301. A liquid surface has surface tension \(0.072\,\text{N m}^{-1}\). The force acting along a line of length \(0.050\,\text{m}\) on the surface is
ⓐ. \(1.44\,\text{N}\)
ⓑ. \(7.2\times10^{-2}\,\text{N}\)
ⓒ. \(2.5\times10^{-3}\,\text{N}\)
ⓓ. \(3.6\times10^{-3}\,\text{N}\)
Correct Answer: \(3.6\times10^{-3}\,\text{N}\)
Explanation: \( \textbf{Given:} \) Surface tension \(T=0.072\,\text{N m}^{-1}\) and length \(l=0.050\,\text{m}\).
\( \textbf{Required:} \) Force \(F\).
Surface tension is:
\[
T=\frac{F}{l}
\]
Rearranging:
\[
F=Tl
\]
Substitute:
\[
F=(0.072)(0.050)
\]
Calculation:
\[
F=0.0036\,\text{N}
\]
Therefore:
\[
F=3.6\times10^{-3}\,\text{N}
\]
The small value is reasonable because the length involved is only \(5.0\,\text{cm}\).
\( \textbf{Final answer:} \) The force is \(3.6\times10^{-3}\,\text{N}\).
302. A student identifies surface tension as “force per unit area.” The best correction is that surface tension is
ⓐ. force per unit area, while pressure is force per unit length
ⓑ. mass per unit volume
ⓒ. force per unit length, while pressure is force per unit area
ⓓ. volume crossing a section per unit time
Correct Answer: force per unit length, while pressure is force per unit area
Explanation: Surface tension and pressure both involve force, but their denominators are different. Surface tension is \(T=\frac{F}{l}\), so it is force per unit length. Pressure is \(P=\frac{F}{A}\), so it is force per unit area. Confusing the two leads to wrong units: \(\text{N m}^{-1}\) for surface tension and \(\text{N m}^{-2}\) for pressure. Surface tension belongs to surface effects, while pressure describes force distributed over an area.
303. The dimensional formula of surface tension is
ⓐ. \([ML^{-1}T^{-2}]\)
ⓑ. \([ML^{-1}T^{-1}]\)
ⓒ. \([MT^{-2}]\)
ⓓ. \([MLT^{-2}]\)
Correct Answer: \([MT^{-2}]\)
Explanation: \( \textbf{Starting relation:} \)
\[
T=\frac{F}{l}
\]
The dimensional formula of force is:
\[
[F]=[MLT^{-2}]
\]
The dimensional formula of length is:
\[
[l]=[L]
\]
Therefore:
\[
[T]=\frac{[MLT^{-2}]}{[L]}
\]
Simplifying:
\[
[T]=[MT^{-2}]
\]
This agrees with the unit \(\text{N m}^{-1}\), since \(\text{N}\) contains one power of metre that cancels with the metre in the denominator.
\( \textbf{Final answer:} \) The dimensional formula is \([MT^{-2}]\).
304. A liquid drop tends to become spherical mainly because surface tension tends to
ⓐ. maximize the surface area for a given volume
ⓑ. minimize the surface area for a given volume
ⓒ. make the liquid density zero
ⓓ. remove all pressure inside the drop
Correct Answer: minimize the surface area for a given volume
Explanation: Surface tension makes a liquid surface tend to contract. For a given volume, the sphere has the least surface area among common shapes. A liquid drop therefore tends to become spherical when other forces are small. This does not mean pressure disappears inside the drop; curved surfaces can have pressure differences, which are studied separately. The surface-minimizing tendency is a direct consequence of surface energy and surface tension.
305. Assertion: Surface tension has the same dimensional formula as surface energy per unit area.
Reason: Surface tension can also be interpreted as work required to increase unit surface area.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Surface tension can be interpreted in two equivalent ways. One way is force per unit length, \(T=\frac{F}{l}\). Another way is surface energy per unit area, \(T=\frac{W}{\Delta A}\). The unit \(\text{J m}^{-2}\) is equivalent to \(\text{N m}^{-1}\) because \(1\,\text{J}=1\,\text{N m}\). The reason explains why surface tension and surface energy per unit area have the same dimensional formula.
