NCERT Solutions · Class 11 Physics Mechanical Properties of Fluids

Bernoulli's equation holds only for smooth, streamline flow — a rapid is turbulent, so the equation breaks down.Bernoulli's equation for a fluid moving along a streamline is \[p + \frac{1}{2}\rho v^2 + \rho g h = \text{constant} \] where \(\displaystyle p\) is the pressure, \(\displaystyle \rho\) is the fluid density, \(\displaystyle v\) is its speed, and \(\displaystyle h\) is its height above some reference level. This result comes from applying the work–energy theorem to a fluid element, and it rests on three assumptions built into the derivation: A rapid in a river is, by definition, a stretch where the water surges over rocks and a steep bed with high speed. Under these conditions the flow becomes turbulent: the water breaks into swirling eddies and whorls, and the velocity at a given point fluctuates rapidly and unpredictably with time instead of following a smooth, well-defined streamline. This is precisely the condition Bernoulli's equation excludes — the "steady, streamline flow" assumption fails.There is also energy loss to consider. In turbulent flow, a large amount of the water's kinetic energy is dissipated as heat through internal friction between layers of fluid moving chaotically against each other (this is why rapids are frothy and noisy). Bernoulli's equation is a statement that mechanical energy per unit volume is conserved along a streamline; once energy is being dissipated by turbulence, that conservation no longer holds.For both reasons — the flow is not steady/streamline, and mechanical energy is not conserved because of turbulent dissipation — Bernoulli's equation cannot be used to describe the flow of water through a rapid.**Answer: No. Bernoulli's equation applies only to steady, streamline (laminar) flow of a non-viscous fluid where mechanical energy is conserved. The flow through a rapid is turbulent — the velocity at each point fluctuates rapidly with time and energy is dissipated through internal friction — so the conditions needed for Bernoulli's equation are not satisfied.

No — because atmospheric pressure is common to both points, it cancels out of the equation and never affects the answer.Bernoulli's equation, applied between any two points $\displaystyle 1$ and $\displaystyle 2$ along a streamline of a flowing fluid of density \(\displaystyle \rho\), reads\[P_1+\tfrac12\rho v_1^{2}+\rho g h_1 = P_2+\tfrac12\rho v_2^{2}+\rho g h_2 \]Here \(\displaystyle P_1\) and \(\displaystyle P_2\) are the absolute pressures at the two points, \(\displaystyle v_1,v_2\) the fluid speeds there, and \(\displaystyle h_1,h_2\) the heights above some reference level.Now write each absolute pressure as gauge pressure plus atmospheric pressure, \(\displaystyle P = P_{gauge}+P_{atm}\), which is simply the definition of gauge pressure (the excess of the absolute pressure over the surrounding atmosphere). Substituting,\[\big(P_{1,gauge}+P_{atm}\big)+\tfrac12\rho v_1^{2}+\rho g h_1 = \big(P_{2,gauge}+P_{atm}\big)+\tfrac12\rho v_2^{2}+\rho g h_2 \]The term \(\displaystyle P_{atm}\) is the same number on both sides of the equation, because both points $\displaystyle 1$ and $\displaystyle 2$ are exposed to (or ultimately connected to) the same atmosphere — the air pressing down on a tank of water, or the air at the open end of a pipe, is the same air. A quantity that is identical on both sides of an equation cancels:\[P_{1,gauge}+\tfrac12\rho v_1^{2}+\rho g h_1 = P_{2,gauge}+\tfrac12\rho v_2^{2}+\rho g h_2 \]This is exactly Bernoulli's equation again, with the absolute pressures simply replaced by gauge pressures. So gauge pressure satisfies Bernoulli's equation just as well as absolute pressure does, and using one or the other changes nothing in the final result for speeds, heights, or pressure differences.Where it would matter. The cancellation relies on \(\displaystyle P_{atm}\) being the same at both points. If the two points being compared were exposed to different atmospheric pressures — for instance one point open to the air at sea level and the other sealed inside a chamber held at a different background pressure — then the two \(\displaystyle P_{atm}\) terms would not be equal, would not cancel, and the choice of gauge versus absolute pressure would then change the equation. In every ordinary situation (a pipe, a tank, a Venturi meter, all open to the same room or the same outside air), this never happens, so gauge pressure can be used freely and is in fact the more convenient choice, since it is what a pressure gauge measures directly.Answer: No, it does not matter — because the same atmospheric pressure appears on both sides of Bernoulli's equation and cancels, using gauge pressure in place of absolute pressure leaves the equation unchanged, provided the atmospheric pressure is the same at the two points being compared (if it differed between the two points, the choice would matter).

