MECHANICAL PROPERTIES OF FLUIDS-Exercise
Physics - Exercise
Q1. Explain why
(a) The blood pressure in humans is greater at the feet than at the brain
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of
its value at the sea level, though the height of the atmosphere is more than
100 km
(c) Hydrostatic pressure is a scalar quantity even though pressure is force
divided by area.
Solution
(a) Blood pressure greater at feet than brain
Blood behaves as a fluid in hydrostatic equilibrium within human blood vessels. Pressure at any point depends on the height of the blood column above it through the relation \(\Delta P = \rho g h\), where \(\rho\) is blood density, \(g\) is gravity, and \(h\) is vertical height difference.
The feet lie about 1.5 m below the heart, while the brain is roughly level with or slightly above it. Thus, the blood column height \(h\) is greater at the feet, producing higher hydrostatic pressure there compared to the brain.
The heart maintains mean arterial pressure around 100 mmHg, but hydrostatic effects add roughly 110 cm of blood height equivalent (about 80 mmHg) at the feet and subtract at the brain.
(b) Atmospheric pressure halves at 6 km despite 100 km height
Unlike constant-density fluids, air density \(\rho\) decreases exponentially with height due to compressibility: \(\rho = \rho_0 e^{-h/H}\), where \(H \approx 8\) km is the scale height.
Atmospheric pressure follows \(P = P_0 e^{-h/H}\). At \(h = 6\) km, \(h/H \approx 0.75\), so \(P \approx P_0 e^{-0.75} \approx 0.47 P_0\), nearly half the sea-level value.
Though total atmosphere extends beyond 100 km, 90% of mass lies below 20 km; density drops so rapidly that higher layers contribute negligibly to surface pressure.
(c) Hydrostatic pressure scalar despite force/area
Hydrostatic pressure arises from random molecular collisions in fluid at rest, isotropic in all directions per Pascal's law.
Though defined as force magnitude per area (\(P = F_\perp / A\)), the effective force direction adjusts normal to any surface, lacking fixed vector direction; pressure exerts equal magnitude perpendicularly everywhere.
Thus, hydrostatic pressure is characterized solely by magnitude, making it scalar, unlike directional shear stress in moving fluids.
Q2. Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water
with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the
same surface tends to form drops. (Put differently, water wets glass while
mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent disolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape
Solution
(a) Angle of contact: mercury obtuse, water acute with glass
The angle of contact \(\theta\) arises from adhesive forces between liquid-glass and cohesive forces within the liquid, balanced against surface tension.
For water-glass, strong hydrogen bonding creates powerful adhesion exceeding water cohesion, minimizing \(\theta\) to acute values (near 0°), pulling meniscus concave upward.
Mercury-glass adhesion is weak due to poor bonding between metal atoms and glass, while strong metallic cohesion dominates, yielding obtuse \(\theta\) (around 140°), convex meniscus.
(b) Water spreads on glass, mercury forms drops
Wetting behavior follows \(\theta\): acute \(\theta < 90^\circ\) indicates liquid adhesion to solid exceeds cohesion, driving spreading to maximize contact area.
Water wets clean glass as its acute \(\theta\) favors thin films; mercury's obtuse \(\theta > 90^\circ\) means cohesion wins, minimizing contact via spherical drops.
Young's equation quantifies: \(\cos\theta = \frac{\sigma_{SG} - \sigma_{SL}}{\sigma_{LG}}\), where high \(\sigma_{SG} - \sigma_{SL}\) for water yields positive cosine (acute \(\theta\)), negative for mercury.
(c) Surface tension independent of surface area
Surface tension \(\gamma\) is force per unit length (\(\gamma = F/L\)) resisting area increase, inherent molecular property from unbalanced attractions at liquid-vapor interface.
Molecules throughout the surface experience identical inward pulls regardless of total patch size; energy per unit area \(dW/dA = \gamma\) stays constant.
Thus, \(\gamma\) depends only on liquid type and temperature, not macroscopic area, as verified by experiments like wire frames on any surface scale.
