The Workbook

1. Surface tension is the force per unit length acting along a liquid surface, perpendicular to any line drawn in the surface. Unit: N/m, which equals J/m\(^2\).

2. The film has two surfaces, front and back, and each pulls the slider with \(Tl\). So \(F = 2Tl = 2 \times 0.03 \times 0.10 = 6 \times 10^{-3}\) N.

1. \(W = F \times d = 6 \times 10^{-3} \times 0.05 = 3 \times 10^{-4}\) J.

2. Each surface grows by \(l \times d = 0.10 \times 0.05 = 5 \times 10^{-3}\) m\(^2\). Two surfaces: \(\Delta A = 10^{-2}\) m\(^2\).

3. \(T\,\Delta A = 0.03 \times 10^{-2} = 3 \times 10^{-4}\) J \(= W\). The work goes into surface energy.

A bubble has two surfaces: \(A = 2 \times 4\pi r^2 = 8\pi \times (0.05)^2 = 2\pi \times 10^{-2}\) m\(^2\).

\(E = TA = 0.03 \times 2\pi \times 10^{-2} = 6\pi \times 10^{-4} \approx 1.9 \times 10^{-3}\) J.

\(W = T\,\Delta A = T \times 8\pi(r_2^2 - r_1^2) = 0.03 \times 8\pi \times (0.01 - 0.0025) = 1.8\pi \times 10^{-3} \approx 5.7 \times 10^{-3}\) J.

1. Volume is conserved: \(8 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi(2)^3\), so \(r^3 = 1\) mm\(^3\) and \(r = 1\) mm.

2. \(\Delta A = 8 \times 4\pi(1)^2 - 4\pi(2)^2 = 32\pi - 16\pi = 16\pi\) mm\(^2 = 16\pi \times 10^{-6}\) m\(^2\).

   \(W = T\,\Delta A = 0.07 \times 16\pi \times 10^{-6} = 1.12\pi \times 10^{-6} \approx 3.5 \times 10^{-6}\) J.

3. Energy must be supplied to make the new surface. If it comes from the water itself, the water cools slightly.

A molecule inside is pulled equally by neighbours on all sides. A molecule at the surface has neighbours only below and beside it, so it feels a net inward pull.

Work must be done against this pull to bring a molecule to the surface, so surface molecules have extra potential energy. The liquid lowers its energy by making its surface as small as it can. The shape with the least surface for a given volume is a sphere.

Surface tension falls as the temperature rises, because the faster molecules are held less strongly by their neighbours. It becomes zero at the critical temperature.

Soap lowers the surface tension of water. The water then spreads more easily into the fibres of cloth and around dirt particles, so it wets and washes better.

Let the radius grow from \(r\) to \(r + dr\). The inside pressure does work \(\Delta P \times\) (area) \(\times dr = \Delta P \times 4\pi r^2\,dr\).

The surface area grows by \(d(4\pi r^2) = 8\pi r\,dr\), needing surface energy \(T \times 8\pi r\,dr\).

Equate: \(\Delta P \times 4\pi r^2\,dr = 8\pi r T\,dr\), so \(\Delta P = \dfrac{2T}{r}\). Smaller drops have larger excess pressure.

1. \(\Delta P = \dfrac{2T}{r} = \dfrac{2 \times 0.07}{10^{-3}} = 140\) Pa.

2. One surface again, water outside: \(\Delta P = \dfrac{2 \times 0.07}{10^{-3}} = 140\) Pa.

3. Two surfaces: \(\Delta P = \dfrac{4T}{r} = \dfrac{4 \times 0.03}{0.01} = 12\) Pa.

A soap bubble is a thin film of liquid with air on both sides, so it has two surfaces, an inner and an outer one. Each surface needs \(\dfrac{2T}{r}\) of pressure difference across it, and the two add to \(\dfrac{4T}{r}\).