Steady flow through a narrow tube is governed by Poiseuille's law, not by Bernoulli's equation — viscosity, not just speed, decides the pressure drop here.Poiseuille's law for the volume flow rate \(\displaystyle Q\) of a viscous fluid through a horizontal tube is \[Q = \frac{\pi p r^{4}}{8\,\eta\, l} \] where \(\displaystyle p\) is the pressure difference between the two ends, \(\displaystyle r\) is the tube's radius, \(\displaystyle \eta\) is the coefficient of viscosity, and \(\displaystyle l\) is the length of the tube.Step $\displaystyle 1$ — Convert the given mass flow rate to a volume flow rate.The data give the mass collected per second, \(\displaystyle \dfrac{dm}{dt} = 4.0\times10^{-3}\ \text{kg s}^{-1}\), but Poiseuille's law needs the volume flow rate. Since mass = density × volume, dividing by the density \(\displaystyle \rho\) converts one to the other: \[Q = \frac{1}{\rho}\frac{dm}{dt} = \frac{4.0\times10^{-3}\ \text{kg s}^{-1}}{1.3\times10^{3}\ \text{kg m}^{-3}} = 3.08\times10^{-6}\ \text{m}^{3}\text{s}^{-1} \]Step $\displaystyle 2$ — Solve Poiseuille's law for \(\displaystyle p\) and substitute.Rearranging, \[p = \frac{8\,\eta\, l\, Q}{\pi r^{4}} \] Here \(\displaystyle l = 1.5\ \text{m}\), \(\displaystyle \eta = 0.83\ \text{Pa s}\), and the radius must be in metres: \(\displaystyle r = 1.0\ \text{cm} = 1.0\times10^{-2}\ \text{m}\), so \(\displaystyle r^{4} = 1.0\times10^{-8}\ \text{m}^{4}\).\[p = \frac{8 \times 0.83\ \text{Pa s} \times 1.5\ \text{m} \times 3.08\times10^{-6}\ \text{m}^{3}\text{s}^{-1}}{\pi \times 1.0\times10^{-8}\ \text{m}^{4}} \]The numerator is \(\displaystyle 8 \times 0.83 \times 1.5 \times 3.08\times10^{-6} = 3.06\times10^{-5}\ \text{Pa m}^{4}\text{s}^{-1}\cdot\text{s}^{-1}\cdot\text{s}\) (units work out to Pa·m\(\displaystyle ^4\)), and the denominator is \(\displaystyle \pi \times 1.0\times10^{-8} = 3.14\times10^{-8}\ \text{m}^{4}\). Dividing, \[p \approx 9.76\times10^{2}\ \text{Pa} \]All five given quantities carry two significant figures, so the answer is rounded to two: \(\displaystyle p \approx 9.8\times10^{2}\ \text{Pa}\) (about $\displaystyle 980$ Pa, roughly one-hundredth of atmospheric pressure).Step $\displaystyle 3$ — Check whether the flow is really laminar.Poiseuille's law assumes laminar (streamline) flow, so that assumption needs checking with the Reynolds number \[N_R = \frac{\rho\, v\, d}{\eta} \] where \(\displaystyle v\) is the average speed of the fluid and \(\displaystyle d = 2r\) is the tube's diameter (not the radius — this is the step people slip on).The average speed comes from the flow rate divided by the cross-sectional area \(\displaystyle A = \pi r^{2}\): \[v = \frac{Q}{\pi r^{2}} = \frac{3.08\times10^{-6}\ \text{m}^{3}\text{s}^{-1}}{\pi \times (1.0\times10^{-2}\ \text{m})^{2}} = \frac{3.08\times10^{-6}}{3.14\times10^{-4}}\ \text{m s}^{-1} \approx 9.8\times10^{-3}\ \text{m s}^{-1} \]Then, with \(\displaystyle d = 2.0\times10^{-2}\ \text{m}\): \[N_R = \frac{(1.3\times10^{3}\ \text{kg m}^{-3})(9.8\times10^{-3}\ \text{m s}^{-1})(2.0\times10^{-2}\ \text{m})}{0.83\ \text{Pa s}} \approx 0.31 \]Flow stays laminar as long as \(\displaystyle N_R\) is well below the critical range (about $\displaystyle 2000$–$\displaystyle 3000$, past which turbulence sets in). Here \(\displaystyle N_R \approx 0.3\), thousands of times smaller than that threshold, so the laminar-flow assumption behind Poiseuille's law is fully justified for this tube.Answer: The pressure difference between the two ends is \(\displaystyle p \approx 9.8\times10^{2}\ \text{Pa}\) (about $\displaystyle 980$ Pa); the flow is confirmed laminar since the Reynolds number \(\displaystyle N_R \approx 0.3\) is far below the critical value of about $\displaystyle 2000$–3000.

