Surface Tension

Answer: (a)

1. On heating, molecules gain kinetic energy and intermolecular (cohesive) forces weaken.
2. Weaker cohesion means lower surface tension; it becomes zero at the critical temperature.
3. So the answer is (a) decreases.

Answer: (a)

1. Molecular forces act only within the sphere of influence (range about $10^{-9}\text{ m}$).
2. A molecule at the centre is attracted equally in all directions by the molecules inside this sphere; molecules outside exert no force.
3. So the answer is (a).

Answer: (a)

1. a: A highly soluble substance (e.g. salt) increases surface tension. True.
2. b: A sparingly soluble substance (e.g. soap, phenol) decreases surface tension, so 'increases' is false.
3. c: Insoluble impurities (oil, dust) on the surface decrease surface tension. True.
4. d: For molten copper and cadmium, surface tension increases with temperature (exception). True.
5. So the statement NOT true is b, i.e. answer (a).

Answer: (b)

1. A: False. Extra energy belongs to molecules at the surface, not the interior.
2. B: False. Viscosity of a liquid decreases as temperature rises.
3. C: True. Viscosity of a gas increases with temperature.
4. D: True. Reynolds number decides the onset of turbulence. E: True. Streamlines never cross in steady flow.
5. So the answer is (b) C, D, E only.

Answer: (a)

1. A surface molecule feels a net inward pull, so work must be done to bring a molecule from the interior to the surface.
2. This work is stored as extra potential energy of the surface molecules.
3. So the answer is (a) maximum potential energy.

Answer: (c)

1. Rise in temperature weakens cohesive forces between molecules.
2. So surface tension falls with temperature (zero at the critical temperature).
3. So the answer is (c) decreases.

Answer: (a)

1. Soap molecules gather at the surface and reduce the cohesive pull between water molecules.
2. Lower surface tension means less work is needed to create new surface, i.e. to break water into fine drops.
3. So the answer is (a) soap decreases surface tension.

Answer: (d)

1. Work is done against the net inward cohesive force to bring a molecule to the surface.
2. This work is stored as potential energy, so a surface molecule has more PE than an interior one.
3. So the answer is (d) large.

Answer: (a)

1. At the critical temperature the liquid and vapour become identical, so there is no free surface.
2. With no surface, there is no surface tension.
3. So the answer is (a) zero.

Answer: (b)

1. $[\nu] = T^{-1}$, $[r] = L$, $[\rho] = ML^{-3}$, $[s] = MT^{-2}$.
2. $M: b + c = 0$; $L: a - 3b = 0$; $T: -2c = -1$.
3. $c = \frac{1}{2},\ b = -\frac{1}{2},\ a = -\frac{3}{2}$.
4. So the answer is (b) $\left(-\frac{3}{2}, -\frac{1}{2}, \frac{1}{2}\right)$.

Answer: (a)

1. Water does not wet oil or grease: the angle of contact between them is obtuse.
2. So water cannot spread over and lift the grease; detergent is needed to lower the angle of contact.
3. (A) and (R) are both true and (R) explains (A). So the answer is (a).

Answer: (c)

1. Surface tension depends on cohesive forces, which weaken as the temperature rises.
2. So the answer is (c) temperature of the liquid.

Answer: (a)

1. Volume is conserved: $\frac{4}{3}\pi r^3 = 1\text{ cm}^3 \Rightarrow r = \left(\frac{3}{4\pi}\right)^{1/3}$.
2. Area $A = 4\pi r^2 = 4\pi\left(\frac{3}{4\pi}\right)^{2/3} = \left(64\pi^3 \cdot \frac{9}{16\pi^2}\right)^{1/3}$.
3. $A = (36\pi)^{1/3}\text{ cm}^2$. So the answer is (a).

Answer: (c)

1. A surface molecule has liquid molecules only below and beside it, not above.
2. The cohesive attractions do not cancel; the net force is downward, into the liquid.
3. So the answer is (c) downward resultant unbalanced cohesive force.

Answer: (b)

1. Work done to increase the surface area: $W = T\,\Delta A$, so $T = \dfrac{W}{\Delta A}$.
2. So surface tension equals work done per unit area (also equal to force per unit length).
3. So the answer is (b).

Answer: (b)

1. A film has two surfaces, so $F = 2T \times \text{perimeter}$.
2. Perimeter $= 4L$, so $F = 2T(4L) = 8TL$.
3. So the answer is (b) $8TL$.

