301. A horizontal light wire of length \(10\,\text{cm}\) forms the movable lower side of a vertical soap-film frame. The soap film has surface tension \(5.0\times10^{-2}\,\text{N m}^{-1}\). The wire carries a small mass and remains in equilibrium, with the downward weight balanced only by the surface-tension force of the two film surfaces. Take \(g=10\,\text{m s}^{-2}\). What is the greatest supported mass? If heating reduces the surface tension by \(20\%\), what mass could then be supported in the same geometry?
ⓐ. \(1.0\,\text{g}\) and \(0.80\,\text{g}\)
ⓑ. \(0.50\,\text{g}\) and \(0.40\,\text{g}\)
ⓒ. \(2.0\,\text{g}\) and \(1.60\,\text{g}\)
ⓓ. \(1.0\,\text{g}\) and \(1.20\,\text{g}\)
Correct Answer: \(1.0\,\text{g}\) and \(0.80\,\text{g}\)
Explanation: The soap film has two surfaces, each contributing a pull
\[
Sl
\]
Thus the total upward film force is
\[
F=2Sl
\]
\(\textbf{Initial supported weight:}\)
\[
F
=
2(5.0\times10^{-2})(0.10)
=
1.0\times10^{-2}\,\text{N}
\]
For equilibrium,
\[
mg=F
\]
Therefore,
\[
m=
\frac{1.0\times10^{-2}}{10}
=
1.0\times10^{-3}\,\text{kg}
\]
Hence,
\[
m=1.0\,\text{g}
\]
\(\textbf{After heating:}\)
The new surface tension is
\[
S'=0.80S
\]
With \(l\) unchanged,
\[
F'=0.80F
\]
Therefore the supported mass also becomes
\[
m'=0.80m=0.80\,\text{g}
\]
This combines the two-surface force with the usual decrease of liquid surface tension on heating.
\(\textbf{Final answer:}\) \(1.0\,\text{g}\) initially and \(0.80\,\text{g}\) after heating.
302. Consider the following statements about surface tension and surface energy.
Statement I: The SI unit of surface tension is \(\text{N m}^{-1}\).
Statement II: The dimensional formula of surface tension is \([MT^{-2}]\).
Statement III: Surface energy per unit area has unit \(\text{J m}^{-2}\), which is dimensionally equivalent to \(\text{N m}^{-1}\).
Statement IV: Surface tension is a force acting normally to the liquid interface at every point.
Which statements are correct?
ⓐ. I and IV only
ⓑ. II and IV only
ⓒ. I, II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Statement I is correct because surface tension is mechanically force per unit length:
\[
S=\frac{F}{l}.
\]
Therefore its SI unit is
\[
\text{N m}^{-1}.
\]
For Statement II,
\[
[\text{N m}^{-1}]
=
[MLT^{-2}]L^{-1}
=
[MT^{-2}],
\]
so it is also correct.
For Statement III,
\[
[\text{J m}^{-2}]
=
[\text{N m}]\,\text{m}^{-2}
=
\text{N m}^{-1}.
\]
Thus surface energy per unit area and surface tension have equivalent dimensions and units in the standard mechanical interpretation.
Statement IV is incorrect.
The surface-tension force associated with a boundary line acts in the plane tangent to the interface and perpendicular to that boundary line.
It is not generally a force normal to the whole liquid surface.
Hence Statements I, II and III only are correct.
303. A square soap film has side \(5.0\,\text{cm}\) and surface tension \(3.0\times10^{-2}\,\text{N m}^{-1}\). It is stretched quasistatically into another square of side \(7.0\,\text{cm}\) without changing the surface tension. What work is required?
ⓐ. \(7.2\times10^{-5}\,\text{J}\)
ⓑ. \(1.44\times10^{-4}\,\text{J}\)
ⓒ. \(2.88\times10^{-4}\,\text{J}\)
ⓓ. \(4.32\times10^{-4}\,\text{J}\)
Correct Answer: \(1.44\times10^{-4}\,\text{J}\)
Explanation: A soap film possesses two surfaces, so the increase in total interfacial area is twice the increase in geometric area.