Faster flow means lower pressure — that pressure difference, multiplied by the wing area, is the lift. This is Bernoulli's principle at work: for a fluid in steady flow, where the speed is higher the pressure is lower, provided the height doesn't change much along the streamline — and across a thin wing it doesn't.Setting up Bernoulli's equationFor the air just above the wing (speed \(\displaystyle v_1 \), pressure \(\displaystyle P_1 \)) and just below the wing (speed \(\displaystyle v_2 \), pressure \(\displaystyle P_2 \)), both streamlines are essentially at the same height, so Bernoulli's equation\[P_1 + \frac{1}{2}\rho v_1^{2} = P_2 + \frac{1}{2}\rho v_2^{2} \]reduces to a balance between pressure and speed alone, with no gravity term. Here \(\displaystyle \rho \) is the density of air.Rearranging for the pressure difference between the lower and upper surfaces:\[P_2 - P_1 = \frac{1}{2}\rho\left(v_1^{2} - v_2^{2}\right) \]The air moves faster over the upper surface (\(\displaystyle v_1 = 70\ \text{m s}^{-1} \)) than under the lower surface (\(\displaystyle v_2 = 63\ \text{m s}^{-1} \)), so \(\displaystyle v_1 > v_2 \) and this quantity is positive: the pressure below the wing is greater than the pressure above it. That excess pressure pushing up is exactly what holds the plane in the air.A step people rush past: it is easy to write \(\displaystyle v_1^2 - v_2^2 \) as \(\displaystyle (v_1-v_2)^2 \) by mistake. It is not — it is the difference of squares, \(\displaystyle (v_1-v_2)(v_1+v_2) \), which is a much bigger number here.Turning pressure difference into a forceThe lift is this pressure difference acting over the wing's area \(\displaystyle A \):\[F_{\text{lift}} = (P_2 - P_1)\,A = \frac{1}{2}\rho\left(v_1^{2}-v_2^{2}\right)A \]Substituting the numbersGiven \(\displaystyle v_1 = 70\ \text{m s}^{-1} \), \(\displaystyle v_2 = 63\ \text{m s}^{-1} \), \(\displaystyle A = 2.5\ \text{m}^{2} \), \(\displaystyle \rho = 1.3\ \text{kg m}^{-3} \):\[v_1^{2} - v_2^{2} = (v_1 - v_2)(v_1 + v_2) = (70 - 63)(70 + 63) = 7 \times 133 = 931\ \text{m}^{2}\text{s}^{-2} \]\[F_{\text{lift}} = \frac{1}{2}\times 1.3\ \text{kg m}^{-3} \times 931\ \text{m}^{2}\text{s}^{-2} \times 2.5\ \text{m}^{2} \]\[F_{\text{lift}} = 0.65 \times 931 \times 2.5\ \text{N} = 605.15 \times 2.5\ \text{N} = 1512.875\ \text{N} \]Rounding for significant figures: every measured quantity here — \(\displaystyle 70\), \(\displaystyle 63\), \(\displaystyle 2.5\), \(\displaystyle 1.3\) — is given to two significant figures, so the answer cannot be stated more precisely than that. Rounding \(\displaystyle 1512.875\ \text{N} \) to two significant figures gives \(\displaystyle 1.5 \times 10^{3}\ \text{N} \).The direction of this force is vertically upward (away from the wing's lower surface, toward its upper surface) — it is this net upward push that is the "lift" keeping the model aeroplane airborne.Answer: The lift on the wing is \(\displaystyle F_{\text{lift}} = \dfrac{1}{2}\rho\left(v_1^{2}-v_2^{2}\right)A \approx 1.5 \times 10^{3}\ \text{N} \) (about $\displaystyle 1513$ N before rounding), directed vertically upward.