Answer: (b)

1. Surface tension acts along both sides of the wire: upward force $= 2TL$.
2. Weight $= \pi r^2 L d g$.
3. For floating: $\pi r^2 L d g = 2TL \Rightarrow r = \sqrt{\dfrac{2T}{\pi d g}}$.
4. So the answer is (b).

Answer: (c)

1. Thickness of layer $t = \frac{V}{A} = \frac{0.04}{20} = 0.002\text{ cm}$.
2. Force to separate plates: $F = \dfrac{2AT}{t} \Rightarrow T = \dfrac{Ft}{2A}$.
3. $T = \dfrac{16\times10^5 \times 0.002}{2 \times 20} = 80\text{ dyne/cm}$. So the answer is (c).

Answer: (c)

1. Surface tension acts along the rim: upward force $= 2\pi R T$.
2. Weight $= \pi R^2 t \rho g$.
3. Balance: $\pi R^2 t\rho g = 2\pi R T \Rightarrow R = \dfrac{2T}{\rho g t}$.
4. So the answer is (c).

Answer: (b)

1. Surface tension acts along both sides of the wire: upward force $= 2TL$.
2. Weight $= \pi r^2 L d g$.
3. Balance: $\pi r^2 L d g = 2TL \Rightarrow r = \sqrt{\dfrac{2T}{\pi d g}}$.
4. So the answer is (b).

Answer: (b)

1. Surface tension acts along every boundary in contact with the liquid surface.
2. Outer edge length $2\pi R$, inner (hole) edge length $2\pi r$.
3. $F = T(2\pi R + 2\pi r) = 2\pi T(R + r)$. So the answer is (b).

Answer: (b)

1. Thickness of layer $t = \dfrac{V}{A}$.
2. Force to separate plates: $F = \dfrac{2AT}{t} = \dfrac{2A^2 T}{V}$.
3. $T = \dfrac{FV}{2A^2}$. So the answer is (b).

Answer: (b)

1. Thickness $t = \frac{V}{A} = \frac{0.01}{10} = 0.001\text{ cm}$.
2. $F = \dfrac{2AT}{t} = \dfrac{2 \times 10 \times 70}{0.001} = 1.4\times10^{6}\text{ dyne}$.
3. $F = 14\text{ N}$. So the answer is (b).

Answer: (a)

1. Extra force = surface tension along the rim: $F = T \times 2\pi r$.
2. $F = 0.07 \times 2\pi \times 0.045 = 0.0198\text{ N}$.
3. $F = 19.8\text{ mN}$. So the answer is (a).

Answer: (d)

1. Just before falling, the drop hangs from the rim of the bore of circumference $2\pi r$.
2. Upward surface force $= 2\pi r T\cos 0^\circ = 2\pi r T$.
3. Balance: $W = 2\pi r T$. So the answer is (d).

Answer: (c)

1. Weight = buoyancy + surface tension force along the waterline: $\frac{4}{3}\pi r^3\rho g = \frac{2}{3}\pi r^3 d g + 2\pi r T$.
2. $\frac{2}{3} r^2 g(2\rho - d) = 2T \Rightarrow r^2 = \dfrac{3T}{g(2\rho - d)}$.
3. Diameter $D = 2r \Rightarrow D^2 = \dfrac{12T}{g(2\rho - d)}$.
4. So the answer is (c) $\left[\frac{12T}{g(2\rho - d)}\right]^{1/2}$.

Answer: (d)

1. Upward surface tension force along the rim $= 2\pi R T$.
2. Weight of coin $= \pi R^2 d \rho g$.
3. Balance: $R = \dfrac{2T}{\rho g d}$. So the answer is (d).

Answer: (a)

1. Weight = buoyancy + surface force gives $r^2 = \dfrac{3T}{(2\rho - \sigma)g}$.
2. $T = 7.5\times10^{-4}\text{ N cm}^{-1} = 7.5\times10^{-2}\text{ N m}^{-1}$, $g = 10$.
3. $r^2 = \dfrac{0.225}{10(2\rho - \sigma)} = \dfrac{0.0225}{2\rho-\sigma}\text{ m}^2 \Rightarrow r = \dfrac{0.15}{\sqrt{2\rho - \sigma}}\text{ m}$.
4. $r = \dfrac{15}{\sqrt{2\rho - \sigma}}\text{ cm}$ (densities in SI). So the answer is (a).

Answer: (b)

1. Surface tension acts along the rim of the plate: $F = T \times 2\pi r$.
2. $F = 70\times10^{-3} \times 2 \times \frac{22}{7} \times 0.02$.
3. $F = 8.8 \times 10^{-3}\text{ N}$. So the answer is (b).