\(\textbf{Initial geometric area:}\)
\[
A_i=(0.050)^2
=
2.5\times10^{-3}\,\text{m}^2
\]
\(\textbf{Final geometric area:}\)
\[
A_f=(0.070)^2
=
4.9\times10^{-3}\,\text{m}^2
\]
Thus,
\[
\Delta A_{\text{geom}}
=
4.9\times10^{-3}
-
2.5\times10^{-3}
=
2.4\times10^{-3}\,\text{m}^2
\]
Because there are two surfaces,
\[
\Delta A_{\text{total}}
=
2(2.4\times10^{-3})
=
4.8\times10^{-3}\,\text{m}^2
\]
\(\textbf{Surface-energy increase:}\)
\[
W=S\Delta A_{\text{total}}
\]
Therefore,
\[
W=
(3.0\times10^{-2})(4.8\times10^{-3})
=
1.44\times10^{-4}\,\text{J}
\]
The factor of two is associated with surface count, not with the square geometry itself.
\(\textbf{Final answer:}\)
\[
\boxed{W=1.44\times10^{-4}\,\text{J}}
\]
304. Assertion: Positive external work is required to increase the area of a liquid surface quasistatically at fixed temperature.
Reason: Molecules transferred from the bulk to the surface occupy a higher-energy molecular environment, so creation of additional surface increases the surface energy.
ⓐ. Assertion is true, but Reason is false
ⓑ. Assertion is false, but Reason is true
ⓒ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓓ. Both Assertion and Reason are true, and Reason correctly explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion
Explanation: The Assertion is correct because increasing an interface area requires work against the tendency of surface forces to contract the interface.
That supplied work appears as increased surface energy in the ideal quasistatic treatment.
The Reason is also correct.
Molecules in the bulk have a more symmetric molecular environment than molecules near a free surface.
Surface molecules consequently possess excess potential energy relative to bulk molecules in the standard description.
Increasing the surface area requires additional molecules to occupy this higher-energy environment.
The surface energy therefore rises when additional interface is created.
That is precisely why external work must be supplied during a quasistatic increase of area.
The Reason provides the molecular explanation for the Assertion.
Hence both statements are true and the Reason correctly explains the Assertion.
305. In a soap-film experiment, the movable wire has length \(6.0\,\text{cm}\). The equilibrium holding forces measured at three temperatures are:
| Temperature | Holding force |
|---|
| \(20^\circ\text{C}\) | \(8.4\times10^{-3}\,\text{N}\) |
| \(40^\circ\text{C}\) | \(7.2\times10^{-3}\,\text{N}\) |
| \(60^\circ\text{C}\) | \(6.0\times10^{-3}\,\text{N}\) |
Which conclusion is correct, and what holding force would be required at \(60^\circ\text{C}\) if the wire length were changed to \(12\,\text{cm}\)?
ⓐ. Surface tension decreases with temperature; the required force would be \(1.20\times10^{-2}\,\text{N}\)
ⓑ. Surface tension increases with temperature; the required force would be \(3.0\times10^{-3}\,\text{N}\)
ⓒ. Surface tension is independent of temperature; the required force would remain \(6.0\times10^{-3}\,\text{N}\)
ⓓ. Surface tension decreases with temperature; the required force would be \(6.0\times10^{-3}\,\text{N}\)
Correct Answer: Surface tension decreases with temperature; the required force would be \(1.20\times10^{-2}\,\text{N}\)
Explanation: For the same soap-film arrangement,
\[
F=2Sl
\]
The wire length is initially fixed.
Therefore the observed decrease of \(F\) with temperature directly indicates a decrease of \(S\).
\(\textbf{Surface tension at 60^\circ\text{C}:}\)
\[
S_{60}
=
\frac{F}{2l}
\]
Thus,
\[
S_{60}
=
\frac{6.0\times10^{-3}}
{2(6.0\times10^{-2})}
\]
Hence,
\[
S_{60}=5.0\times10^{-2}\,\text{N m}^{-1}
\]
\(\textbf{Double the wire length at the same temperature:}\)
The new length is
\[
l'=12\,\text{cm}=0.12\,\text{m}
\]
Therefore,
\[
F'=2S_{60}l'
\]
Substitution gives
\[
F'
=
2(5.0\times10^{-2})(0.12)
=
1.20\times10^{-2}\,\text{N}
\]
The temperature change determines the material value of surface tension, while the wire length determines how that surface tension translates into total force.
\(\textbf{Final answer:}\) Surface tension decreases with temperature, and the new holding force is \(1.20\times10^{-2}\,\text{N}\).
