Chapter 9: Mechanical Properties of Fluids

The total weight of the upper body is supported by the two femurs taken together.

Total cross-sectional area of the two femurs:

$$A = 2 \times 10 \, \mathrm{cm^2} = 20 \, \mathrm{cm^2} = 20 \times 10^{-4} \, \mathrm{m^2} = 2 \times 10^{-3} \, \mathrm{m^2}$$

The downward force on the bones is the weight of the upper body:

$$F = M g = 40 \times 9.8 = 392 \, \mathrm{N}$$

The average pressure sustained by the femurs is therefore

$$P = \frac{F}{A} = \frac{392}{2 \times 10^{-3}} = 1.96 \times 10^{5} \, \mathrm{Pa}$$

This pressure is about twice the atmospheric pressure — much smaller than the breaking stress of bone (~ $$10^{8} \, \mathrm{Pa}$$), so the femurs can comfortably bear it.

The total (absolute) pressure on the swimmer at depth $$h$$ is the sum of the atmospheric pressure $$P_a$$ acting on the lake surface and the hydrostatic pressure of the water column above her:

$$P = P_a + \rho g h$$

Substituting $$P_a = 1.01 \times 10^{5} \, \mathrm{Pa}$$, $$\rho = 1000 \, \mathrm{kg \, m^{-3}}$$, $$g = 9.8 \, \mathrm{m \, s^{-2}}$$ and $$h = 10 \, \mathrm{m}$$:

$$P = 1.01 \times 10^{5} + 1000 \times 9.8 \times 10$$

$$P = 1.01 \times 10^{5} + 0.98 \times 10^{5} = 1.99 \times 10^{5} \, \mathrm{Pa}$$

This is roughly twice the atmospheric pressure on the surface. At $$h = 10 \, \mathrm{m}$$, the pressure on the swimmer's eardrums is about $$2 \, \mathrm{atm}$$.

If the air had a uniform density $$\rho$$ from the ground up to a height $$h$$, the atmospheric pressure at the ground would simply be the weight per unit area of this air column:

$$P_a = \rho g h$$

Rearranging for the height:

$$h = \frac{P_a}{\rho g}$$

Putting $$P_a = 1.013 \times 10^{5} \, \mathrm{Pa}$$, $$\rho = 1.29 \, \mathrm{kg \, m^{-3}}$$ and $$g = 9.8 \, \mathrm{m \, s^{-2}}$$:

$$h = \frac{1.013 \times 10^{5}}{1.29 \times 9.8} = \frac{1.013 \times 10^{5}}{12.642} \approx 8.01 \times 10^{3} \, \mathrm{m}$$

So under the (false) assumption of constant density, the atmosphere would end at about $$8 \, \mathrm{km}$$. In reality the density of air decreases with altitude (in fact exponentially), so the atmosphere extends far beyond this. The simple estimate above is roughly the “scale height” over which the pressure falls by a factor of $$1/e$$.

Given: $$h = 1000 \, \mathrm{m}$$, $$\rho = 1.03 \times 10^{3} \, \mathrm{kg \, m^{-3}}$$, $$g = 10 \, \mathrm{m \, s^{-2}}$$, $$P_a = 1.01 \times 10^{5} \, \mathrm{Pa}$$.

(a) Absolute pressure.

$$P = P_a + \rho g h$$

$$P = 1.01 \times 10^{5} + (1.03 \times 10^{3})(10)(1000)$$

$$P = 1.01 \times 10^{5} + 1.03 \times 10^{7}$$

$$P = 0.0101 \times 10^{7} + 1.03 \times 10^{7} = 1.04 \times 10^{7} \, \mathrm{Pa}$$

$$P \approx 104 \, \mathrm{atm}$$

(b) Gauge pressure.

The gauge pressure is the pressure in excess of the atmospheric pressure:

$$P_g = P - P_a = \rho g h = 1.03 \times 10^{7} \, \mathrm{Pa} \approx 103 \, \mathrm{atm}$$

(c) Force on the window.

