Pressure at the bottom of the spirit arm mercury surface.

Let the mercury level difference be $2x$ (water arm mercury is lower by $x$, spirit arm is higher by $x$).

Pressure balance at the lower mercury level (water side):
$$\rho_w g (25 + x) + \rho_{Hg} g (2x) = \rho_s g (27.5 - x) + \rho_{Hg} g (2x)$$

Wait, let me redo this carefully.

Let mercury in the water arm go down by $x$ cm and mercury in the spirit arm go up by $x$ cm. The difference in mercury levels = $2x$.

Pressure at the bottom of the U-tube must be equal from both sides. Taking the reference at the lower mercury surface (water arm side):

Left side (water arm): Pressure = $\rho_w g (25 + x) + P_{\text{atm}}$

Right side (spirit arm): Pressure = $\rho_s g (27.5 - x) + \rho_{Hg} g (2x) + P_{\text{atm}}$

Equating:
$$\rho_w (25 + x) = \rho_s (27.5 - x) + \rho_{Hg}(2x)$$

Substituting $\rho_w = 1$, $\rho_s = 0.8$, $\rho_{Hg} = 13.6$ (in g/cm³, $g$ cancels):
$$1 \times (25 + x) = 0.8 \times (27.5 - x) + 13.6 \times (2x)$$
$$25 + x = 22 - 0.8x + 27.2x$$
$$25 + x = 22 + 26.4x$$
$$25 - 22 = 26.4x - x$$
$$3 = 25.4x$$
$$x = \frac{3}{25.4} \approx 0.118\ \text{cm}$$

Difference in mercury levels:
$$2x = 2 \times 0.118 \approx 0.236\ \text{cm}$$

$$\boxed{\Delta h = 2x \approx 0.24\ \text{cm}}$$

The mercury level in the spirit arm is higher than in the water arm by approximately 0.24 cm.

Answer: No, Bernoulli's equation cannot be used to describe the flow of water through a rapid in a river.

Reason:
Bernoulli's equation is valid only under the following conditions:
1. The flow must be steady (streamlined/laminar).
2. The fluid must be non-viscous (ideal fluid).
3. The fluid must be incompressible.
4. The flow must be along a streamline.

In a river rapid, the flow is highly turbulent — the water moves in irregular, chaotic patterns with eddies and whirlpools. The flow is neither steady nor along streamlines. Also, viscous effects and energy dissipation are significant in turbulent flow.

Therefore, Bernoulli's equation, which is based on conservation of energy for steady, non-viscous, incompressible flow, cannot be applied to describe the flow of water through a rapid.

Answer: No, it does not matter whether one uses gauge pressure or absolute pressure in Bernoulli's equation, as long as the same reference is used consistently throughout.

Explanation:

Bernoulli's equation is:
$$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$$

Gauge pressure is defined as: $P_{\text{gauge}} = P_{\text{absolute}} - P_{\text{atm}}$

If we substitute $P = P_{\text{gauge}} + P_{\text{atm}}$ at both points:
$$(P_{\text{gauge},1} + P_{\text{atm}}) + \frac{1}{2}\rho v_1^2 + \rho g h_1 = (P_{\text{gauge},2} + P_{\text{atm}}) + \frac{1}{2}\rho v_2^2 + \rho g h_2$$

The $P_{\text{atm}}$ cancels from both sides:
$$P_{\text{gauge},1} + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_{\text{gauge},2} + \frac{1}{2}\rho v_2^2 + \rho g h_2$$

Since $P_{\text{atm}}$ is a constant and appears on both sides, it cancels out. Therefore, using gauge pressures gives the same result as using absolute pressures. It does not matter, provided the same type of pressure is used at both points.