(d) Detergent lowers water-glass contact angle
Detergents as surfactants reduce \(\gamma\) by positioning hydrophobic tails outward, hydrophilic heads inward, disrupting surface molecules.
Lower \(\gamma\) shifts Young's balance toward smaller \(\theta\), enhancing glass wetting for easier spreading in cleaning or emulsification.
Practical angles drop from ~0° (pure water) to near-complete wetting, explaining soap's superior rinse and stain removal on surfaces.
(e) Liquid drop spherical without external forces
Surface tension minimizes surface free energy proportional to area for fixed volume; sphere encloses maximum volume per area, lowest energy configuration.
Isotropic \(\gamma\) pulls uniformly in all directions, yielding spherical symmetry absent gravity or other fields.
Mathematically, for volume \(V\), sphere radius \(r = (3V/4\pi)^{1/3}\) gives minimal \(A = 4\pi r^2\), confirmed by pendant drop shapes distorting only under weight.
Q3. Fill in the blanks using the word(s) from the list appended with each statement:
(a) Surface tension of liquids generally ... with temperatures (increases / decreases)
(b) Viscosity of gases ... with temperature, whereas viscosity of liquids ... with
temperature (increases / decreases)
(c) For solids with elastic modulus of rigidity, the shearing force is proportional
to ... , while for fluids it is proportional to ... (shear strain / rate of shear
strain)
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows
(conservation of mass / Bernoulli’s principle)
(e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for
turbulence for an actual plane (greater / smaller)
Solution
Q3. Fill in the blanks
(a) Surface tension of liquids generally decreases with temperature
Higher temperature increases molecular kinetic energy, weakening intermolecular attractions at the surface. Thus, force per unit length \(\gamma\) drops, as observed in \(\gamma \propto (T_c - T)^n\) near critical point.
(b) Viscosity of gases increases with temperature, whereas viscosity of liquids decreases with temperature
Gas viscosity arises from momentum transfer between layers; higher speed enhances collisions, raising \(\eta\). Liquid viscosity stems from cohesive forces slowing flow; thermal agitation breaks these bonds, lowering \(\eta\).
(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
Solids obey Hooke's law: \(\tau = G \gamma\), where \(\gamma\) is shear strain, \(G\) rigidity modulus. Fluids follow Newton's law: \(\tau = \eta \dot{\gamma}\), \(\dot{\gamma} = d\gamma/dt\) rate of strain.
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows conservation of mass
Steady incompressible flow requires constant volume flux: \(A_1 v_1 = A_2 v_2\). Narrower constriction (\(A_2 < A_1\)) demands higher \(v_2\), directly from mass continuity.
(e) For the model of a plane in a wind tunnel, turbulence occurs at a greater speed for turbulence for an actual plane
Reynolds number \(Re = \rho v L / \eta\) governs transition; model has smaller length scale \(L_m < L_p\). Same fluid means turbulence at higher \(v_m> v_p\) to match critical \(Re\).
Q4. Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it
(b) When we try to close a water tap with our fingers, fast jets of water gush
through the openings between our fingers
(c) The size of the needle of a syringe controls flow rate better than the thumb
pressure exerted by a doctor while administering an injection
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on
the vessel
(e) A spinning cricket ball in air does not follow a parabolic trajectory
Solution
Q4. Explanations based on fluid dynamics principles
(a) Blow over paper to keep horizontal, not under
Blowing under adds upward pressure supporting weight; air stagnates below, higher \(P\). Blowing over accelerates air above paper, lowering pressure via Bernoulli.
Resulting pressure imbalance \(P_\text{below} > P_\text{above}\) creates upward lift force exceeding weight, raising paper toward faster stream.
(b) Water jets gush through finger gaps when closing tap
Fingers create multiple narrow constrictions around main flow. Continuity demands \(A v =\) constant; tiny gaps mean very high jet speeds.
Bernoulli confirms: same pressure drop yields \(v \propto 1/\sqrt{A}\), so minuscule openings produce fast, forceful jets.
(c) Syringe needle size controls flow better than thumb pressure
Poiseuille's law governs laminar flow: \(Q = \frac{\pi r^4 \Delta P}{8 \eta L}\). Discharge \(Q \propto r^4\), tiny radius changes dominate over moderate \(\Delta P\) from thumb.