The window experiences pressure $$P$$ from outside (sea) and $$P_a$$ from inside (cabin). The net inward pressure equals the gauge pressure $$P_g$$.

$$A = 20 \, \mathrm{cm} \times 20 \, \mathrm{cm} = 400 \, \mathrm{cm^2} = 4 \times 10^{-2} \, \mathrm{m^2}$$

$$F = P_g \, A = (1.03 \times 10^{7})(4 \times 10^{-2})$$

$$F = 4.12 \times 10^{5} \, \mathrm{N}$$

The submarine's window must withstand a force of about $$4 \times 10^{5} \, \mathrm{N}$$ from the surrounding sea water.

Let the radii of the small and large pistons be $$r_1 = 0.5 \, \mathrm{cm}$$ and $$r_2 = 1.5 \, \mathrm{cm}$$. Their cross-sectional areas are

$$A_1 = \pi r_1^2, \quad A_2 = \pi r_2^2, \quad \frac{A_2}{A_1} = \left(\frac{r_2}{r_1}\right)^2 = \left(\frac{1.5}{0.5}\right)^2 = 9$$

(a) Force on the larger piston.

By Pascal's law the pressure transmitted through the water is the same at both pistons:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies F_2 = F_1 \cdot \frac{A_2}{A_1}$$

$$F_2 = 10 \times 9 = 90 \, \mathrm{N}$$

(b) Distance moved by the larger piston.

Water is incompressible, so the volume swept by the small piston equals the volume swept by the large piston:

$$A_1 \, L_1 = A_2 \, L_2 \implies L_2 = L_1 \cdot \frac{A_1}{A_2}$$

$$L_2 = 6.0 \times \frac{1}{9} = 0.67 \, \mathrm{cm}$$

Atmospheric pressure acts equally on both pistons and so does not enter the balance.

Let the small and large piston radii be $$r_1 = 5.0 \, \mathrm{cm} = 0.05 \, \mathrm{m}$$ and $$r_2 = 15 \, \mathrm{cm} = 0.15 \, \mathrm{m}$$.

Pascal's law gives equal pressure on both pistons:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$

The large piston has to support the car's weight, so $$F_2 = M g$$:

$$F_1 = M g \cdot \frac{A_1}{A_2} = M g \cdot \left(\frac{r_1}{r_2}\right)^2$$

$$F_1 = 1350 \times 9.8 \times \left(\frac{0.05}{0.15}\right)^2 = 13230 \times \frac{1}{9}$$

$$F_1 \approx 1470 \, \mathrm{N}$$

The pressure required is

$$P = \frac{F_1}{A_1} = \frac{F_1}{\pi r_1^2} = \frac{1470}{\pi (0.05)^2} = \frac{1470}{7.854 \times 10^{-3}}$$

$$P \approx 1.87 \times 10^{5} \, \mathrm{Pa} \approx 1.9 \times 10^{5} \, \mathrm{Pa}$$

This is about $$1.9 \, \mathrm{atm}$$ — easily produced by an ordinary air compressor.

(a) Pressure difference.

In steady level flight the total upward lift on the wings balances the weight of the aircraft:

$$(P_1 - P_2) A = M g$$

where $$P_1$$ and $$P_2$$ are pressures on the lower and upper surfaces of the wing.

$$\Delta P = \frac{M g}{A} = \frac{3.3 \times 10^{5} \times 9.8}{500}$$

$$\Delta P = \frac{3.234 \times 10^{6}}{500} \approx 6.47 \times 10^{3} \, \mathrm{Pa}$$

(b) Fractional speed increase on top.

Apply Bernoulli's principle at the two surfaces (height difference of the wing is negligible):

$$P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2$$

$$P_1 - P_2 = \tfrac{1}{2}\rho (v_2^2 - v_1^2) = \tfrac{1}{2}\rho (v_2 - v_1)(v_2 + v_1)$$

The plane's airspeed is $$v = 960 \, \mathrm{km/h} = \dfrac{960 \times 1000}{3600} = 266.7 \, \mathrm{m \, s^{-1}}$$. Since $$v_1$$ and $$v_2$$ are both close to $$v$$, we may approximate $$v_1 + v_2 \approx 2 v$$:

$$\Delta P \approx \rho v (v_2 - v_1)$$

$$\frac{v_2 - v_1}{v} \approx \frac{\Delta P}{\rho v^2} = \frac{6.47 \times 10^{3}}{1.2 \times (266.7)^2}$$

$$\frac{v_2 - v_1}{v} \approx \frac{6.47 \times 10^{3}}{8.53 \times 10^{4}} \approx 0.076$$

So the air on the upper surface flows about $$7.6\%$$ faster than on the lower surface.