Given:
- Length of tube, $l = 1.5\ \text{m}$
- Radius of tube, $r = 1.0\ \text{cm} = 1.0 \times 10^{-2}\ \text{m}$
- Mass flow rate, $\dot{m} = 4.0 \times 10^{-3}\ \text{kg s}^{-1}$
- Density of glycerine, $\rho = 1.3 \times 10^3\ \text{kg m}^{-3}$
- Viscosity of glycerine, $\eta = 0.83\ \text{Pa s}$

Step 1: Find volume flow rate $Q$:
$$Q = \frac{\dot{m}}{\rho} = \frac{4.0 \times 10^{-3}}{1.3 \times 10^3} = 3.077 \times 10^{-6}\ \text{m}^3\text{s}^{-1}$$

Step 2: Apply Poiseuille's formula:
$$Q = \frac{\pi r^4 \Delta P}{8 \eta l}$$

Solving for $\Delta P$:
$$\Delta P = \frac{8 \eta l Q}{\pi r^4}$$

$$\Delta P = \frac{8 \times 0.83 \times 1.5 \times 3.077 \times 10^{-6}}{\pi \times (1.0 \times 10^{-2})^4}$$

$$\Delta P = \frac{8 \times 0.83 \times 1.5 \times 3.077 \times 10^{-6}}{\pi \times 10^{-8}}$$

Numerator: $8 \times 0.83 \times 1.5 \times 3.077 \times 10^{-6} = 8 \times 3.831 \times 10^{-6} = 3.065 \times 10^{-5}$

Denominator: $\pi \times 10^{-8} = 3.1416 \times 10^{-8}$

$$\Delta P = \frac{3.065 \times 10^{-5}}{3.1416 \times 10^{-8}} \approx 975.6\ \text{Pa}$$

$$\boxed{\Delta P \approx 9.76 \times 10^2\ \text{Pa}}$$

Check for laminar flow (Reynolds number):

Average velocity: $v = \frac{Q}{\pi r^2} = \frac{3.077 \times 10^{-6}}{\pi \times (10^{-2})^2} = \frac{3.077 \times 10^{-6}}{3.1416 \times 10^{-4}} \approx 9.8 \times 10^{-3}\ \text{m s}^{-1}$

$$Re = \frac{\rho v d}{\eta} = \frac{1.3 \times 10^3 \times 9.8 \times 10^{-3} \times 2 \times 10^{-2}}{0.83}$$
$$Re = \frac{1.3 \times 10^3 \times 9.8 \times 10^{-3} \times 2 \times 10^{-2}}{0.83} = \frac{0.2548}{0.83} \approx 0.3$$

Since $Re \approx 0.3 \ll 2000$, the flow is well within the laminar regime. The assumption is correct.

Given:
- Speed on upper surface, $v_1 = 70\ \text{m s}^{-1}$
- Speed on lower surface, $v_2 = 63\ \text{m s}^{-1}$
- Area of wing, $A = 2.5\ \text{m}^2$
- Density of air, $\rho = 1.3\ \text{kg m}^{-3}$

Concept: By Bernoulli's principle (at the same height for both surfaces):
$$P_2 - P_1 = \frac{1}{2}\rho(v_1^2 - v_2^2)$$

where $P_1$ is pressure on upper surface and $P_2$ is pressure on lower surface.

Pressure difference:
$$\Delta P = P_2 - P_1 = \frac{1}{2} \times 1.3 \times (70^2 - 63^2)$$
$$= \frac{1}{2} \times 1.3 \times (4900 - 3969)$$
$$= \frac{1}{2} \times 1.3 \times 931$$
$$= 0.65 \times 931 = 605.15\ \text{Pa}$$

Lift force:
$$F = \Delta P \times A = 605.15 \times 2.5$$
$$\boxed{F \approx 1512.9\ \text{N} \approx 1.51 \times 10^3\ \text{N}}$$

The lift on the wing is approximately 1513 N.

Answer: Figure (a) is incorrect.

Reasoning:

In steady flow of a non-viscous, incompressible liquid, the equation of continuity states:
$$Av = \text{constant}$$

At a constriction (narrower cross-section), the velocity of flow increases.