In steady flow, a narrower cross-section must carry faster-moving, lower-pressure fluid — never the reverse — and that is exactly where Figure $\displaystyle 9.20$(a) goes wrong.For a non-viscous liquid in steady flow, two conservation laws fix what must happen at a constriction, and together they rule out one of the two pictures.Step $\displaystyle 1$: The equation of continuity fixes the speed.For an incompressible liquid, the volume of fluid entering a tube of flow per second must equal the volume leaving it per second. If \(\displaystyle A_1\) and \(\displaystyle v_1\) are the cross-sectional area and speed at a wide part of the tube, and \(\displaystyle A_2\), \(\displaystyle v_2\) are the same quantities at the constriction, mass conservation gives \[A_1 v_1 = A_2 v_2 . \] Since the constriction has the smaller area, \(\displaystyle A_2 < A_1\), the speed there must be larger: \[v_2 = \frac{A_1}{A_2}\,v_1 > v_1 . \] A narrow neck always speeds the flow up — this follows from mass conservation alone, it is common to both figures, and it is not what distinguishes them.Step $\displaystyle 2$: Bernoulli's equation fixes the pressure.For steady, non-viscous, incompressible flow along a horizontal streamline (so the height \(\displaystyle h\) does not change), Bernoulli's principle states \[p + \tfrac{1}{2}\rho v^{2} = \text{constant along the streamline,} \] where \(\displaystyle p\) is the pressure, \(\displaystyle \rho\) the density of the liquid, and \(\displaystyle v\) the flow speed at that point. Applying this between the wide section ($\displaystyle 1$) and the constriction ($\displaystyle 2$), \[p_1 + \tfrac12 \rho v_1^{2} = p_2 + \tfrac12 \rho v_2^{2}. \] Because \(\displaystyle v_2 > v_1\) (Step $\displaystyle 1$), the kinetic-energy term on the right is larger, so the pressure term must be smaller to keep the total unchanged: \[p_2 = p_1 - \tfrac12\rho\left(v_2^{2}-v_1^{2}\right) < p_1 . \] The pressure at the constriction must be lower than the pressure in the wide sections on either side of it — not higher. This is the step people get backwards: it feels natural to think a squeezed tube should show "more pressure," the way squeezing a solid does, but for a flowing liquid the continuity equation forces the speed up first, and it is that higher speed which then pulls the pressure down through Bernoulli's equation.Step $\displaystyle 3$: Read the two figures against this conclusion.Both figures show the same shape of tube — flow entering wide, narrowing to a constriction, then widening again — with the fluid pressure marked at the wide and narrow sections. Only one ordering of pressures is physically possible for steady, non-viscous flow: lower pressure exactly where the tube is narrowest, by Steps $\displaystyle 1$ and $\displaystyle 2$ together. A figure that instead shows the pressure rising at the constriction is asking the liquid to move faster and push harder at the very same point, which would need an external supply of kinetic energy from nowhere — a direct violation of the energy-conservation statement built into Bernoulli's equation.Figure $\displaystyle 9.20$(a) is drawn with the pressure at the constriction higher than in the wider parts of the tube — the impossible ordering. Figure $\displaystyle 9.20$(b) draws the pressure correctly falling at the constriction, consistent with the derivation above.Answer: Figure $\displaystyle 9.20$(a) is incorrect. At a constriction, the equation of continuity, \(\displaystyle A_1v_1 = A_2v_2\), forces the flow speed to increase because the area is smaller; Bernoulli's equation, \(\displaystyle p+\tfrac12\rho v^2=\text{constant}\), then forces the pressure to decrease there, since the fluid is assumed incompressible. Figure $\displaystyle 9.20$(a) shows the pressure rising at the narrow section instead of falling, which cannot happen in steady, non-viscous flow.