Because the block moves at constant speed, it is in equilibrium. The pulling force (the weight of the hanging $$m = 0.010 \, \mathrm{kg}$$ mass transmitted through the string) is exactly balanced by the viscous drag of the liquid film:

$$F = m g = 0.010 \times 9.8 = 0.098 \, \mathrm{N}$$

The viscous force on the block from a Newtonian liquid film of thickness $$l$$ is

$$F = \eta \, A \, \frac{v}{l}$$

where the velocity gradient across the film is $$v/l$$ (top moving at $$v$$, bottom stuck to the table).

Solving for $$\eta$$:

$$\eta = \frac{F \, l}{A \, v}$$

Substituting $$A = 0.10 \, \mathrm{m^2}$$, $$v = 0.085 \, \mathrm{m \, s^{-1}}$$, $$l = 0.30 \times 10^{-3} \, \mathrm{m}$$:

$$\eta = \frac{0.098 \times 0.30 \times 10^{-3}}{0.10 \times 0.085}$$

$$\eta = \frac{2.94 \times 10^{-5}}{8.5 \times 10^{-3}} \approx 3.46 \times 10^{-3} \, \mathrm{Pa \, s}$$

At terminal velocity the net force on the sphere is zero. Equating the apparent weight (gravity $$-$$ buoyancy) to the Stokes' drag:

$$\tfrac{4}{3}\pi r^3 (\rho - \sigma) g = 6 \pi \eta r v_t$$

Solving for $$\eta$$ gives the standard terminal-velocity result:

$$\eta = \frac{2}{9} \cdot \frac{r^2 (\rho - \sigma) g}{v_t}$$

Substituting $$r = 2.0 \times 10^{-3} \, \mathrm{m}$$, $$\rho - \sigma = (8.9 - 1.5) \times 10^{3} = 7.4 \times 10^{3} \, \mathrm{kg \, m^{-3}}$$, $$g = 9.8 \, \mathrm{m \, s^{-2}}$$ and $$v_t = 6.5 \times 10^{-2} \, \mathrm{m \, s^{-1}}$$:

$$\eta = \frac{2}{9} \cdot \frac{(2.0 \times 10^{-3})^2 \times 7.4 \times 10^{3} \times 9.8}{6.5 \times 10^{-2}}$$

$$\eta = \frac{2}{9} \cdot \frac{4.0 \times 10^{-6} \times 7.4 \times 10^{3} \times 9.8}{6.5 \times 10^{-2}}$$

$$\eta = \frac{2}{9} \cdot \frac{0.2901}{6.5 \times 10^{-2}} = \frac{2}{9} \times 4.464$$

$$\eta \approx 0.99 \, \mathrm{Pa \, s}$$

For a hemispherical bubble at the lower end of the tube, the bubble's radius equals the tube's radius:

$$r = \frac{2.00 \, \mathrm{mm}}{2} = 1.00 \, \mathrm{mm} = 1.00 \times 10^{-3} \, \mathrm{m}$$

The bubble has only one air–water interface (it is an air bubble blown into water), so the excess pressure across that surface is

$$\Delta P = \frac{2 S}{r} = \frac{2 \times 7.30 \times 10^{-2}}{1.00 \times 10^{-3}} = 146 \, \mathrm{Pa}$$

The water pressure just outside the bubble (which sits at a depth $$h = 8.00 \, \mathrm{cm} = 0.08 \, \mathrm{m}$$) is

$$P_{\mathrm{out}} = P_a + \rho g h$$

$$P_{\mathrm{out}} = 1.01 \times 10^{5} + 1000 \times 9.80 \times 0.08$$

$$P_{\mathrm{out}} = 1.01 \times 10^{5} + 784 = 1.0178 \times 10^{5} \, \mathrm{Pa}$$