306. Match List I with List II.
| List I | List II |
|---|
| P. One liquid-air interface enlarged by area \(\Delta A\) | 1. \(2Sl\) |
| Q. Soap film of geometric area increased by \(\Delta A\) | 2. \(S\Delta A\) |
| R. Resultant surface-tension force on a movable wire of length \(l\) bounding a soap film | 3. Generally decreases on heating a liquid |
| S. Surface tension of a liquid | 4. \(2S\Delta A\) |
ⓐ. P-4, Q-2, R-1, S-3
ⓑ. P-2, Q-1, R-4, S-3
ⓒ. P-2, Q-4, R-1, S-3
ⓓ. P-3, Q-4, R-2, S-1
Correct Answer: P-2, Q-4, R-1, S-3
Explanation: For P, increasing the area of a single interface by \(\Delta A\) increases the surface energy by
\[
\Delta U=S\Delta A.
\]
Thus P matches 2.
For Q, a soap film has two surfaces.
A geometric area increase \(\Delta A\) therefore creates total interfacial area
\[
2\Delta A,
\]
so
\[
\Delta U=2S\Delta A.
\]
Thus Q matches 4.
For R, both surfaces of a soap film pull on the movable wire.
The resultant force is
\[
F=2Sl.
\]
Thus R matches 1.
For S, the surface tension of a liquid generally decreases as temperature increases.
Thus S matches 3.
The matching simultaneously distinguishes one versus two interfaces, energy versus force relations, and the temperature dependence of the material property.
\(\textbf{Final answer:}\) P-2, Q-4, R-1, S-3.
307. Water on clean glass forms an acute angle of contact, whereas mercury on clean glass forms an obtuse angle of contact. Which statement correctly connects angle of contact with the corresponding capillary behaviour?
ⓐ. Water wets glass and rises in a capillary; mercury does not wet glass and is depressed
ⓑ. Water does not wet glass and is depressed, whereas mercury wets glass and rises
ⓒ. Both liquids must rise because surface tension always pulls liquids upward
ⓓ. Both liquids must be depressed because atmospheric pressure acts downward
Correct Answer: Water wets glass and rises in a capillary; mercury does not wet glass and is depressed
Explanation: The angle of contact is measured through the liquid between the tangent to the liquid surface and the solid surface.
For a wetting liquid such as water on clean glass, the angle of contact is acute:
\[
\theta\lt 90^\circ.
\]
Therefore,
\[
\cos\theta\gt 0.
\]
In the capillary-rise relation,
\[
h=\frac{2S\cos\theta}{\rho gr},
\]
this gives positive \(h\), corresponding to capillary rise.
Mercury on clean glass has an obtuse angle of contact:
\[
\theta\gt 90^\circ,
\]
so
\[
\cos\theta\lt 0.
\]
The corresponding \(h\) is negative, representing capillary depression.
Thus surface tension does not imply upward motion in every liquid-solid combination; the contact angle determines the vertical component.
308. A liquid in a narrow capillary initially has angle of contact \(60^\circ\) and rises through height \(h\). The capillary radius, liquid density, surface tension and \(g\) are unchanged, but surface treatment changes the angle of contact to \(120^\circ\). What happens to the liquid level relative to the outside liquid surface?
ⓐ. It rises to \(2h\)
ⓑ. It remains at height \(h\)
ⓒ. It becomes depressed by a magnitude \(h\)
ⓓ. It becomes depressed by a magnitude \(2h\)
Correct Answer: It becomes depressed by a magnitude \(h\)
Explanation: Capillary displacement is
\[
h=\frac{2S\cos\theta}{\rho gr}.
\]
Initially,
\[
\cos60^\circ=\frac12.
\]
Hence the original displacement is positive and the liquid rises.
After surface treatment,
\[
\cos120^\circ=-\frac12.
\]
All remaining factors in the capillary relation are unchanged.
Therefore the new displacement is
\[
h'=
\frac{\cos120^\circ}{\cos60^\circ}h.
\]
Thus,
\[
h'=
\frac{-1/2}{1/2}h=-h.
\]
The magnitude is unchanged, but the sign reverses.
This changed-condition result shows that contact angle controls whether the surface-tension contribution produces rise or depression.
\(\textbf{Final answer:}\) The liquid becomes depressed by magnitude \(h\).
309. A capillary tube of radius \(0.50\,\text{mm}\) is dipped vertically into a liquid of density \(1000\,\text{kg m}^{-3}\), surface tension \(0.075\,\text{N m}^{-1}\), and angle of contact \(60^\circ\). Take \(g=10\,\text{m s}^{-2}\) and \(\pi=3.14\). Neglect the mass of the meniscus. What are the capillary rise and the upward vertical component of surface-tension force?