The principle here is conservation of mass for an incompressible liquid, i.e., the equation of continuity: the volume of liquid entering per second must equal the volume leaving per second, however many holes it splits into.For steady, incompressible flow through a tube that branches into several exits, \[A_{1}v_{1} = A_{2}v_{2} \] where \(\displaystyle A_1\) is the cross-sectional area of the wide entry section, \(\displaystyle v_1\) is the speed of the liquid there, and \(\displaystyle A_2\), \(\displaystyle v_2\) are the total area and speed of the liquid at the exit — here, at the $\displaystyle 40$ holes taken together. The equation of continuity always compares total areas, so the right-hand side must use the combined area of all $\displaystyle 40$ holes, not one hole.Step $\displaystyle 1$: Put every quantity in SI units.Cross-section of the tube: \[A_1 = 8.0\ \text{cm}^2 = 8.0\times10^{-4}\ \text{m}^2 \]Speed of flow inside the tube — this is given per minute, and it is easy to forget to convert to seconds: \[v_1 = 1.5\ \text{m min}^{-1} = \frac{1.5}{60}\ \text{m s}^{-1} = 0.025\ \text{m s}^{-1} \]Each hole has diameter \(\displaystyle 1.0\ \text{mm}\), so its radius is \[r = 0.5\ \text{mm} = 0.5\times10^{-3}\ \text{m} \]Step $\displaystyle 2$: Find the total area of the $\displaystyle 40$ holes.Area of one circular hole: \[a = \pi r^{2} = \pi\left(0.5\times10^{-3}\ \text{m}\right)^{2} = \pi \times 0.25\times10^{-6}\ \text{m}^2 = 7.85\times10^{-7}\ \text{m}^2 \]This is the step people most often get wrong — using the diameter in \(\displaystyle \pi r^2\) instead of the radius, which overstates the area by a factor of 4. Now multiply by the number of holes, since all $\displaystyle 40$ act in parallel to carry the flow out: \[A_2 = 40\,a = 40 \times 7.85\times10^{-7}\ \text{m}^2 = 3.14\times10^{-5}\ \text{m}^2 \]Step $\displaystyle 3$: Apply the equation of continuity.\[v_2 = \frac{A_1 v_1}{A_2} \]First find the volume flow rate through the tube, since it is what stays constant: \[A_1 v_1 = \left(8.0\times10^{-4}\ \text{m}^2\right)\left(0.025\ \text{m s}^{-1}\right) = 2.0\times10^{-5}\ \text{m}^3\text{s}^{-1} \]Then divide by the total hole area: \[v_2 = \frac{2.0\times10^{-5}\ \text{m}^3\text{s}^{-1}}{3.14\times10^{-5}\ \text{m}^2} = 0.6366\ \text{m s}^{-1} \]Step $\displaystyle 4$: Round to the correct number of significant figures.Every given quantity — \(\displaystyle 8.0\ \text{cm}^2\), \(\displaystyle 1.0\ \text{mm}\), \(\displaystyle 1.5\ \text{m min}^{-1}\) — carries $\displaystyle 2$ significant figures, so the answer is stated to $\displaystyle 2$ significant figures, not carried out to four digits as the raw division gives: \[v_2 \approx 0.64\ \text{m s}^{-1} \]The liquid speeds up sharply on the way out (from \(\displaystyle 0.025\ \text{m s}^{-1}\) to \(\displaystyle 0.64\ \text{m s}^{-1}\), roughly $\displaystyle 25$ times faster) because the same volume per second is being forced through a much smaller total opening — the $\displaystyle 40$ fine holes together present far less area than the tube itself.Answer: The liquid is ejected from the holes at a speed of about \(\displaystyle 0.64\ \text{m s}^{-1}\).