The pressure inside the bubble (which is the pressure required inside the capillary tube to keep the bubble hemispherical) is

$$P_{\mathrm{in}} = P_{\mathrm{out}} + \Delta P = 1.0178 \times 10^{5} + 146$$

$$P_{\mathrm{in}} \approx 1.019 \times 10^{5} \, \mathrm{Pa}$$

Equivalently, the gauge pressure required is

$$P_{\mathrm{in}} - P_a = \rho g h + \frac{2 S}{r} = 784 + 146 = 930 \, \mathrm{Pa}$$

The whole weight of the girl rests on a single circular heel of radius $$r = 0.5 \, \mathrm{cm} = 5.0 \times 10^{-3} \, \mathrm{m}$$.

Area of the heel:

$$A = \pi r^2 = \pi (5.0 \times 10^{-3})^2 = 7.854 \times 10^{-5} \, \mathrm{m^2}$$

Weight on the heel:

$$F = m g = 50 \times 9.8 = 490 \, \mathrm{N}$$

Pressure exerted on the floor:

$$P = \frac{F}{A} = \frac{490}{7.854 \times 10^{-5}} \approx 6.24 \times 10^{6} \, \mathrm{Pa}$$

This is about $$60 \, \mathrm{atm}$$ — far greater than the pressure under an ordinary flat shoe, which is why stiletto heels can dent soft floors.

In a barometer the atmospheric pressure $$P_a$$ is balanced by the hydrostatic pressure of the fluid column of height $$h$$ and density $$\rho$$:

$$P_a = \rho g h \implies h = \frac{P_a}{\rho g}$$

For normal atmospheric pressure $$P_a = 1.013 \times 10^{5} \, \mathrm{Pa}$$, with $$\rho = 984 \, \mathrm{kg \, m^{-3}}$$ and $$g = 9.8 \, \mathrm{m \, s^{-2}}$$:

$$h = \frac{1.013 \times 10^{5}}{984 \times 9.8} = \frac{1.013 \times 10^{5}}{9643.2}$$

$$h \approx 10.5 \, \mathrm{m}$$

Compare with a mercury barometer (density $$13.6 \times 10^{3} \, \mathrm{kg \, m^{-3}}$$), where the same atmospheric pressure supports a column of only $$0.76 \, \mathrm{m}$$. A wine barometer therefore has to be impractically tall — this is precisely why Torricelli used mercury.

The maximum (gauge) pressure on the structure will be at the sea bed. With $$\rho_{\text{sea}} \approx 1.03 \times 10^{3} \, \mathrm{kg \, m^{-3}}$$, $$g = 9.8 \, \mathrm{m \, s^{-2}}$$ and $$h = 3 \, \mathrm{km} = 3 \times 10^{3} \, \mathrm{m}$$,

$$P = \rho g h = (1.03 \times 10^{3})(9.8)(3 \times 10^{3})$$

$$P \approx 3.03 \times 10^{7} \, \mathrm{Pa}$$

Comparing with the design strength $$10^{9} \, \mathrm{Pa}$$:

$$\frac{P}{P_{\max}} = \frac{3.03 \times 10^{7}}{10^{9}} \approx 0.030$$

The maximum stress at the depth is only about $$3\%$$ of what the structure is designed to take. The structure is therefore suitable for installation on the oil well.

By Pascal's law the pressure throughout the hydraulic fluid is the same, so the pressure that the small piston must apply equals the pressure exerted on the load piston by the car's weight.

$$F = m g = 3000 \times 9.8 = 2.94 \times 10^{4} \, \mathrm{N}$$

$$A = 425 \, \mathrm{cm^2} = 425 \times 10^{-4} \, \mathrm{m^2} = 4.25 \times 10^{-2} \, \mathrm{m^2}$$

$$P_{\max} = \frac{F}{A} = \frac{2.94 \times 10^{4}}{4.25 \times 10^{-2}}$$

$$P_{\max} \approx 6.92 \times 10^{5} \, \mathrm{Pa}$$

The two mercury columns are at the same level, so the pressure at the water–mercury interface equals the pressure at the spirit–mercury interface (they lie on the same horizontal line in a single connected mercury body).