ⓐ. \(1.5\,\text{cm}\) and \(1.18\times10^{-4}\,\text{N}\)
ⓑ. \(3.0\,\text{cm}\) and \(2.36\times10^{-4}\,\text{N}\)
ⓒ. \(1.5\,\text{cm}\) and \(2.36\times10^{-4}\,\text{N}\)
ⓓ. \(3.0\,\text{cm}\) and \(1.18\times10^{-4}\,\text{N}\)
Correct Answer: \(1.5\,\text{cm}\) and \(1.18\times10^{-4}\,\text{N}\)
Explanation: \(\textbf{Capillary-rise relation:}\)
\[
h=\frac{2S\cos\theta}{\rho gr}
\]
Substituting,
\[
h=
\frac{2(0.075)(\cos60^\circ)}
{(1000)(10)(0.50\times10^{-3})}
\]
Using
\[
\cos60^\circ=\frac12,
\]
\[
h=
\frac{0.075}{5.0}
=
1.5\times10^{-2}\,\text{m}
\]
Therefore,
\[
h=1.5\,\text{cm}
\]
\(\textbf{Vertical surface-tension force around the rim:}\)
The tube circumference is
\[
2\pi r
\]
Hence the upward component is
\[
F=2\pi rS\cos\theta
\]
Substitution gives
\[
F=
2(3.14)(0.50\times10^{-3})(0.075)\left(\frac12\right)
\]
Thus,
\[
F\approx1.18\times10^{-4}\,\text{N}
\]
This equals the weight of the supported liquid column under the stated approximation.
\(\textbf{Final answer:}\)
\[
\boxed{h=1.5\,\text{cm}\quad\text{and}\quad F\approx1.18\times10^{-4}\,\text{N}}
\]
310. Two capillary tubes P and Q are dipped in the same wetting liquid. Tube P has radius \(r\) and angle of contact \(60^\circ\). Tube Q has radius \(4r\) and angle of contact \(0^\circ\). If the liquid rises through height \(h_P\) in P and \(h_Q\) in Q, what is \(h_Q/h_P\)?
ⓐ. \(1/4\)
ⓑ. \(1/2\)
ⓒ. \(1\)
ⓓ. \(2\)
Correct Answer: \(1/2\)
Explanation: For the same liquid and gravitational field,
\[
h\propto\frac{\cos\theta}{r}
\]
Therefore,
\[
\frac{h_Q}{h_P}
=
\frac{\cos\theta_Q}{\cos\theta_P}
\frac{r_P}{r_Q}
\]
The given angles give
\[
\cos0^\circ=1,
\qquad
\cos60^\circ=\frac12
\]
The radius ratio is
\[
\frac{r_P}{r_Q}=\frac14
\]
Hence,
\[
\frac{h_Q}{h_P}
=
\frac{1}{1/2}\cdot\frac14
\]
Therefore,
\[
\frac{h_Q}{h_P}
=
2\cdot\frac14
=
\frac12
\]
The better wetting orientation in Q increases the capillary tendency, but its fourfold larger radius more than compensates.
\(\textbf{Final answer:}\)
\[
\boxed{h_Q/h_P=1/2}
\]
311. A particular liquid rises \(3.0\,\text{cm}\) in a capillary tube on Earth where \(g=10\,\text{m s}^{-2}\). The same liquid and tube are transferred to a planet where \(g=4.0\,\text{m s}^{-2}\). The tube is then replaced by another tube of twice the original radius, with all other properties unchanged. What is the final capillary rise?
ⓐ. \(3.0\,\text{cm}\)
ⓑ. \(7.5\,\text{cm}\)
ⓒ. \(6.0\,\text{cm}\)
ⓓ. \(3.75\,\text{cm}\)
Correct Answer: \(3.75\,\text{cm}\)
Explanation: For unchanged liquid and contact angle,
\[
h\propto\frac{1}{gr}
\]
Let the original rise be
\[
h_1=3.0\,\text{cm}
\]
at
\[
g_1=10\,\text{m s}^{-2},
\qquad
r_1=r
\]
The final conditions are
\[
g_2=4.0\,\text{m s}^{-2},
\qquad
r_2=2r
\]
Therefore,
\[
\frac{h_2}{h_1}
=
\frac{g_1r_1}{g_2r_2}
\]
Substitution gives
\[
\frac{h_2}{h_1}
=
\frac{10r}{(4.0)(2r)}
=
\frac54
\]
Hence,
\[
h_2=
\frac54(3.0\,\text{cm})
=
3.75\,\text{cm}
\]
The weaker gravity tends to increase the rise, whereas the larger tube radius tends to decrease it.