Halving needle radius reduces \(Q\) by factor of 16; doubling pressure only doubles \(Q\). Thus, gauge size precisely tunes rate.
(d) Fluid jet from hole causes backward thrust on vessel
Outgoing fluid carries forward momentum flux \(\rho A v^2\). Container imparts this via reaction, yielding equal backward force by momentum conservation.
Like rocket thrust, continuous ejection produces steady recoil propelling vessel oppositely, independent of hole direction if symmetric.
(e) Spinning cricket ball deviates from parabola
Ball spin generates asymmetric boundary layer: faster side delays separation, slower side advances it, yielding lateral Magnus force \(\vec{F}_M \propto \vec{\omega} \times \vec{v}\).
This sideways deflection plus gravity curves trajectory unpredictably, explaining swing bowling versus symmetric parabolic flight.
Q5. A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor ?
Solution
Given: Mass of girl \(m = 50\) kg, diameter of circular heel \(d = 1.0\) cm \(= 0.01\) m, so radius \(r = 0.5\) cm \(= 0.005\) m, \(g = 9.8\) m/s².
Weight (force) exerted by girl on floor,
$$\begin{aligned} F &= mg \\ &= 50 \times 9.8 \\ &= 490~\text{N} \end{aligned}$$Area of contact (one heel),
$$\begin{aligned} A &= \pi r^2 \\ &= \frac{22}{7} \times (0.005)^2 \\ &= 7.85 \times 10^{-5}~\text{m}^2 \end{aligned}$$Pressure exerted by heel on floor,
$$\begin{aligned} P &= \frac{F}{A} \\ &= \frac{490}{7.85 \times 10^{-5}} \\ &= 6.24 \times 10^6~\text{N m}^{-2} \end{aligned}$$Thus, the pressure is approximately \(6.2 \times 10^6\) Pa, demonstrating why high heels concentrate force effectively.
Q6. Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m–3. Determine the height of the wine column for normal atmospheric pressure.
Solution
Given: Atmospheric pressure \(P = 1.01 \times 10^5\) Pa, density of French wine \(\rho = 984\) kg/m³, \(g = 9.8\) m/s².
Pressure balance at base of column,
$$\begin{aligned} P &= h \rho g \\ h &= \frac{P}{\rho g} \\ &= \frac{1.01 \times 10^5}{984 \times 9.8} \end{aligned}$$Compute denominator,
$$\begin{aligned} 984 \times 9.8 &= 9643.2 \\ h &= \frac{1.01 \times 10^5}{9643.2} \\ &= 10.47~\text{m} \end{aligned}$$Thus, wine column height is approximately 10.5 m for normal atmospheric pressure, much taller than mercury's 0.76 m due to lower density.
Q7. A vertical off-shore structure is built to withstand a maximum stress of \(\mathrm{10^9}\) Pa. Is the structure suitable for putting up on top of an oil well in the ocean ? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.
Solution
Given: Maximum stress capacity \(10^9\) Pa, ocean depth \(h = 3\) km \(= 3 \times 10^3\) m, seawater density \(\rho = 1.03 \times 10^3\) kg/m³, \(g = 9.8\) m/s², \(P_0 = 1.01 \times 10^5\) Pa.
Hydrostatic pressure at ocean bottom,
$$\begin{aligned} P &= P_0 + h \rho g \\ &= 1.01 \times 10^5 + (3 \times 10^3) \times (1.03 \times 10^3) \times 9.8 \end{aligned}$$Compute hydrostatic term,
$$\begin{aligned} h \rho g &= 3000 \times 1030 \times 9.8 \\ &= 3.09 \times 10^7~\text{Pa} \\ P &= 1.01 \times 10^5 + 3.09 \times 10^7 \\ &= 3.10 \times 10^7~\text{Pa} \approx 0.31 \times 10^9~\text{Pa} \end{aligned}$$Since \(P \approx 0.31 \times 10^9\) Pa \(< 10^9\) Pa, the structure can safely withstand bottom pressure, making it suitable atop the oil well.