Above each mercury column, only atmospheric pressure plus the column of the lighter liquid contributes:

$$P_a + \rho_w g h_w = P_a + \rho_s g h_s$$

$$\rho_w h_w = \rho_s h_s$$

$$\frac{\rho_s}{\rho_w} = \frac{h_w}{h_s} = \frac{10.0}{12.5} = 0.8$$

The specific gravity of spirit (relative to water) is therefore

$$\boxed{\rho_s / \rho_w = 0.8}$$

After adding $$15.0 \, \mathrm{cm}$$ to each arm, the columns of liquid above the mercury become

$$h_w = 10.0 + 15.0 = 25.0 \, \mathrm{cm} \quad (\text{water arm})$$

$$h_s = 12.5 + 15.0 = 27.5 \, \mathrm{cm} \quad (\text{spirit arm})$$

From the previous problem the specific gravities (in g/cm$${}^3$$) are: water $$1.00$$, spirit $$0.80$$, mercury $$13.6$$. Since the water column is heavier than the spirit column, mercury will be pushed down in the water arm and up in the spirit arm.

Let the mercury level on the spirit side be higher than on the water side by $$h$$. Consider the horizontal level passing through the (lower) mercury surface in the water arm. On the water arm at that level:

$$P_{\text{water side}} = P_a + \rho_w g \, h_w$$

On the spirit arm at the same horizontal level there is, above this level, a column of mercury of height $$h$$ (the level difference) topped by the spirit column:

$$P_{\text{spirit side}} = P_a + \rho_s g \, h_s + \rho_{Hg} g \, h$$

Equating the two:

$$\rho_w h_w = \rho_s h_s + \rho_{Hg} h$$

Dividing by $$\rho_w$$ and using specific gravities:

$$1.00 \times 25.0 = 0.80 \times 27.5 + 13.6 \, h$$

$$25.0 = 22.0 + 13.6 \, h$$

$$h = \frac{3.0}{13.6} \approx 0.221 \, \mathrm{cm}$$

The mercury level on the spirit side is about $$0.22 \, \mathrm{cm}$$ higher than on the water side.

Bernoulli's equation

$$P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{const.}$$

is derived under the assumptions that the flow is steady, incompressible, non-viscous, and irrotational (streamline flow).

In a river rapid the water is extremely turbulent: streamlines do not exist, the velocity at each point fluctuates rapidly in magnitude and direction, eddies form and dissipate mechanical energy as heat, and viscous dissipation is no longer negligible. None of Bernoulli's assumptions is satisfied.

So Bernoulli's equation cannot be used to describe the flow of water through a rapid.

Bernoulli's equation is generally applied as a relation between two points:

$$P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2$$

If we replace each $$P_i$$ by the corresponding gauge pressure $$P_i - P_a$$, the atmospheric pressure $$P_a$$ subtracts out from both sides and the equation is unchanged. So as long as the same reference (absolute everywhere, or gauge everywhere) is used at both points, Bernoulli's equation gives the same result. But the two must not be mixed in one equation.

An exception is when one point is open to the atmosphere (and therefore at $$P = P_a$$, gauge pressure $$= 0$$) — here gauge pressure is particularly convenient.

Volume flow rate from the given mass flow rate:

$$Q = \frac{\dot{m}}{\rho} = \frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}} = 3.08 \times 10^{-6} \, \mathrm{m^3 \, s^{-1}}$$

For laminar flow of a viscous liquid through a horizontal pipe, Poiseuille's formula gives the pressure drop:

$$Q = \frac{\pi r^4 \, \Delta P}{8 \eta L} \implies \Delta P = \frac{8 \eta L Q}{\pi r^4}$$

With $$r = 1.0 \, \mathrm{cm} = 10^{-2} \, \mathrm{m}$$, $$L = 1.5 \, \mathrm{m}$$, $$\eta = 0.83 \, \mathrm{Pa \, s}$$:

$$\Delta P = \frac{8 \times 0.83 \times 1.5 \times 3.08 \times 10^{-6}}{\pi \times (10^{-2})^4}$$

$$\Delta P = \frac{3.066 \times 10^{-5}}{3.1416 \times 10^{-8}}$$

$$\Delta P \approx 975 \, \mathrm{Pa} \approx 9.8 \times 10^{2} \, \mathrm{Pa}$$

Check on laminar flow. The mean speed in the tube is

$$v = \frac{Q}{\pi r^2} = \frac{3.08 \times 10^{-6}}{\pi (10^{-2})^2} \approx 9.80 \times 10^{-3} \, \mathrm{m \, s^{-1}}$$