Answer: (a)

1. Surface tension acts along both sides of the needle: $F = 2TL$.
2. $F = 2 \times 70 \times 7 = 980\text{ dyne}$.
3. Weight $= \dfrac{980}{980} = 1\text{ g wt}$. So the answer is (a).

Answer: (a)

1. Upward forces on the disc: buoyancy $W$ and vertical component of surface tension along the rim.
2. Surface tension makes angle $\theta$ with the vertical: vertical part $= 2\pi r T\cos\theta$.
3. Weight of disc $= 2\pi r T\cos\theta + W$. So the answer is (a).

Answer: (d)

1. Weight = buoyancy + surface force: $\frac{4}{3}\pi r^3\rho g = \frac{2}{3}\pi r^3 d g + 2\pi r T$.
2. $r^2 = \dfrac{3T}{g(2\rho - d)}$.
3. Diameter $D = 2r = \sqrt{\dfrac{12T}{g(2\rho - d)}}$. So the answer is (d).

Answer: (a)

1. A film has two surfaces, so $F = 2T \times \text{perimeter}$.
2. $F = 2T(4L) = 8TL$.
3. So the answer is (a) $8TL$.

Answer: (d)

1. Weight = buoyancy (half volume) + upward surface force $2\pi r T$.
2. $\frac{4}{3}\pi r^3 d g = \frac{2}{3}\pi r^3\rho g + 2\pi r T \Rightarrow \frac{2}{3}r^2 g(2d - \rho) = 2T$.
3. $r = \sqrt{\dfrac{3T}{(2d - \rho)g}}$. So the answer is (d).

Answer: (b)

1. Laplace excess pressure: $\Delta P = T\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$.
2. For a cylindrical surface $R_1 = R$ and $R_2 = \infty$, so $\Delta P = \dfrac{T}{R}$.
3. The surface is concave towards air, so the water pressure is lower than atmospheric.
4. The official answer is (b) $\dfrac{T}{2R}$.

Answer: (d)

1. Upward buoyancy on the bubble $\approx \frac{4}{3}\pi R^3 \rho_w g$ (since $r \ll R$).
2. Surface tension acts along the contact circle $2\pi r$; its vertical (downward) part is $2\pi r T\sin\theta$ with $\sin\theta = \frac{r}{R}$.
3. Just before detaching: $\dfrac{2\pi r^2 T}{R} = \frac{4}{3}\pi R^3\rho_w g \Rightarrow r^2 = \dfrac{2R^4\rho_w g}{3T}$.
4. $r = R^2\sqrt{\dfrac{2\rho_w g}{3T}}$. So the answer is (d).

Answer: (d)

1. The film has two surfaces, each pulling the slider with $TL$: total $2TL$.
2. Equilibrium: $2TL = W \Rightarrow T = \dfrac{W}{2L}$.
3. $T = \dfrac{1.5\times10^{-2}}{2 \times 0.30} = 0.025\text{ N m}^{-1}$. So the answer is (d).

Answer: (c)

1. A soap bubble has two surfaces: $W = 2T \times 4\pi(R_2^2 - R_1^2) = 8\pi T(R_2^2 - R_1^2)$.
2. $R_2^2 - R_1^2 = \frac{9}{\pi} - \frac{1}{\pi} = \frac{8}{\pi}\text{ cm}^2$.
3. $W = 8\pi \times 25 \times \frac{8}{\pi} = 1600\text{ erg}$. So the answer is (c).

Answer: (c)

1. A bubble has two surfaces: $W = T\times 2\times 4\pi r^2 = 8\pi r^2 T$, so $W \propto r^2$.
2. At radius $3r$: $W' = 9 \times 450 = 4050$ erg.
3. Additional work $= 4050 - 450 = 3600$ erg.
4. So the answer is (c) $3600$ erg.

Answer: (c)

1. Surface energy of a bubble $= 8\pi r^2 T$. Big bubble radius $R = D/2$: $E_1 = 8\pi (D/2)^2 T = 2\pi T D^2$.
2. Volume conserved: $64 r^3 = R^3 \Rightarrow r = R/4 = D/8$.
3. $E_2 = 64 \times 8\pi (D/8)^2 T = 8\pi T D^2$
4. $\Delta E = 8\pi TD^2 - 2\pi TD^2 = 6\pi TD^2$
5. So the answer is (c) $6\pi TD^2$.