A soap film has two free surfaces, so the surface tension pulls along both of them — treat the slider as being pulled by two liquid surfaces, not one.The soap film is stretched between the two arms of the U-shaped wire and the slider. Because the film is a thin sheet of liquid open to air on both faces, it has two surfaces, and each surface exerts a surface-tension force along the length of the slider that touches it.Step $\displaystyle 1$: Note the given quantities.Weight supported (equal to the total surface-tension force holding the slider in equilibrium): \[F = 1.5 \times 10^{-2}\ \text{N} \]Length of the slider: \[l = 30\ \text{cm} = 30 \times 10^{-2}\ \text{m} = 0.30\ \text{m} \]Step $\displaystyle 2$: Write the force balance.For the slider to be in equilibrium, the weight it supports must equal the total upward pull of the film. Surface tension \(\displaystyle S\) is defined as force per unit length of the line along which the surface acts: \[S = \frac{F}{L} \] where \(\displaystyle L\) is the total length of the film edge in contact with the slider.Since the film has two surfaces, each of length \(\displaystyle l\), the total length pulling on the slider is \[L = 2l \]This factor of $\displaystyle 2$ is the step most people miss — a single free liquid surface (like on top of water in a beaker) has only one surface, but a film (soap film, bubble wall) always has two, front and back.Step $\displaystyle 3$: Substitute and solve for \(\displaystyle S\).\[S = \frac{F}{2l} = \frac{1.5 \times 10^{-2}\ \text{N}}{2 \times 0.30\ \text{m}} \]\[S = \frac{1.5 \times 10^{-2}\ \text{N}}{0.60\ \text{m}} \]\[S = 2.5 \times 10^{-2}\ \text{N/m} \]Step $\displaystyle 4$: Check significant figures.Both given values (\(\displaystyle 1.5 \times 10^{-2}\) N and $\displaystyle 30$ cm) carry $\displaystyle 2$ significant figures, so the result is correctly quoted to $\displaystyle 2$ significant figures: \(\displaystyle 2.5 \times 10^{-2}\ \text{N/m}\). This value is close to the surface tension of soap solution reported in the textbook's own tables, which is a useful check that the factor of $\displaystyle 2$ was applied correctly.Answer: The surface tension of the soap film is \(\displaystyle 2.5 \times 10^{-2}\ \text{N/m}\) (i.e., \(\displaystyle 0.025\ \text{N/m}\)).