By Bernoulli's principle:
$$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$$

At the same height, if velocity increases, pressure must decrease.

In figure (a): The streamlines at the constriction are shown wider apart (suggesting lower velocity) and the pressure gauge shows higher pressure at the constriction. This contradicts both the continuity equation (velocity should be higher at constriction) and Bernoulli's principle (pressure should be lower where velocity is higher).

In figure (b): The streamlines are closer together at the constriction (indicating higher velocity) and the pressure is lower there — consistent with Bernoulli's principle.

Therefore, figure (a) is incorrect because it shows increased pressure at the constriction, whereas the pressure should decrease at the constriction where flow speed increases.

Given:
- Cross-sectional area of tube, $A_1 = 8.0\ \text{cm}^2 = 8.0 \times 10^{-4}\ \text{m}^2$
- Number of holes, $n = 40$
- Diameter of each hole, $d = 1.0\ \text{mm} = 1.0 \times 10^{-3}\ \text{m}$
- Radius of each hole, $r = 0.5 \times 10^{-3}\ \text{m}$
- Speed of liquid in tube, $v_1 = 1.5\ \text{m min}^{-1} = \frac{1.5}{60}\ \text{m s}^{-1} = 0.025\ \text{m s}^{-1}$

Area of each hole:
$$a = \pi r^2 = \pi \times (0.5 \times 10^{-3})^2 = \pi \times 0.25 \times 10^{-6}\ \text{m}^2$$
$$a = 7.854 \times 10^{-7}\ \text{m}^2$$

Total area of all 40 holes:
$$A_2 = 40 \times a = 40 \times 7.854 \times 10^{-7} = 3.1416 \times 10^{-5}\ \text{m}^2$$

By equation of continuity:
$$A_1 v_1 = A_2 v_2$$
$$v_2 = \frac{A_1 v_1}{A_2} = \frac{8.0 \times 10^{-4} \times 0.025}{3.1416 \times 10^{-5}}$$
$$v_2 = \frac{2.0 \times 10^{-5}}{3.1416 \times 10^{-5}}$$
$$\boxed{v_2 \approx 0.637\ \text{m s}^{-1}}$$

The speed of ejection of liquid through the holes is approximately 0.64 m s⁻¹.

Given:
- Weight supported by the film, $W = 1.5 \times 10^{-2}\ \text{N}$
- Length of slider, $l = 30\ \text{cm} = 0.30\ \text{m}$

Concept: A soap film has two surfaces. The surface tension force acts along both surfaces of the film.

Total length of contact = $2l$ (two surfaces)

Force due to surface tension:
$$F = S \times 2l$$

At equilibrium, this force balances the weight:
$$S \times 2l = W$$

Solving for surface tension $S$:
$$S = \frac{W}{2l} = \frac{1.5 \times 10^{-2}}{2 \times 0.30}$$
$$S = \frac{1.5 \times 10^{-2}}{0.60}$$
$$\boxed{S = 2.5 \times 10^{-2}\ \text{N m}^{-1}}$$

The surface tension of the soap film is $2.5 \times 10^{-2}$ N m⁻¹.

Given:
- Weight supported by film in figure (a), $W_a = 4.5 \times 10^{-2}\ \text{N}$
- Same liquid, same temperature in figures (b) and (c).

Key Concept: Surface tension $S$ is a property of the liquid at a given temperature and is independent of the area or shape of the surface. The surface tension (force per unit length) remains the same.