306. Study the table and identify the row that is incorrectly matched.
| Row | Quantity | Unit |
|---|
| P | Surface tension | \(\text{N m}^{-1}\) |
| Q | Surface energy per unit area | \(\text{J m}^{-2}\) |
| R | Pressure | \(\text{N m}^{-2}\) |
| S | Coefficient of viscosity | \(\text{kg m}^{-3}\) |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row S
ⓓ. Row R
Correct Answer: Row S
Explanation: Surface tension has unit \(\text{N m}^{-1}\), so row P is correct. Surface energy per unit area has unit \(\text{J m}^{-2}\), which is equivalent to \(\text{N m}^{-1}\), so row Q is also correct. Pressure has unit \(\text{N m}^{-2}\), so row R is correct. Row S is incorrect because coefficient of viscosity has unit \(\text{Pa s}\) or \(\text{N s m}^{-2}\), not \(\text{kg m}^{-3}\). The unit \(\text{kg m}^{-3}\) belongs to density.
307. Work \(W\) is needed to increase the free surface area of a liquid by \(\Delta A\). The surface tension is
ⓐ. \(T=\frac{W}{\Delta A}\)
ⓑ. \(T=W\Delta A\)
ⓒ. \(T=\frac{\Delta A}{W}\)
ⓓ. \(T=\rho g\Delta A\)
Correct Answer: \(T=\frac{W}{\Delta A}\)
Explanation: Surface tension can be interpreted as surface energy per unit area. If work \(W\) is done to increase the surface area by \(\Delta A\), then \(T=\frac{W}{\Delta A}\). This relation is useful when surface tension is discussed through energy rather than force. The numerator is work or energy, and the denominator is increase in surface area. The unit becomes \(\text{J m}^{-2}\), which is equivalent to \(\text{N m}^{-1}\).
308. A liquid surface is increased by \(2.0\times10^{-3}\,\text{m}^2\), and the work done against surface tension is \(1.4\times10^{-4}\,\text{J}\). The surface tension is
ⓐ. \(7.0\times10^{-2}\,\text{N m}^{-1}\)
ⓑ. \(2.8\times10^{-7}\,\text{N m}^{-1}\)
ⓒ. \(1.4\times10^{-1}\,\text{N m}^{-1}\)
ⓓ. \(2.0\times10^{-3}\,\text{N m}^{-1}\)
Correct Answer: \(7.0\times10^{-2}\,\text{N m}^{-1}\)
Explanation: \( \textbf{Given:} \) Work \(W=1.4\times10^{-4}\,\text{J}\) and increase in area \(\Delta A=2.0\times10^{-3}\,\text{m}^2\).
\( \textbf{Required:} \) Surface tension \(T\).
Energy interpretation:
\[
T=\frac{W}{\Delta A}
\]
Substitute:
\[
T=\frac{1.4\times10^{-4}}{2.0\times10^{-3}}
\]
Separate numbers and powers:
\[
T=\frac{1.4}{2.0}\times10^{-1}
\]
\[
T=0.70\times10^{-1}=7.0\times10^{-2}\,\text{N m}^{-1}
\]
The unit \(\text{J m}^{-2}\) is the same as \(\text{N m}^{-1}\).
\( \textbf{Final answer:} \) The surface tension is \(7.0\times10^{-2}\,\text{N m}^{-1}\).
309. Use the graph description below.
For a liquid surface at constant temperature, a graph is plotted with work done \(W\) on the vertical axis and increase in surface area \(\Delta A\) on the horizontal axis. The graph is a straight line through the origin.
The slope of the graph represents
ⓐ. static pressure
ⓑ. fluid density
ⓒ. coefficient of viscosity
ⓓ. surface tension
Correct Answer: surface tension
Explanation: \( \textbf{Energy relation:} \)
\[
T=\frac{W}{\Delta A}
\]
Rearranging:
\[
W=T\Delta A
\]
This has the graph form:
\[
y=mx
\]
Here \(W\) is on the vertical axis and \(\Delta A\) is on the horizontal axis.
Therefore, the slope is:
\[
T
\]
A steeper graph means more work is needed for the same increase in surface area, so the liquid has larger surface tension.
\( \textbf{Final answer:} \) The slope represents \(T\).