Surface tension is a property of the liquid–air interface at a given temperature — it depends only on the length of the boundary the film pulls on, never on the film's area or the shape of the frame holding it.A soap or liquid film has two free surfaces (front and back), so if it is stretched across a wire frame and pulls on a straight slider of length \(\displaystyle l \) with surface tension \(\displaystyle T \), the total pulling force — and hence the weight it can hold in equilibrium — is\[W = 2Tl \]Figure (a). Here the film supports \(\displaystyle W_a = 4.5 \times 10^{-2}\ \text{N} \) on a slider of length \(\displaystyle l \). This fixes the surface tension of the liquid at this temperature:\[T = \frac{W_a}{2l} = \frac{4.5 \times 10^{-2}\ \text{N}}{2l} \]Figures (b) and (c). The liquid is the same liquid, at the same temperature, so \(\displaystyle T \) has exactly the value found above — surface tension does not change with the size or shape of the film, only with the liquid and temperature (this is the point people miss: it is easy to assume a bigger or oddly-shaped film "holds more," but \(\displaystyle T \) is a boundary force per unit length, not something that scales with area). What matters is only the length of the slider wire actually in contact with the film. In figures (b) and (c) the frames are drawn with different outlines, but the movable slider that bears the weight has the same length \(\displaystyle l \) as in (a).Since both \(\displaystyle T \) and \(\displaystyle l \) are unchanged,\[W_b = W_c = 2Tl = 2\left(\frac{4.5\times10^{-2}\ \text{N}}{2l}\right)l = 4.5 \times 10^{-2}\ \text{N} \]The weight supported is exactly the same in all three cases. Physically: the film's ability to hold up a weight comes from the tension acting along the edge of the slider, not from how much liquid surface is stretched behind it — a large film and a small film of the same liquid, pulling on sliders of equal length, exert identical forces.Answer: The weight supported in Figs. (b) and (c) is the same as in Fig. (a), namely \(\displaystyle 4.5 \times 10^{-2}\ \text{N}\), because surface tension depends only on the liquid and temperature (unchanged here) and on the length of the slider in contact with the film (also unchanged), not on the area or shape of the film.

A liquid drop has just one free surface, so its excess pressure is \(\displaystyle \dfrac{2S}{r} \), not \(\displaystyle \dfrac{4S}{r}\) — that formula is for a soap bubble, which has two surfaces (inner and outer film).Because the surface film is stretched like a curved membrane, it squeezes the liquid inside a drop to a pressure higher than the pressure outside. For a spherical drop of radius \(\displaystyle r\) and surface tension \(\displaystyle S\), this excess pressure is\[\Delta P = \frac{2S}{r} \]Step $\displaystyle 1$: List the given values, in SI units. Step $\displaystyle 2$: Find the excess pressure inside the drop.\[\Delta P = \frac{2S}{r} = \frac{2 \times 0.465\ \text{N m}^{-1}}{3.00 \times 10^{-3}\ \text{m}} \]\[\Delta P = \frac{0.930}{3.00 \times 10^{-3}}\ \text{Pa} = 3.10 \times 10^{2}\ \text{Pa} = 310\ \text{Pa} \]Step $\displaystyle 3$: Find the total pressure inside the drop.The pressure just inside the drop is the atmospheric pressure acting on its outer surface plus this excess pressure due to the curved surface film:\[P_{\text{inside}} = P_0 + \Delta P \]\[P_{\text{inside}} = 1.01 \times 10^{5}\ \text{Pa} + 3.10 \times 10^{2}\ \text{Pa} \]\[P_{\text{inside}} = (1.01 \times 10^{5} + 0.0310 \times 10^{5})\ \text{Pa} = 1.0131 \times 10^{5}\ \text{Pa} \]Since \(\displaystyle P_0\) is known to $\displaystyle 3$ significant figures, the sum is rounded to the same precision:\[P_{\text{inside}} \approx 1.013 \times 10^{5}\ \text{Pa} \]A common slip here is to report only the excess pressure as "the pressure inside the drop" — the question asks for the actual pressure inside, which means the atmospheric pressure must be added on, since the drop sits in air at \(\displaystyle 1.01 \times 10^{5}\ \text{Pa}\) to begin with.Answer: The excess pressure inside the drop is $\displaystyle 310$ Pa, and the total pressure inside the drop is \(\displaystyle 1.013 \times 10^{5}\ \text{Pa}\).