1. Bubble \(A\). The excess pressure \(\dfrac{4T}{r}\) is larger for the smaller radius.

2. Air flows from the higher pressure in \(A\) to the lower pressure in \(B\). \(A\) shrinks and \(B\) grows, until \(A\) is only a flat film across the end of the tube.

1. The smaller bubble has the higher pressure, so it pushes the common wall outward, into the larger bubble.

2. Pressure in the small bubble: \(P_0 + \dfrac{4T}{r_1}\); in the large one: \(P_0 + \dfrac{4T}{r_2}\). The common wall is a soap film of radius \(r\) with the small bubble on its concave side. The pressure difference across it is \(\dfrac{4T}{r}\).

   \(\dfrac{4T}{r} = \dfrac{4T}{r_1} - \dfrac{4T}{r_2}\), so \(\dfrac{1}{r} = \dfrac{1}{r_1} - \dfrac{1}{r_2}\) and \(r = \dfrac{r_1 r_2}{r_2 - r_1} = \dfrac{2 \times 4}{4 - 2} = 4\) cm.

In a vacuum the pressure inside a bubble is just \(\dfrac{4T}{r}\). Isothermal: \(PV\) of the air is conserved, \(P_1V_1 + P_2V_2 = PV\).

\(\dfrac{4T}{r_1} \cdot \dfrac{4}{3}\pi r_1^3 + \dfrac{4T}{r_2} \cdot \dfrac{4}{3}\pi r_2^3 = \dfrac{4T}{R} \cdot \dfrac{4}{3}\pi R^3\), which gives \(r_1^2 + r_2^2 = R^2\).

\(R = \sqrt{9 + 16} = 5\) cm. (The surface energy is conserved: \(T \times 8\pi R^2 = T \times 8\pi(r_1^2 + r_2^2)\).)

Pressure of the water at the bubble: \(P_0 + h\rho g = 10^{5} + 0.10 \times 1000 \times 10 = 1.01 \times 10^{5}\) Pa.

Excess pressure inside: \(\dfrac{2T}{r} = \dfrac{2 \times 0.07}{10^{-4}} = 1400\) Pa.

Air pressure \(= 101000 + 1400 = 102400\) Pa \(= 1.024 \times 10^{5}\) Pa.

Drop: \(\dfrac{2 \times 0.07}{r} = \dfrac{0.14}{r}\). Bubble: \(\dfrac{4 \times 0.03}{r} = \dfrac{0.12}{r}\). Ratio \(= \dfrac{0.14}{0.12} = \dfrac{7}{6}\).

1. The surface pulls along both sides of the needle, a total length \(2l = 0.10\) m. The largest upward force is when the surface is vertical at the needle: \(F = 2Tl = 2 \times 0.07 \times 0.05 = 7 \times 10^{-3}\) N.

2. A \(0.5\) g needle weighs \(5 \times 10^{-3}\) N, less than \(7 \times 10^{-3}\) N: it floats. A \(1\) g needle weighs \(10^{-2}\) N, more than the surface can hold: it sinks.

1. The film clings to the inside and the outside of the ring: total length \(2 \times 2\pi r = 4\pi r = 4 \times 3.14 \times 0.02 \approx 0.25\) m.

2. Extra force \(= T \times 4\pi r = 0.07 \times 0.251 \approx 1.8 \times 10^{-2}\) N, about the weight of \(1.8\) g.

\(2Tl = 4 \times 10^{-3}\), so \(T = \dfrac{4 \times 10^{-3}}{2 \times 0.05} = 0.04\) N/m.

\(T = \dfrac{\text{force}}{\text{length}}\): \(\dfrac{\mathrm{MLT^{-2}}}{\mathrm{L}} = \mathrm{MT^{-2}}\).

Energy per area: \(\dfrac{\mathrm{ML^2T^{-2}}}{\mathrm{L^2}} = \mathrm{MT^{-2}}\), the same. Physically, the work to enlarge a surface by \(\Delta A\) against the force \(T\) per unit length is \(T\,\Delta A\) (Question 2). So \(T\) is also the energy stored per unit area.