Q8. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is \(\mathrm{425\ cm^2}\). What maximum pressure would the smaller piston have to bear ?
Solution
Given: Maximum mass \(m = 3000\) kg, area of large piston \(A = 425\) cm² \(= 425 \times 10^{-4}\) m² \(= 0.0425\) m², \(g = 9.8\) m/s².
Weight of car (force on large piston),
$$\begin{aligned} F &= mg \\ &= 3000 \times 9.8 \\ &= 2.94 \times 10^4~\text{N} \end{aligned}$$Pressure on large piston,
$$\begin{aligned} P &= \frac{F}{A} \\ &= \frac{2.94 \times 10^4}{0.0425} \\ &= 6.92 \times 10^5~\text{Pa} \end{aligned}$$By Pascal's law, pressure transmits undiminished throughout incompressible fluid, so smaller piston bears identical maximum pressure of \(6.92 \times 10^5\) Pa.
Q9. A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit ?
Solution
Given: Height of water column \(h_1 = 10.0\) cm, height of spirit column \(h_2 = 12.5\) cm, relative density of water \(\rho_1 = 1\), mercury levels equal in both arms.
Pressure balance at mercury interface level,
$$\begin{aligned} h_1 \rho_1 g &= h_2 \rho_2 g \\ 10 \times 1 &= 12.5 \times S \\ S &= \frac{10 \times 1}{12.5} \end{aligned}$$where \(S\) is specific gravity (relative density) of spirit.
Compute,
$$\begin{aligned} S &= \frac{10}{12.5} \\ &= 0.8 \end{aligned}$$Thus, the specific gravity of methylated spirit is 0.8.
Q10. In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6)
Solution
Given: Additional 15.0 cm water and spirit poured; initial from previous: water 10.0 cm, spirit 12.5 cm (specific gravity \(S = 0.8\)); total water height \(h_w = 25.0\) cm, spirit height \(h_s = 27.5\) cm; mercury specific gravity 13.6.
Pressure balance (water side = spirit side + mercury difference),
$$\begin{aligned} \rho_w g h_w &= \rho_s g h_s + \rho_m g h_m \\ 1 \times 25 &= 0.8 \times 27.5 + 13.6 \times h_m \end{aligned}$$Solve for mercury level difference \(h_m\),
$$\begin{aligned} 25 &= 22 + 13.6 h_m \\ 13.6 h_m &= 3 \\ h_m &= \frac{3}{13.6} \\ &= 0.221~\text{cm} \end{aligned}$$Mercury level is lower by about 0.22 cm on the Spirit side.
Q11. Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain.
Solution
Bernoulli's equation \(P + \rho g h + \frac{1}{2} \rho v^2 =\) constant applies strictly to ideal, steady, incompressible, non-viscous flow along streamlines.
Rapids feature highly turbulent, chaotic water motion with eddies and vortices, violating the steady streamline assumption and ideal fluid conditions.
Significant viscous dissipation converts mechanical energy to heat, destroying the total head required for Bernoulli conservation. Thus, the equation cannot accurately describe rapid flow.
Q12. Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain.
Solution
No, it does not matter whether gauge or absolute pressures are used in Bernoulli's equation.
Consider Bernoulli's equation between two points:
$$\begin{aligned} P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 &= P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2 \end{aligned}$$Using absolute pressures \(P_1 = P_{g1} + P_\text{atm}\), \(P_2 = P_{g2} + P_\text{atm}\):
$$\scriptsize\begin{aligned} (P_{g1} + P_\text{atm}) + \frac{1}{2}\rho v_1^2 + \rho g h_1 &= (P_{g2} + P_\text{atm}) + \frac{1}{2}\rho v_2^2 + \rho g h_2 \end{aligned}$$Atmospheric pressure \(P_\text{atm}\) cancels from both sides:
$$\scriptsize\begin{aligned} P_{g1} + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_{g2} + \frac{1}{2}\rho v_2^2 + \rho g h_2 \end{aligned}$$Thus, gauge pressures yield identical relations as absolute pressures; only pressure differences matter in Bernoulli applications.