310. The equivalence \(\text{J m}^{-2}=\text{N m}^{-1}\) is valid because
ⓐ. \(1\,\text{J}=1\,\text{N m}\)
ⓑ. \(1\,\text{J}=1\,\text{N m}^{-1}\)
ⓒ. \(1\,\text{N}=1\,\text{J m}^{-2}\)
ⓓ. \(1\,\text{m}^2=1\,\text{m}^{-1}\)
Correct Answer: \(1\,\text{J}=1\,\text{N m}\)
Explanation: Surface tension can be written as energy per unit area, \(\text{J m}^{-2}\). Since \(1\,\text{J}=1\,\text{N m}\), we get:
\[
\text{J m}^{-2}=(\text{N m})\text{m}^{-2}
\]
Simplifying the powers of metre:
\[
\text{J m}^{-2}=\text{N m}^{-1}
\]
This is why the energy definition and force-per-length definition give the same unit. The equality comes from unit algebra, not from treating area and length as the same quantity.
311. A student says, “If surface area increases, surface energy decreases because the liquid surface stretches.” The correct statement is that for positive surface tension, increasing surface area generally
ⓐ. increases surface energy
ⓑ. decreases surface energy
ⓒ. leaves surface energy always zero
ⓓ. changes surface energy into density
Correct Answer: increases surface energy
Explanation: Creating more liquid surface requires work against surface tension. The increase in surface energy is given by \(W=T\Delta A\) when \(T\) is constant. If \(\Delta A\) is positive and \(T\) is positive, the work required is positive. Therefore, the surface energy increases when extra surface is created. This is why a liquid tends to minimize its surface area when no other strong effect prevents it.
312. A wire frame experiment increases a liquid film’s surface area by \(5.0\times10^{-4}\,\text{m}^2\). If the liquid has surface tension \(0.060\,\text{N m}^{-1}\), the work done against surface tension is
ⓐ. \(8.3\times10^{-3}\,\text{J}\)
ⓑ. \(1.2\times10^{2}\,\text{J}\)
ⓒ. \(3.0\times10^{-5}\,\text{J}\)
ⓓ. \(5.0\times10^{-4}\,\text{J}\)
Correct Answer: \(3.0\times10^{-5}\,\text{J}\)
Explanation: \( \textbf{Given:} \) \(T=0.060\,\text{N m}^{-1}\) and \(\Delta A=5.0\times10^{-4}\,\text{m}^2\).
\( \textbf{Required:} \) Work done \(W\).
Surface energy relation:
\[
W=T\Delta A
\]
Substitute:
\[
W=(0.060)(5.0\times10^{-4})
\]
Calculate:
\[
0.060\times5.0=0.300
\]
Therefore:
\[
W=0.300\times10^{-4}\,\text{J}
\]
\[
W=3.0\times10^{-5}\,\text{J}
\]
The small work value is expected because the area increase is very small.
\( \textbf{Final answer:} \) The work done is \(3.0\times10^{-5}\,\text{J}\).
313. Consider the following statements about surface tension and surface energy.
Statement I: Surface tension may be treated as force per unit length.
Statement II: Surface tension may be treated as surface energy per unit area.
Statement III: Surface tension has the same unit as pressure.
ⓐ. II and III only
ⓑ. I and II only
ⓒ. I and III only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Statement I is true because \(T=\frac{F}{l}\) defines surface tension as force per unit length. Statement II is also true because \(T=\frac{W}{\Delta A}\) gives the surface-energy interpretation. Statement III is false because surface tension has unit \(\text{N m}^{-1}\), while pressure has unit \(\text{N m}^{-2}\). The two quantities may both involve force, but they are distributed over different geometrical measures. This distinction in units prevents mixing surface effects with pressure effects.
314. A liquid with larger surface tension requires more work to create the same additional surface area because
ⓐ. \(W=\frac{\Delta A}{T}\)
ⓑ. \(W=T\Delta A\)
ⓒ. \(W=\rho gh\)
ⓓ. \(W=\frac{P}{A}\)
Correct Answer: \(W=T\Delta A\)
Explanation: Surface tension can be interpreted as the energy required per unit increase in surface area. If the increase in area \(\Delta A\) is fixed, then \(W=T\Delta A\) shows that work is directly proportional to surface tension. A liquid with larger \(T\) needs more work to create the same new surface area. This is the energy form of the same tendency that makes a liquid surface contract. The relation is about surface creation, not hydrostatic pressure or viscous flow.