1. \(1000 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi R^3\), so \(R = 10r = 1\) mm.

2. Area lost \(= 1000 \times 4\pi(0.1)^2 - 4\pi(1)^2 = 40\pi - 4\pi = 36\pi\) mm\(^2 = 36\pi \times 10^{-6}\) m\(^2\).

   Energy released \(= T\,\Delta A = 0.07 \times 36\pi \times 10^{-6} \approx 7.9 \times 10^{-6}\) J. The drop warms very slightly.

Surface tension pulls a drop into a sphere; gravity flattens it. For a small drop the surface energy (proportional to \(r^2\)) matters more than the gravitational energy (proportional to \(r^4\)). The drop stays round. For a large drop gravity wins and the drop spreads.

Two small bubbles have more surface area than one bubble of the same total volume. Merging lowers the surface energy, so the small bubbles join into larger ones.

1. Draw the tangent to the liquid surface where it meets the solid. The angle of contact is the angle between this tangent and the solid surface, measured inside the liquid.

   For water it is the small angle between the glass and the meniscus, below the surface. For mercury it is the obtuse angle, again measured through the liquid.

2. Water wets glass (angle less than \(90^\circ\)). Mercury has an angle of about \(135^\circ\), greater than \(90^\circ\), and does not wet glass.

Cohesion is the attraction between molecules of the liquid; adhesion is the attraction between the liquid and the glass.

For water the adhesion to glass is stronger than the cohesion. The water near the wall is pulled up the glass, and the surface curves down towards the centre. For mercury the cohesion is much stronger, so the mercury pulls away from the glass and its surface curves the other way.

Detergent lowers the surface tension and makes water wet glass better, so the angle of contact falls. Warming also lowers the surface tension and lowers the angle of contact.

1. The surface meets the glass along the rim of the tube, a circle of length \(2\pi r\). The surface tension pulls along the surface, at angle \(\theta\) to the wall, so its upward component is \(T\cos\theta\) per unit length.

   Total upward pull \(= 2\pi r T\cos\theta\).

2. Weight of the column \(= \pi r^2 h\rho g\). Equilibrium: \(2\pi r T\cos\theta = \pi r^2 h\rho g\), so \(h = \dfrac{2T\cos\theta}{r\rho g}\).

\(h = \dfrac{2T}{r\rho g} = \dfrac{2 \times 0.07}{10^{-4} \times 1000 \times 10} = \dfrac{0.14}{1} = 0.14\) m \(= 14\) cm.

\(h \propto \dfrac{1}{r}\): five times the radius gives one fifth of the rise.

1. \(r = 0.5\) mm: \(h = \dfrac{14}{5} = 2.8\) cm.

2. \(r = 1\) mm: \(h = \dfrac{14}{10} = 1.4\) cm.

\(h = \dfrac{2T\cos\theta}{r\rho g} = \dfrac{2 \times 0.5 \times (-0.7)}{5 \times 10^{-4} \times 13600 \times 10} = \dfrac{-0.7}{68} \approx -0.0103\) m.

The mercury inside stands about \(1\) cm below the outside level: a capillary depression.

1. \(h \propto \dfrac{1}{r}\), so doubling the radius halves the rise: \(h_2 = 3.5\) cm.

2. \(h_1 = \dfrac{2 \times 0.07}{2 \times 10^{-4} \times 1000 \times 10} = 0.07\) m \(= 7\) cm, as read from the figure.

\(m = \pi r^2 h\rho = \pi r^2\rho \times \dfrac{2T\cos\theta}{r\rho g} = \dfrac{2\pi r T\cos\theta}{g}\). The mass raised grows with \(r\), although the height falls.

For \(r = 0.1\) mm: \(m = \dfrac{2 \times 3.14 \times 10^{-4} \times 0.07}{10} = 4.4 \times 10^{-6}\) kg, about \(4\) mg.