Q13. Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is \(\mathrm{4.0 × 10^{–3}\ kg\ s^{–1}}\), what is the pressure difference between the two ends of the tube ? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].
Solution
Since glycerine flows steadily through a horizontal tube, the motion is governed by Poiseuille’s law for viscous flow. The given mass flow rate is first converted into volume flow rate using the density of glycerine.
\[ \begin{aligned} \dot{m} &= 4.0 \times 10^{-3}\,\text{kg s}^{-1}, \\ \rho &= 1.3 \times 10^{3}\,\text{kg m}^{-3} \end{aligned} \]
\[ \begin{aligned} Q &= \frac{\dot{m}}{\rho}\\ &= \frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}}\\ &= 3.08 \times 10^{-6}\,\text{m}^{3}\text{s}^{-1} \end{aligned} \]
For laminar flow through a tube of radius \(r\) and length \(l\), Poiseuille’s equation is
\[ \begin{aligned} Q = \frac{\pi r^{4}\Delta P}{8\eta l} \end{aligned} \]
Solving for the pressure difference \(\Delta P\),
\[ \begin{aligned} \Delta P &= \frac{8\eta l Q}{\pi r^{4}} \end{aligned} \]
Substituting the given values \(\small \eta = 0.83\,\text{Pa s},\; l = 1.5\,\text{m},\; r = 0.01\,\text{m} \)
\[ \begin{aligned} \Delta P &=\scriptsize \frac{8 \times 0.83 \times 1.5 \times 3.08 \times 10^{-6}} {\pi \times (0.01)^{4}} \\ &\approx 9.8 \times 10^{2}\,\text{Pa} \end{aligned} \]
Hence, the pressure difference between the two ends of the tube is approximately
\[ \boxed{\Delta P \approx 1.0 \times 10^{3}\,\text{Pa}} \]
To verify the assumption of laminar flow, the average speed of flow is calculated as
\[ \begin{aligned} v &= \frac{Q}{\pi r^{2}}\\ &= \frac{3.08 \times 10^{-6}}{\pi \times (0.01)^{2}}\\ &\approx 9.8 \times 10^{-3}\,\text{m s}^{-1} \end{aligned} \]
The Reynolds number is then
\[ \begin{aligned} Re &= \frac{\rho v D}{\eta}\\ &=\scriptsize \frac{(1.3 \times 10^{3})(9.8 \times 10^{-3})(0.02)}{0.83}\\ &\approx 0.3 \end{aligned} \]
Since \(Re \ll 2000\), the flow is well within the laminar regime, and the use of Poiseuille’s law is fully justified.
Q14. In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s–1and \(\mathrm{63\ m\ s^{-1}}\) respectively. What is the lift on the wing if its area is \(\mathrm{2.5\ m^2}\) ? Take the density of air to be \(\mathrm{1.3\ kg\ m^{–3}}\).
Solution
Given: Upper surface speed \(v_1 = 70\) m/s, lower surface speed \(v_2 = 63\) m/s, wing area \(A = 2.5\) m², air density \(\rho = 1.3\) kg/m³.
Pressure difference (Bernoulli, same height),
$$\begin{aligned} \Delta P &= \frac{1}{2} \rho (v_1^2 - v_2^2) \\ v_1^2 - v_2^2 &= 70^2 - 63^2 \\ &= 4900 - 3969 \\&= 931~\text{m}^2\text{/s}^2 \\ \Delta P &= \frac{1}{2} \times 1.3 \times 931 \\ &= 605~\text{Pa} \end{aligned}$$Lift force,
$$\begin{aligned} L &= \Delta P \cdot A \\ &= 605 \times 2.5 \\ &= 1513~\text{N} \approx 1.5 \times 10^3~\text{N} \end{aligned}$$The lift acts upward as lower pressure above wing.
Q15. Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why
Solution
Figure (b) is incorrect.
In steady incompressible flow, continuity requires \(A_1 v_1 = A_2 v_2\). Figure (b) shows symmetric narrowing with \(v_1 > v_2\), violating mass conservation since smaller \(A_2\) demands higher \(v_2 > v_1\).