\(\textbf{Final answer:}\)
\[
\boxed{h_2=3.75\,\text{cm}}
\]
312. A liquid drop, an air bubble inside a liquid, and a soap bubble all have the same radius \(r\) and the same relevant surface tension \(S\). Neglect hydrostatic variation across their diameters. What is the ratio of their excess internal pressures in the order liquid drop : air bubble in liquid : soap bubble?
ⓐ. \(1:1:1\)
ⓑ. \(1:2:2\)
ⓒ. \(1:1:2\)
ⓓ. \(2:1:4\)
Correct Answer: \(1:1:2\)
Explanation: A liquid drop has one liquid-gas interface.
Its excess pressure is
\[
\Delta P_{\text{drop}}=\frac{2S}{r}
\]
An air bubble inside a liquid also has one liquid-gas interface.
Therefore,
\[
\Delta P_{\text{air bubble}}=\frac{2S}{r}
\]
A soap bubble is a thin liquid film with two surfaces.
Its excess pressure is therefore
\[
\Delta P_{\text{soap bubble}}=\frac{4S}{r}
\]
Comparing the three expressions,
\[
\frac{2S}{r}:\frac{2S}{r}:\frac{4S}{r}
=
1:1:2
\]
The factor of two for the soap bubble arises from the presence of two curved interfaces, not from its being filled with air.
Hence option C is correct.
313. A soap bubble of radius \(2.0\,\text{mm}\) has excess internal pressure \(120\,\text{Pa}\). The same liquid is then used to form a spherical liquid drop of radius \(1.0\,\text{mm}\). What are the surface tension and the excess pressure inside the drop?
ⓐ. \(0.030\,\text{N m}^{-1}\) and \(60\,\text{Pa}\)
ⓑ. \(0.060\,\text{N m}^{-1}\) and \(60\,\text{Pa}\)
ⓒ. \(0.060\,\text{N m}^{-1}\) and \(120\,\text{Pa}\)
ⓓ. \(0.120\,\text{N m}^{-1}\) and \(240\,\text{Pa}\)
Correct Answer: \(0.060\,\text{N m}^{-1}\) and \(120\,\text{Pa}\)
Explanation: \(\textbf{Infer surface tension from the soap bubble:}\)
For a soap bubble,
\[
\Delta P=\frac{4S}{r}
\]
Therefore,
\[
S=\frac{\Delta Pr}{4}
\]
Substitution gives
\[
S=
\frac{(120)(2.0\times10^{-3})}{4}
\]
Hence,
\[
S=0.060\,\text{N m}^{-1}
\]
\(\textbf{Apply the one-interface relation to the liquid drop:}\)
For a liquid drop,
\[
\Delta P_{\text{drop}}=\frac{2S}{r_d}
\]
Thus,
\[
\Delta P_{\text{drop}}
=
\frac{2(0.060)}
{1.0\times10^{-3}}
\]
Therefore,
\[
\Delta P_{\text{drop}}=120\,\text{Pa}
\]
The smaller radius of the drop exactly compensates for its having only one interface instead of the soap bubble's two.
\(\textbf{Final answer:}\)
\[
\boxed{S=0.060\,\text{N m}^{-1}\quad\text{and}\quad \Delta P_{\text{drop}}=120\,\text{Pa}}
\]
314. Two soap bubbles of radii \(1.0\,\text{mm}\) and \(2.0\,\text{mm}\) are connected by a narrow tube. They are made from the same soap solution of surface tension \(0.030\,\text{N m}^{-1}\), and external pressure is the same around both. Initially, what is the pressure difference between their interiors, and in which direction does air tend to flow?
ⓐ. \(30\,\text{Pa}\), from the larger bubble to the smaller bubble
ⓑ. \(120\,\text{Pa}\), from the smaller bubble to the larger bubble
ⓒ. \(60\,\text{Pa}\), from the larger bubble to the smaller bubble
ⓓ. \(60\,\text{Pa}\), from the smaller bubble to the larger bubble
Correct Answer: \(60\,\text{Pa}\), from the smaller bubble to the larger bubble
Explanation: For a soap bubble,
\[
\Delta P=\frac{4S}{r}
\]
\(\textbf{Excess pressure in the smaller bubble:}\)
\[
\Delta P_s
=
\frac{4(0.030)}
{1.0\times10^{-3}}
=
120\,\text{Pa}
\]
\(\textbf{Excess pressure in the larger bubble:}\)
\[
\Delta P_L
=
\frac{4(0.030)}
{2.0\times10^{-3}}
=
60\,\text{Pa}
\]
Since the same external pressure surrounds both bubbles, their internal-pressure difference is
\[
P_s-P_L
=
120-60
=
60\,\text{Pa}
\]
Air tends to move from higher pressure to lower pressure.