1. No. The water rises to the top and stops. The meniscus flattens until the pressure difference across it, \(\dfrac{2T}{R}\), is just enough to hold the shorter column. It cannot push water out, because a flat or less curved surface gives less lift, never more.

2. The excess pressure \(\dfrac{2T}{R}\) must equal \(h'\rho g\) with \(h' = 7\) cm. Since \(hR\) is constant for a given liquid, \(R = r \times \dfrac{14}{7} = 2r = 0.2\) mm, twice the tube radius.

In the lift the effective gravity is \(g + a = 20\) m/s\(^2\). \(h \propto \dfrac{1}{g_{\text{eff}}}\), so the rise halves to \(7\) cm.

In free fall \(g_{\text{eff}} = 0\) and the formula gives an infinite height. The water rises to the top of the tube and fills it, whatever its length.

\(h \propto \dfrac{1}{g}\). With \(g\) six times smaller the rise is six times larger: \(6 \times 14 = 84\) cm.

The pressure balance fixes the vertical height of the column, which stays \(14\) cm.

Length along the tube \(= \dfrac{h}{\cos 60^\circ} = \dfrac{14}{0.5} = 28\) cm.

\(h \propto \cos\theta\) and \(\cos 60^\circ = 0.5\). The rise is half that of Question 5: \(7\) cm.

1. The film of width \(w\) touches both plates along its top edge: total length \(2w\), so the upward pull is \(2wT\) (with \(\theta = 0\)).

2. Weight of the film \(= w d h\rho g\). Equilibrium: \(2wT = wdh\rho g\), so \(h = \dfrac{2T}{d\rho g}\).

   \(h = \dfrac{2 \times 0.07}{2 \times 10^{-4} \times 1000 \times 10} = 0.07\) m \(= 7\) cm.

A tube of diameter \(d\) has radius \(\dfrac{d}{2}\): \(h_{\text{tube}} = \dfrac{2T}{(d/2)\rho g} = \dfrac{4T}{d\rho g}\), twice the rise between the plates.

The tube pulls the water up along a circle all round, while the plates pull only along two lines. So the tube raises the water twice as high.

1. The water touches the outer wall of the inner tube and the inner wall of the outer tube: total length \(2\pi r_1 + 2\pi r_2\). Upward pull \(= 2\pi(r_1 + r_2)T\).

   Weight \(= \pi(r_2^2 - r_1^2)h\rho g = \pi(r_2 - r_1)(r_2 + r_1)h\rho g\). Equating and cancelling \((r_1 + r_2)\): \(h = \dfrac{2T}{(r_2 - r_1)\rho g}\).

2. \(r_2 - r_1 = 0.2\) mm: \(h = \dfrac{2 \times 0.07}{2 \times 10^{-4} \times 10^{4}} = 0.07\) m \(= 7\) cm, the same as two plates \(0.2\) mm apart.

1. Gap \(= \alpha x = 0.01 \times 0.02 = 2 \times 10^{-4}\) m \(= 0.2\) mm. Between plates \(0.2\) mm apart the rise is \(7\) cm (Question 15).

2. At \(x = 4\) cm the gap is \(0.4\) mm and the rise halves: \(3.5\) cm.

   \(h = \dfrac{2T}{\alpha x\rho g}\), so \(hx\) is constant: the top edge of the water is a rectangular hyperbola.

\(T = \dfrac{h r\rho g}{2} = \dfrac{0.07 \times 10^{-4} \times 800 \times 10}{2} = 0.028\) N/m.

\(h \propto T\): \(h = 14 \times \dfrac{0.03}{0.07} = 6\) cm.

Blotting paper and a towel soak up water: the tiny gaps between the fibres act as capillaries. Oil climbs the wick of a lamp: the gaps between the threads are the tubes.

Water rises through soil to the roots of plants: the spaces between soil grains are the capillaries. Ink spreads along a nib for the same reason.