Figure (a) correctly depicts converging section where \(v_2 > v_1\) as required, consistent with Bernoulli's principle lowering pressure where speed rises.
Q16. The cylindrical tube of a spray pump has a cross-section of \(\mathrm{8.0\ cm^2}\) one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is \(\mathrm{1.5\ m\ min^{–1}}\), what is the speed of ejection of the liquid through the holes ?
Solution
Given: Tube cross-section \(A_1 = 8.0\) cm² \(= 8.0 \times 10^{-4}\) m², speed inside tube \(v_1 = 1.5\) m/min \(= 1.5 / 60 = 0.025\) m/s, 40 holes each diameter 1.0 mm so radius \(r = 0.5 \times 10^{-3}\) m.
Area of one hole,
$$\begin{aligned} a &= \pi r^2 \\ &= \pi \left( \frac{1}{2} \times 10^{-3} \right)^2 \\ &= 7.85 \times 10^{-7}~\text{m}^2 \end{aligned}$$Total holes area \(\scriptsize A_2 = 40 \times 7.85 \times 10^{-7} = 3.14 \times 10^{-5}\) m².
By continuity \(A_1 v_1 = A_2 v_2\),
$$\begin{aligned} v_2 &= \frac{A_1 v_1}{A_2} \\ &= \frac{8.0 \times 10^{-4} \times 0.025}{3.14 \times 10^{-5}} \\ &= \frac{2.0 \times 10^{-5}}{3.14 \times 10^{-5}} \\ &= 0.637~\text{m/s} \approx 0.64~\text{m/s} \end{aligned}$$Liquid ejects through holes at 0.64 m/s.
Q17. A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of \(\mathrm{1.5 × 10^{–2}\ N}\) (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film ?
Solution
Given: Weight supported \(F = 1.5 \times 10^{-2}\) N (includes slider), slider length \(L = 30\) cm \(= 0.30\) m.
Soap film has two surfaces (front and back), total length along which tension acts,
$$\begin{aligned} l &= 2L \\ &= 2 \times 0.30 \\ &= 0.60~\text{m} \end{aligned}$$Surface tension,
$$\begin{aligned} T &= \frac{F}{l} \\ &= \frac{1.5 \times 10^{-2}}{0.60} \\ &= 2.5 \times 10^{-2}~\text{N/m} \end{aligned}$$Thus, surface tension of film is \(2.5 \times 10^{-2}\) N/m.
Q18. Figure 9.21 (a) shows a thin liquid film supporting a small weight = \(\mathrm{4.5 × 10^{–2}}\) N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c) ? Explain your answer physically.
Solution
In Fig. (a), the film of width 40 cm supports a weight \(W_a = 4.5 \times 10^{-2}\ \text{N}\). Since a soap film has two free surfaces; the supporting force is \(F = 2 T L\), where \(T\) is the surface tension and \(L\) is the length of the horizontal edge (40 cm).
From Fig. (a), the surface tension is
$$\begin{aligned} W_a &= 2 T L \\ 4.5 \times 10^{-2} &= 2 T \times 0.40 \\ T &= \frac{4.5 \times 10^{-2}}{0.80}\\ &= 5.625 \times 10^{-2}\ \text{N m}^{-1}. \end{aligned}$$In Fig. (b), the width of the film at the bottom is again 40 cm, so the effective length \(L\) in contact with the slider is still 0.40 m. Hence, the supporting force remains
$$\begin{aligned} W_b &= 2 T L \\ &= 2 \times 5.625 \times 10^{-2} \times 0.40 \\ &= 4.5 \times 10^{-2}\ \text{N}. \end{aligned}$$In Fig. (c), the bottom edge in contact with the slider is also 40 cm, so the same reasoning applies: only the length of the line of contact with the slider matters, not the shape of the rest of the film. Therefore,
$$\begin{aligned} W_c &= 2 T L \\ &= 2 \times 5.625 \times 10^{-2} \times 0.40 \\ &= 4.5 \times 10^{-2}\ \text{N}. \end{aligned}$$Thus, each film in Fig. (a), (b) and (c) supports the same weight \(4.5 \times 10^{-2}\ \text{N}\), because surface tension force depends only on surface tension and the length of the slider in contact with the film, which is identical in all three cases.