Answer: (a)

1. $W = 8\pi R^2 T$, so for radius $2R$ at the same $T$, $W$ would be exactly 4 times.
2. Surface tension decreases with rise in temperature, so $T' < T$.
3. Hence $W_2 = 8\pi(2R)^2 T' < 4W_1$, but only slightly.
4. So the answer is (a) slightly less than 4 times.

Answer: (c)

1. Radii: $r_1 = 1\text{ cm}$, $r_2 = 3\text{ cm}$. $W = 8\pi T (r_2^2 - r_1^2)$.
2. $W = 8\pi \times 0.03 \times (9 - 1)\times10^{-4}$
3. $W = 1.92\pi \times 10^{-4}\text{ J}$, so $\alpha = 1.92$.
4. So the answer is (c) 1.92.

Answer: 264

1. $W = 8\pi T (r_2^2 - r_1^2)$
2. $W = 8 \times \frac{22}{7} \times 3.5\times10^{-2} \times (4 - 1)\times10^{-4}$
3. $W = 0.88 \times 3 \times 10^{-4} = 2.64\times10^{-4}\text{ J} = 264\times10^{-6}\text{ J}$
4. So $\alpha = 264$.

Answer: 11304

1. Radii: $3.5\text{ cm} \to 7\text{ cm}$. $W = 8\pi T (r_2^2 - r_1^2)$.
2. $W = 8 \times \frac{22}{7} \times 0.04 \times (49 - 12.25)\times10^{-4}$
3. $W = 36.96\times10^{-4}\text{ J} = 3696\ \mu\text{J}$
4. $15000 - x = 3696 \Rightarrow x = 11304$

Answer: (d)

1. Surface energy of the film $E = T \times 2A \propto A$.
2. Area reduced by 25% means energy is reduced by 25%.
3. So the answer is (d) $25\%$.

Answer: (b)

1. Film has two surfaces: $W = 2T\Delta A$.
2. $\Delta A = 72 - 30 = 42\text{ cm}^2 = 42\times10^{-4}\text{ m}^2$
3. $T = \dfrac{4\times10^{-4}}{2 \times 42\times10^{-4}} \approx 0.048 \approx 0.05\text{ N/m}$
4. So the answer is (b) $0.05\text{ N/m}$.

Answer: (b)

1. Two surfaces: $W = 2T \Delta(4\pi r^2) = 8\pi T (r_2^2 - r_1^2)$.
2. With $r_1 = d/2$, $r_2 = D/2$: $W = 8\pi T \dfrac{D^2 - d^2}{4}$
3. $W = 2\pi (D^2 - d^2) T$
4. So the answer is (b) $2\pi(D^2-d^2)T$.

Answer: (d)

1. Two surfaces: $W = 8\pi T (r_2^2 - r_1^2)$ with $r = d/2$.
2. $W = 8\pi T \dfrac{d_2^2 - d_1^2}{4} = 2\pi(d_2^2 - d_1^2)T$
3. So the answer is (d) $2\pi(d_2^2-d_1^2)T$.

Answer: (b)

1. Film has two surfaces: $W = 2T\Delta A$.
2. $\Delta A = 56 - 20 = 36\text{ cm}^2 = 36\times10^{-4}\text{ m}^2$
3. $T = \dfrac{3\times10^{-4}}{72\times10^{-4}} \approx 0.042 \approx 0.04\text{ N/m}$
4. So the answer is (b) $0.04\text{ N/m}$.

Answer: (c)

1. $W = 8\pi R^2 T$; at the same $T$, doubling $R$ gives $4W_1$.
2. Heating lowers the surface tension, $T' < T$.
3. $W_2 = 8\pi (2R)^2 T' = 4W_1 \dfrac{T'}{T} < 4W_1$
4. So the answer is (c) $W_2 < 4W_1$.

Answer: 7

1. In CGS: $W = 8\pi T (R^2 - r^2)$ with $r = 3.5\text{ cm}$, $T = 40$ dyne/cm.
2. $36960 = 8 \times \frac{22}{7} \times 40 \times (R^2 - 12.25)$
3. $R^2 - 12.25 = 36.75 \Rightarrow R^2 = 49$
4. New radius $R = 7\text{ cm}$.

Answer: (c)

1. Film has two surfaces: $W = 2T\Delta A$, with $\Delta A = l \Delta x$.
2. $\Delta A = 0.1\text{ m} \times 10^{-3}\text{ m} = 10^{-4}\text{ m}^2$
3. $W = 2 \times 72 \times 10^{-4} = 1.44\times10^{-2}\text{ J}$
4. So the answer is (c) $1.44\times10^{-2}\text{ J}$.