315. A small insect can stand on the surface of clean water mainly because the water surface
ⓐ. has much larger density near the insect
ⓑ. exerts upward force without surface deformation
ⓒ. acts like a stretched membrane
ⓓ. has no cohesive effect at the surface
Correct Answer: acts like a stretched membrane
Explanation: Surface tension makes the free surface of water behave somewhat like a stretched membrane. If the insect is light and its legs do not break the surface, the surface-tension forces can help support it. This does not mean the insect has no weight; gravity still acts downward on it. The water surface deforms slightly and provides an upward effect through surface forces. The explanation depends on the liquid surface property, not on the density of water becoming zero.
316. A clean steel needle may float on water if placed carefully, although steel is denser than water. The best explanation is that
ⓐ. the needle’s density becomes less than water after touching it
ⓑ. buoyant force is always greater than weight for every steel object
ⓒ. surface tension supports an unbroken surface
ⓓ. water pressure is zero above the needle
Correct Answer: surface tension supports an unbroken surface
Explanation: A steel needle is denser than water, so density comparison alone suggests that it should sink if fully wetted and immersed. However, if it is placed carefully, the water surface can remain unbroken and slightly depressed. Surface tension then provides an upward supporting effect along the contact region. This support can balance the needle’s weight when the needle is light enough and the surface is clean. The floating here is not the same as ordinary floating of a wooden block due only to displacement.
317. Detergent is added to water on which small insects were able to stand. The insects may sink more easily because detergent
ⓐ. increases water density enormously
ⓑ. removes gravitational force on the water
ⓒ. makes water incompressible for the first time
ⓓ. reduces the surface tension of water
Correct Answer: reduces the surface tension of water
Explanation: Detergents reduce the surface tension of water. A lower surface tension weakens the stretched-membrane-like support at the free surface. Small insects or light objects that depend partly on surface tension then become harder to support. The water does not need a large density change for this effect to occur. The key change is in the surface property, not in hydrostatic pressure or incompressibility.
318. Tiny liquid drops tend to become nearly spherical when other forces are small. This happens because a sphere
ⓐ. has the greatest surface area for a given volume
ⓑ. has the least surface area for a given volume
ⓒ. has no surface tension
ⓓ. has zero pressure inside it
Correct Answer: has the least surface area for a given volume
Explanation: Surface tension tends to reduce the free surface area of a liquid. For a given volume, a spherical shape has the smallest surface area. Therefore, small drops tend to become spherical when gravity and other external effects are not dominant. The drop still has pressure inside it, especially because its surface is curved. The spherical tendency is a surface-energy effect, not an absence of internal pressure.
319. Water drops placed on a waxy surface often form rounded beads. This observation is mainly connected with
ⓐ. zero atmospheric pressure above the wax
ⓑ. poor wetting and surface tension
ⓒ. absence of gravity near the surface
ⓓ. water becoming a gas immediately
Correct Answer: poor wetting and surface tension
Explanation: A waxy surface is not easily wetted by water. Adhesion between water and wax is relatively weak compared with the cohesive attraction among water molecules. As a result, water tends to reduce its contact with the wax and forms rounded beads. Surface tension pulls the water surface into a compact shape. The effect is not due to atmospheric pressure becoming zero or water losing its liquid state.
320. Study the table and identify the row that gives the best surface-tension explanation.
| Row | Observation | Explanation |
|---|
| P | Small insects stand on water. | Surface tension helps support them. |
| Q | Detergent added to water. | Surface tension always becomes infinite. |
| R | Small drops become spherical. | Surface area is maximized for fixed volume. |
| S | Needle floats carefully on water. | Water density becomes zero. |
ⓐ. Row Q
ⓑ. Row R
ⓒ. Row S
ⓓ. Row P
Correct Answer: Row P
Explanation: Row P correctly connects insects standing on water with surface tension. Row Q is wrong because detergent usually reduces surface tension rather than making it infinite. Row R is wrong because drops tend to minimize surface area for a given volume, not maximize it. Row S is wrong because the density of water does not become zero when a needle is placed on it. The correct examples all depend on surface behaviour, but the physical reason must be stated carefully.