Q19. What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20 °C) is \(\mathrm{4.65 × 10^{–1}\ N\ m^{–1}}\). The atmospheric pressure is \(\mathrm{1.01 × 10^5\ Pa}\). Also give the excess pressure inside the drop.
Solution
Given: Radius of mercury drop \(r = 3.00~\text{mm} = 3.0 \times 10^{-3}~\text{m}\), surface tension \(T = 4.65 \times 10^{-1}~\text{N m}^{-1}\), atmospheric pressure \(P_0 = 1.01 \times 10^{5}~\text{Pa}\).
Excess pressure inside a spherical drop is
$$\begin{aligned} \Delta P &= \frac{2T}{r} \\ &= \frac{2 \times 4.65 \times 10^{-1}}{3.0 \times 10^{-3}} \\ &= \frac{0.93}{3.0 \times 10^{-3}} \\ &= 3.1 \times 10^{2}~\text{Pa}. \end{aligned}$$Therefore, pressure inside the drop is
$$\begin{aligned} P &= P_0 + \Delta P \\ &= 1.01 \times 10^{5} + 3.1 \times 10^{2} \\ &= 1.0131 \times 10^{5}~\text{Pa}. \end{aligned}$$So, the excess pressure is \(3.1 \times 10^{2}~\text{Pa}\) and the absolute pressure inside the drop is \(1.0131 \times 10^{5}~\text{Pa}\).
Q20. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is \(\mathrm{2.50 ×10^{–2}\ N\ m^{–1}}\)? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is \(\mathrm{1.01 × 10^5\ Pa}\)).
Solution
Given: Radius of soap bubble \(r = 5.00~\text{mm} = 5.0 \times 10^{-3}~\text{m}\), surface tension \(T = 2.50 \times 10^{-2}~\text{N m}^{-1}\).
For a soap bubble (two liquid–air surfaces), excess pressure is
$$\begin{aligned} \Delta P &= \frac{4T}{r} \\ &= \frac{4 \times 2.50 \times 10^{-2}}{5.0 \times 10^{-3}} \\ &= \frac{0.10}{5.0 \times 10^{-3}} \\ &= 2.0 \times 10^{1}~\text{Pa} \\ &= 20~\text{Pa}. \end{aligned}$$Thus, excess pressure inside soap bubble in air is \(20~\text{Pa}\).
Now, for an air bubble of same radius at depth \(h = 40.0~\text{cm} = 0.40~\text{m}\) in soap solution of relative density 1.20, density \(\rho = 1.20 \times 10^{3}~\text{kg m}^{-3}\). Outside pressure at that depth is
$$\begin{aligned} P_{\text{out}} &= P_0 + h \rho g \\ &=\scriptsize 1.01 \times 10^{5} + 0.40 \times 1.20 \times 10^{3} \times 9.8 \\ &= 1.01 \times 10^{5} + 4.70 \times 10^{3} \\ &= 1.057 \times 10^{5}~\text{Pa}. \end{aligned}$$For an air bubble in liquid (one liquid–air surface), excess pressure is \(\Delta P' = \dfrac{2T}{r}\),
$$\begin{aligned} \Delta P' &= \frac{2 \times 2.50 \times 10^{-2}}{5.0 \times 10^{-3}} \\ &= \frac{0.05}{5.0 \times 10^{-3}} \\ &= 1.0 \times 10^{1}~\text{Pa} \\ &= 10~\text{Pa}. \end{aligned}$$Therefore, pressure inside the air bubble is
$$\begin{aligned} P_{\text{in}} &= P_{\text{out}} + \Delta P' \\ &= 1.057 \times 10^{5} + 10 \\ &= 1.058 \times 10^{5}~\text{Pa}. \end{aligned}$$So, excess pressure in the soap bubble is \(20~\text{Pa}\), and pressure inside the air bubble at 40 cm depth is \(1.058 \times 10^{5}~\text{Pa}\).