A soap bubble has two liquid surfaces (inner and outer), so its excess pressure is \(\displaystyle 4S/r\); an air bubble sitting inside a liquid has only one surface, so its excess pressure is \(\displaystyle 2S/r\) — mixing these two up is the usual mistake here.Part $\displaystyle 1$ — excess pressure inside the soap bubbleFor a soap bubble in air, the soap film has two air–liquid interfaces (the surface facing outward and the surface facing the trapped air inside), and each contributes surface tension pulling the film inward. This doubles the usual excess-pressure formula, giving\[\Delta p_{\text{soap bubble}} = \frac{4S}{r} \]where \(\displaystyle S\) is the surface tension of the soap solution and \(\displaystyle r\) is the bubble's radius.Here \(\displaystyle S = 2.50 \times 10^{-2}\ \text{N m}^{-1}\) and \(\displaystyle r = 5.00\ \text{mm} = 5.00 \times 10^{-3}\ \text{m}\). Substituting,\[\Delta p_{\text{soap bubble}} = \frac{4 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}} = \frac{0.100}{5.00 \times 10^{-3}} = 20.0\ \text{Pa} \]Part $\displaystyle 2$ — pressure inside an air bubble formed at depth $\displaystyle 40.0$ cm in the soap solutionThis bubble is air trapped inside the liquid, so it has only one curved surface (liquid on the outside, air on the inside). Its excess pressure (the extra pressure the surface tension adds over whatever pressure already exists in the liquid around it) is therefore only\[\Delta p_{\text{air bubble}} = \frac{2S}{r} = \frac{2 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}} = \frac{0.0500}{5.00\times10^{-3}} = 10.0\ \text{Pa} \]The bubble also has to support the weight of the liquid column above it, so I need the pressure that already exists in the liquid at that depth. By the hydrostatic pressure law,\[p_{\text{liquid at depth } h} = p_{\text{atm}} + \rho g h \]where \(\displaystyle \rho\) is the density of the soap solution, \(\displaystyle g\) is the acceleration due to gravity, and \(\displaystyle h\) is the depth. The relative density (specific gravity) of the solution is $\displaystyle 1.20$, so its actual density is\[\rho = 1.20 \times 1000\ \text{kg m}^{-3} = 1200\ \text{kg m}^{-3} \](Relative density has no units — you must multiply by the density of water, \(\displaystyle 1000\ \text{kg m}^{-3}\), to get a usable density; forgetting this step is a common slip.)With \(\displaystyle h = 40.0\ \text{cm} = 0.400\ \text{m}\) and \(\displaystyle g = 9.8\ \text{m s}^{-2}\),\[\rho g h = 1200 \times 9.8 \times 0.400 = 4704\ \text{Pa} \approx 4.70 \times 10^{3}\ \text{Pa} \]So the pressure in the liquid surrounding the bubble, before accounting for surface tension, is\[p_{\text{atm}} + \rho g h = 1.01 \times 10^{5} + 4.704\times10^{3} = 1.0570 \times 10^{5}\ \text{Pa} \]The pressure inside the air bubble is this surrounding (absolute) liquid pressure plus the extra push from the curved surface — the same way gauge pressure and absolute pressure add up:\[p_{\text{inside}} = p_{\text{atm}} + \rho g h + \frac{2S}{r} \]\[p_{\text{inside}} = 1.01\times10^{5} + 4.704\times10^{3} + 10.0 = 1.05714 \times 10^{5}\ \text{Pa} \]All three given quantities (radius, surface tension, atmospheric pressure) carry three significant figures, so the final answer is rounded to three significant figures:\[p_{\text{inside}} \approx 1.06 \times 10^{5}\ \text{Pa} \]Answer: Excess pressure inside the soap bubble = $\displaystyle 20.0$ Pa; pressure inside the air bubble at $\displaystyle 40.0$ cm depth = $\displaystyle 1.06$ × $\displaystyle 10$⁵ Pa