Therefore it flows from the smaller bubble toward the larger bubble.
As that happens, the smaller bubble tends to shrink and the larger one tends to grow.
\(\textbf{Final answer:}\) Pressure difference \(=60\,\text{Pa}\), with air flowing from smaller to larger bubble.
315. Two spherical air bubbles are present in water of density \(1000\,\text{kg m}^{-3}\). Bubble A has radius \(1.0\,\text{mm}\) and is \(1.0\,\text{m}\) below the free surface. Bubble B has radius \(2.0\,\text{mm}\) and is \(2.0\,\text{m}\) below the free surface. The water-air surface tension is \(0.050\,\text{N m}^{-1}\), atmospheric pressure is \(1.00\times10^{5}\,\text{Pa}\), and \(g=10\,\text{m s}^{-2}\). Which pair correctly gives the internal absolute pressures?
ⓐ. \(P_A=1.101\times10^{5}\,\text{Pa},\ P_B=1.2005\times10^{5}\,\text{Pa}\)
ⓑ. \(P_A=1.1005\times10^{5}\,\text{Pa},\ P_B=1.201\times10^{5}\,\text{Pa}\)
ⓒ. \(P_A=1.201\times10^{5}\,\text{Pa},\ P_B=1.1005\times10^{5}\,\text{Pa}\)
ⓓ. \(P_A=1.101\times10^{5}\,\text{Pa},\ P_B=1.101\times10^{5}\,\text{Pa}\)
Correct Answer: \(P_A=1.101\times10^{5}\,\text{Pa},\ P_B=1.2005\times10^{5}\,\text{Pa}\)
Explanation: Each bubble's internal pressure must include both the hydrostatic external pressure at its depth and its surface-tension excess pressure.
\(\textbf{Bubble A external pressure:}\)
\[
P_{\text{ext},A}
=
P_a+\rho gh_A
\]
Thus,
\[
P_{\text{ext},A}
=
1.00\times10^{5}
+
(1000)(10)(1.0)
=
1.10\times10^{5}\,\text{Pa}
\]
For an air bubble in liquid,
\[
\Delta P_A=\frac{2S}{r_A}
\]
Hence,
\[
\Delta P_A
=
\frac{2(0.050)}{1.0\times10^{-3}}
=
100\,\text{Pa}
\]
Therefore,
\[
P_A=1.101\times10^{5}\,\text{Pa}
\]
\(\textbf{Bubble B external pressure:}\)
\[
P_{\text{ext},B}
=
1.00\times10^{5}
+
(1000)(10)(2.0)
=
1.20\times10^{5}\,\text{Pa}
\]
Its surface excess is
\[
\Delta P_B
=
\frac{2(0.050)}{2.0\times10^{-3}}
=
50\,\text{Pa}
\]
Thus,
\[
P_B
=
1.2005\times10^{5}\,\text{Pa}
\]
Although the smaller bubble has the larger curvature-induced excess pressure, the greater depth of B dominates the absolute-pressure comparison.
\(\textbf{Final answer:}\) \(P_A=1.101\times10^{5}\,\text{Pa}\), \(P_B=1.2005\times10^{5}\,\text{Pa}\).
316. A soap bubble of surface tension \(0.025\,\text{N m}^{-1}\) is expanded quasistatically from radius \(1.0\,\text{cm}\) to \(2.0\,\text{cm}\). Take \(\pi=3.14\). What are the increase in surface energy and the final excess pressure?
ⓐ. \(9.42\times10^{-5}\,\text{J}\) and \(10\,\text{Pa}\)
ⓑ. \(1.884\times10^{-4}\,\text{J}\) and \(10\,\text{Pa}\)
ⓒ. \(3.768\times10^{-4}\,\text{J}\) and \(5\,\text{Pa}\)
ⓓ. \(1.884\times10^{-4}\,\text{J}\) and \(5\,\text{Pa}\)
Correct Answer: \(1.884\times10^{-4}\,\text{J}\) and \(5\,\text{Pa}\)
Explanation: A soap bubble has two surfaces.