The Reynolds number (using diameter $$d = 2r = 2 \times 10^{-2} \, \mathrm{m}$$) is

$$\mathrm{Re} = \frac{\rho v d}{\eta} = \frac{1.3 \times 10^{3} \times 9.80 \times 10^{-3} \times 2 \times 10^{-2}}{0.83} \approx 0.31$$

This is far smaller than the critical value ($$\mathrm{Re}_c \sim 2000$$), so the assumption of laminar (streamline) flow is very well satisfied.

Apply Bernoulli's equation between the upper ($$v_u = 70 \, \mathrm{m \, s^{-1}}$$) and lower ($$v_l = 63 \, \mathrm{m \, s^{-1}}$$) surfaces of the wing. The vertical separation of the two surfaces is small, so the gravitational term can be neglected:

$$P_l + \tfrac{1}{2} \rho v_l^2 = P_u + \tfrac{1}{2} \rho v_u^2$$

$$P_l - P_u = \tfrac{1}{2} \rho (v_u^2 - v_l^2)$$

Compute the squared-speed difference:

$$v_u^2 - v_l^2 = 70^2 - 63^2 = 4900 - 3969 = 931 \, \mathrm{m^2 \, s^{-2}}$$

$$P_l - P_u = \tfrac{1}{2} \times 1.3 \times 931 = 605.15 \, \mathrm{Pa}$$

The net upward lift is this pressure difference times the wing area:

$$F = (P_l - P_u) \, A = 605.15 \times 2.5$$

$$F \approx 1.51 \times 10^{3} \, \mathrm{N}$$

Figure 9.20(a) is incorrect.

In a steady flow through a pipe of varying cross-section, the equation of continuity demands

$$A_1 v_1 = A_2 v_2$$

So where the pipe narrows, the fluid speed must increase, which means the streamlines should crowd closer together at the constriction. By Bernoulli's principle the increased speed at the narrow section implies a decrease in pressure there.

Figure 9.20(a) shows the streamlines as equally spaced through the constriction (and/or the pressure being higher at the narrow part), which violates both the continuity equation and Bernoulli's principle. Figure 9.20(b) correctly depicts the streamlines as crowded together at the narrower section, consistent with a higher flow speed (and a lower pressure) there.

Convert the inside-tube speed to SI units:

$$v_1 = 1.5 \, \mathrm{m \, min^{-1}} = \frac{1.5}{60} \, \mathrm{m \, s^{-1}} = 2.5 \times 10^{-2} \, \mathrm{m \, s^{-1}}$$

Tube cross-section:

$$A_1 = 8.0 \, \mathrm{cm^2} = 8.0 \times 10^{-4} \, \mathrm{m^2}$$

Each hole has radius $$r = 0.5 \, \mathrm{mm} = 5.0 \times 10^{-4} \, \mathrm{m}$$, so the combined area of $$N = 40$$ holes is

$$A_2 = N \pi r^2 = 40 \times \pi \times (5.0 \times 10^{-4})^2$$

$$A_2 = 40 \times \pi \times 2.5 \times 10^{-7} = 3.14 \times 10^{-5} \, \mathrm{m^2}$$