The weight supported by the film depends on the surface tension and the length of contact:
$$W = S \times 2l$$

Since the same liquid is used at the same temperature, $S$ is the same. If the length of the slider/contact line $l$ is the same in all three figures (which is the case as shown in the figures — all have the same length of contact), then:

$$W_b = W_c = W_a = 4.5 \times 10^{-2}\ \text{N}$$

Physical Explanation:
Surface tension is an intrinsic property of the liquid surface — it does not depend on the area of the film or its shape. The restoring force due to surface tension depends only on the length of the contact line (perimeter) and the surface tension coefficient. Since the contact length is the same in all three cases and the liquid and temperature are identical, the surface tension force (and hence the weight supported) is the same in all three figures.

$$\boxed{W_b = W_c = 4.5 \times 10^{-2}\ \text{N}}$$

Given:
- Radius of mercury drop, $r = 3.00\ \text{mm} = 3.00 \times 10^{-3}\ \text{m}$
- Surface tension of mercury, $S = 4.65 \times 10^{-1}\ \text{N m}^{-1}$
- Atmospheric pressure, $P_0 = 1.01 \times 10^5\ \text{Pa}$

Excess pressure inside a liquid drop:

For a liquid drop (one surface), the excess pressure is:
$$\Delta P = \frac{2S}{r}$$

$$\Delta P = \frac{2 \times 4.65 \times 10^{-1}}{3.00 \times 10^{-3}}$$
$$\Delta P = \frac{9.30 \times 10^{-1}}{3.00 \times 10^{-3}}$$
$$\boxed{\Delta P = 310\ \text{Pa}}$$

Total pressure inside the drop:
$$P_{\text{inside}} = P_0 + \Delta P = 1.01 \times 10^5 + 310$$
$$P_{\text{inside}} = 1.01 \times 10^5 + 0.00310 \times 10^5$$
$$\boxed{P_{\text{inside}} \approx 1.0131 \times 10^5\ \text{Pa}}$$

Summary:
- Excess pressure inside the drop = 310 Pa
- Total pressure inside the drop ≈ 1.0131 × 10⁵ Pa

Given:
- Radius of bubble, $r = 5.00\ \text{mm} = 5.00 \times 10^{-3}\ \text{m}$
- Surface tension of soap solution, $S = 2.50 \times 10^{-2}\ \text{N m}^{-1}$
- Depth of air bubble, $h = 40.0\ \text{cm} = 0.40\ \text{m}$
- Relative density of soap solution = 1.20, so $\rho = 1.20 \times 10^3\ \text{kg m}^{-3}$
- Atmospheric pressure, $P_0 = 1.01 \times 10^5\ \text{Pa}$
- $g = 9.8\ \text{m s}^{-2}$

---

Part 1: Excess pressure inside a soap bubble:

A soap bubble has two liquid surfaces (inner and outer), so:
$$\Delta P = \frac{4S}{r} = \frac{4 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}}$$
$$\Delta P = \frac{10^{-1}}{5.00 \times 10^{-3}} = 20\ \text{Pa}$$

$$\boxed{\Delta P_{\text{soap bubble}} = 20\ \text{Pa}}$$

---

Part 2: Pressure inside an air bubble at depth 40 cm in soap solution:

An air bubble in a liquid has only one surface, so:
$$\Delta P_{\text{air bubble}} = \frac{2S}{r} = \frac{2 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}} = \frac{5.00 \times 10^{-2}}{5.00 \times 10^{-3}} = 10\ \text{Pa}$$

Pressure outside the air bubble (at depth $h = 0.40$ m):
$$P_{\text{outside}} = P_0 + \rho g h$$
$$= 1.01 \times 10^5 + 1.20 \times 10^3 \times 9.8 \times 0.40$$
$$= 1.01 \times 10^5 + 4704$$
$$= 1.01 \times 10^5 + 0.04704 \times 10^4$$
$$= 1.01 \times 10^5 + 4.704 \times 10^3$$
$$\approx 1.0571 \times 10^5\ \text{Pa}$$

Pressure inside the air bubble:
$$P_{\text{inside}} = P_{\text{outside}} + \Delta P = 1.0571 \times 10^5 + 10$$
$$\boxed{P_{\text{inside}} \approx 1.06 \times 10^5\ \text{Pa}}$$