\(\textbf{Increase in total interfacial area:}\)
For radius \(r\), total surface area is
\[
A=2(4\pi r^2)=8\pi r^2
\]
Therefore,
\[
\Delta A
=
8\pi(r_2^2-r_1^2)
\]
Substitution gives
\[
\Delta A
=
8(3.14)
\left[
(2.0\times10^{-2})^2
-
(1.0\times10^{-2})^2
\right]
\]
Thus,
\[
\Delta A
=
8(3.14)(3.0\times10^{-4})
\]
The surface-energy increase is
\[
\Delta U=S\Delta A
\]
Hence,
\[
\Delta U
=
(0.025)(8)(3.14)(3.0\times10^{-4})
=
1.884\times10^{-4}\,\text{J}
\]
\(\textbf{Final excess pressure:}\)
For a soap bubble,
\[
\Delta P=\frac{4S}{r_2}
\]
Therefore,
\[
\Delta P
=
\frac{4(0.025)}{2.0\times10^{-2}}
=
5\,\text{Pa}
\]
As the bubble grows, its total surface energy increases even though its curvature-induced excess pressure decreases.
\(\textbf{Final answer:}\) \(\Delta U=1.884\times10^{-4}\,\text{J}\) and final excess pressure \(=5\,\text{Pa}\).
317. Eight identical liquid drops, each of radius \(1.0\,\text{mm}\), coalesce to form one spherical drop. The liquid has surface tension \(0.050\,\text{N m}^{-1}\). Take \(\pi=3.14\). Neglect any energy loss other than the decrease in surface energy. What are the radius of the final drop, the energy released, and its excess pressure?
ⓐ. \(2.0\,\text{mm},\ 2.512\times10^{-6}\,\text{J},\ 50\,\text{Pa}\)
ⓑ. \(4.0\,\text{mm},\ 5.024\times10^{-6}\,\text{J},\ 25\,\text{Pa}\)
ⓒ. \(2.0\,\text{mm},\ 5.024\times10^{-6}\,\text{J},\ 100\,\text{Pa}\)
ⓓ. \(8.0\,\text{mm},\ 2.512\times10^{-6}\,\text{J},\ 12.5\,\text{Pa}\)
Correct Answer: \(2.0\,\text{mm},\ 2.512\times10^{-6}\,\text{J},\ 50\,\text{Pa}\)
Explanation: \(\textbf{Final radius from volume conservation:}\)
\[
8\left(\frac43\pi r^3\right)
=
\frac43\pi R^3
\]
Therefore,
\[
R^3=8r^3,
\]
so
\[
R=2r=2.0\,\text{mm}
\]
\(\textbf{Decrease in surface area:}\)
Initial area:
\[
A_i=8(4\pi r^2)=32\pi r^2
\]
Final area:
\[
A_f=4\pi(2r)^2=16\pi r^2
\]
Thus,
\[
A_i-A_f=16\pi r^2
\]
The released surface energy is
\[
E=S(A_i-A_f)
\]
Hence,
\[
E
=
(0.050)(16)(3.14)(1.0\times10^{-3})^2
\]
Therefore,
\[
E=2.512\times10^{-6}\,\text{J}
\]
\(\textbf{Excess pressure in the final liquid drop:}\)
\[
\Delta P=\frac{2S}{R}
\]
Thus,
\[
\Delta P=
\frac{2(0.050)}{2.0\times10^{-3}}
=
50\,\text{Pa}
\]
\(\textbf{Final answer:}\) \(R=2.0\,\text{mm}\), energy released \(=2.512\times10^{-6}\,\text{J}\), and \(\Delta P=50\,\text{Pa}\).
318. A spherical liquid drop of radius \(3.0\,\text{mm}\) is broken into \(27\) identical smaller drops. Surface tension is \(0.040\,\text{N m}^{-1}\), and \(\pi=3.14\). Which pair correctly gives the ratio of final total surface area to initial surface area and the work required to create the additional surface?
ⓐ. \(3\) and \(9.04\times10^{-6}\,\text{J}\)
ⓑ. \(9\) and \(4.52\times10^{-6}\,\text{J}\)
ⓒ. \(3\) and \(1.81\times10^{-5}\,\text{J}\)
ⓓ. \(27\) and \(9.04\times10^{-6}\,\text{J}\)
Correct Answer: \(3\) and \(9.04\times10^{-6}\,\text{J}\)
Explanation: \(\textbf{Radius of each daughter drop:}\)
Volume conservation gives
\[
27r^3=R^3
\]
Therefore,
\[
r=\frac{R}{3}
\]
\(\textbf{Total-area ratio:}\)
Initial area is
\[
A_i=4\pi R^2
\]
Final total area is
\[
A_f=27(4\pi r^2)
\]
Substituting
\[
r=\frac R3,
\]
\[
A_f
=
27(4\pi)\frac{R^2}{9}
=
3(4\pi R^2)
\]
Thus,
\[
\frac{A_f}{A_i}=3
\]
The increase in area is therefore
\[
\Delta A=A_f-A_i=2A_i=8\pi R^2
\]
\(\textbf{Work required:}\)
\[
W=S\Delta A
\]
Hence,
\[
W
=
(0.040)(8)(3.14)(3.0\times10^{-3})^2
\]
Thus,
\[
W\approx9.04\times10^{-6}\,\text{J}
\]
Each smaller drop also has three times the original excess pressure because its radius is one-third as large.