Conservation of mass (continuity): the same volume of liquid per second passing through the tube must pass through the holes:

$$A_1 v_1 = A_2 v_2 \implies v_2 = \frac{A_1 v_1}{A_2}$$

$$v_2 = \frac{8.0 \times 10^{-4} \times 2.5 \times 10^{-2}}{3.14 \times 10^{-5}} = \frac{2.0 \times 10^{-5}}{3.14 \times 10^{-5}}$$

$$v_2 \approx 0.637 \, \mathrm{m \, s^{-1}}$$

A soap film has two free surfaces (front and back), and both pull the slider upward. Hence the total upward force from surface tension acting on the slider of length $$L$$ is

$$F = 2 \, S \, L$$

In equilibrium this balances the suspended weight $$W$$:

$$W = 2 S L \implies S = \frac{W}{2 L}$$

With $$W = 1.5 \times 10^{-2} \, \mathrm{N}$$ and $$L = 30 \, \mathrm{cm} = 0.30 \, \mathrm{m}$$:

$$S = \frac{1.5 \times 10^{-2}}{2 \times 0.30} = \frac{1.5 \times 10^{-2}}{0.60}$$

$$S = 2.5 \times 10^{-2} \, \mathrm{N \, m^{-1}}$$

The vertical force exerted by a soap film on the slider depends only on the surface tension $$S$$ (set by the liquid and temperature) and the length of the slider $$L$$ along which the two surfaces of the film pull:

$$W = 2 \, S \, L$$

It does not depend on the height (or area) of the film, because surface tension is force per unit length, not force per unit area. So as long as the slider has the same length and the same liquid is used at the same temperature, the weight supported is identical to that in (a):

$$W_{(b)} = W_{(c)} = W_{(a)} = 4.5 \times 10^{-2} \, \mathrm{N}$$

Physically, the film in (b) is larger and in (c) tilted, but in each case the slider still bounds the same length of film with two surfaces, and the same $$2 S L$$ acts vertically on it.

A mercury drop has a single liquid–air interface, so the excess pressure inside the drop over that of the surrounding air is

$$\Delta P = \frac{2 S}{r}$$

With $$S = 4.65 \times 10^{-1} \, \mathrm{N \, m^{-1}}$$ and $$r = 3.00 \, \mathrm{mm} = 3.00 \times 10^{-3} \, \mathrm{m}$$:

$$\Delta P = \frac{2 \times 4.65 \times 10^{-1}}{3.00 \times 10^{-3}} = \frac{0.930}{3.00 \times 10^{-3}}$$

$$\Delta P = 310 \, \mathrm{Pa}$$

The absolute pressure inside the drop is the atmospheric pressure plus this excess:

$$P_{\mathrm{in}} = P_a + \Delta P = 1.01 \times 10^{5} + 310$$

$$P_{\mathrm{in}} \approx 1.0131 \times 10^{5} \, \mathrm{Pa}$$

The excess pressure is only about $$0.3\%$$ of atmospheric.

(i) Soap bubble in air.

A soap bubble has two air–liquid surfaces (the outer and inner faces of the soap film). The excess pressure inside it is therefore

$$\Delta P_1 = \frac{4 S}{r}$$

With $$S = 2.50 \times 10^{-2} \, \mathrm{N \, m^{-1}}$$ and $$r = 5.00 \, \mathrm{mm} = 5.00 \times 10^{-3} \, \mathrm{m}$$:

$$\Delta P_1 = \frac{4 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}} = \frac{0.10}{5.00 \times 10^{-3}} = 20 \, \mathrm{Pa}$$

(ii) Air bubble in soap solution at depth $$h = 40 \, \mathrm{cm}$$.

An air bubble inside the liquid has only a single liquid–air interface, so the excess pressure inside the bubble over the pressure of the surrounding liquid is

$$\Delta P_2 = \frac{2 S}{r} = \frac{2 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}} = 10 \, \mathrm{Pa}$$

The pressure of the soap solution at depth $$h = 0.40 \, \mathrm{m}$$ (density $$\rho = 1.20 \times 10^{3} \, \mathrm{kg \, m^{-3}}$$) is

$$P_{\mathrm{liq}} = P_a + \rho g h = 1.01 \times 10^{5} + 1.20 \times 10^{3} \times 9.8 \times 0.40$$

$$P_{\mathrm{liq}} = 1.01 \times 10^{5} + 4.704 \times 10^{3} = 1.0570 \times 10^{5} \, \mathrm{Pa}$$

Pressure inside the air bubble:

$$P_{\mathrm{in}} = P_{\mathrm{liq}} + \Delta P_2 = 1.0570 \times 10^{5} + 10$$

$$P_{\mathrm{in}} \approx 1.058 \times 10^{5} \, \mathrm{Pa}$$