\(\textbf{Final answer:}\) Area ratio \(=3\) and \(W\approx9.04\times10^{-6}\,\text{J}\).
319. A spherical liquid drop of radius \(1.0\,\text{mm}\) is held at a depth of \(3.0\,\text{m}\) below the free surface of water. The interfacial surface tension is \(0.040\,\text{N m}^{-1}\). Take atmospheric pressure \(1.00\times10^{5}\,\text{Pa}\), water density \(1000\,\text{kg m}^{-3}\), and \(g=10\,\text{m s}^{-2}\). Neglect the density of the drop in determining the surrounding hydrostatic pressure. What is the absolute pressure inside the drop?
ⓐ. \(1.3000\times10^{5}\,\text{Pa}\)
ⓑ. \(1.3008\times10^{5}\,\text{Pa}\)
ⓒ. \(1.3080\times10^{5}\,\text{Pa}\)
ⓓ. \(1.8000\times10^{5}\,\text{Pa}\)
Correct Answer: \(1.3008\times10^{5}\,\text{Pa}\)
Explanation: The pressure inside the drop exceeds the surrounding water pressure because of surface curvature.
\(\textbf{External water pressure at the stated depth:}\)
\[
P_{\text{out}}
=
P_a+\rho gh
\]
Thus,
\[
P_{\text{out}}
=
1.00\times10^{5}
+
(1000)(10)(3.0)
\]
Therefore,
\[
P_{\text{out}}
=
1.30\times10^{5}\,\text{Pa}
\]
\(\textbf{Curvature-induced excess pressure:}\)
For a liquid drop,
\[
\Delta P=\frac{2S}{r}
\]
Hence,
\[
\Delta P
=
\frac{2(0.040)}
{1.0\times10^{-3}}
=
80\,\text{Pa}
\]
Therefore,
\[
P_{\text{inside}}
=
P_{\text{out}}+\Delta P
\]
Thus,
\[
P_{\text{inside}}
=
1.30\times10^{5}+80
=
1.3008\times10^{5}\,\text{Pa}
\]
The hydrostatic and surface-tension pressure contributions must both be included.
\(\textbf{Final answer:}\)
\[
\boxed{P_{\text{inside}}=1.3008\times10^{5}\,\text{Pa}}
\]
320. A liquid rises \(1.44\,\text{cm}\) in a capillary tube of radius \(0.50\,\text{mm}\). Its surface tension is \(0.072\,\text{N m}^{-1}\), density is \(1000\,\text{kg m}^{-3}\), and \(g=10\,\text{m s}^{-2}\). What is the angle of contact? If the radius is then halved without changing the liquid-solid pair, what is the new rise?
ⓐ. \(30^\circ\) and \(2.88\,\text{cm}\)
ⓑ. \(90^\circ\) and zero rise
ⓒ. \(60^\circ\) and \(0.72\,\text{cm}\)
ⓓ. \(60^\circ\) and \(2.88\,\text{cm}\)
Correct Answer: \(60^\circ\) and \(2.88\,\text{cm}\)
Explanation: \(\textbf{Infer the contact angle from the observed rise:}\)
\[
h=\frac{2S\cos\theta}{\rho gr}
\]
Therefore,
\[
\cos\theta
=
\frac{h\rho gr}{2S}
\]
Substitution gives
\[
\cos\theta
=
\frac{(1.44\times10^{-2})(1000)(10)(0.50\times10^{-3})}
{2(0.072)}
\]
The numerator is
\[
0.072
\]
The denominator is
\[
0.144
\]
Hence,
\[
\cos\theta=\frac12
\]
Therefore,
\[
\theta=60^\circ
\]
\(\textbf{Halve the tube radius:}\)
For the unchanged liquid-solid pair,
\[
h\propto\frac1r
\]
Thus,
\[
h'=2h=2.88\,\text{cm}
\]
\(\textbf{Final answer:}\) \(\theta=60^\circ\) and the new rise is \(2.88\,\text{cm}\).
