Q.Is surface tension a vector?

Showing the 12 most recent of 23 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Calculate the amount of work done in spraying a drop of a liquid of radius 1 mm into million identical droplets under isothermal conditions. [surface tension of liquid $= 550 \times 10^{-3}Nm^{-1}$] (A) $6.84 \times 10^{-4}J$ (B) $5.50 \times 10^{-4}J$ (C) $7.25 \times 10^{-4}J$ (D) $3.42 \times 10^{-4}J$

   ›Reveal solutionSolution

   Tests the energy cost of increasing surface area when a drop is broken into many smaller droplets. Answer: $6.84\times10^{-4}$ J.

   Concept and Intuition

   Surface tension means the surface of a liquid stores energy proportional to its area. Breaking one large drop into many tiny droplets dramatically increases the total surface area (surface-to-volume ratio grows as size shrinks), and the work done against surface tension to create this new surface equals $\sigma\times(\text{increase in area})$.

   Step-by-Step Solution

   1. Volume is conserved: $\dfrac43\pi R^3=n\times\dfrac43\pi r^3 \Rightarrow R^3=nr^3$.
   2. With $n=10^6$: $r=\dfrac{R}{n^{1/3}}=\dfrac{R}{100}=\dfrac{10^{-3}}{100}=10^{-5}$ m.
   3. Total initial surface area: $A_i=4\pi R^2=4\pi(10^{-3})^2=4\pi\times10^{-6}\ m^2$.
   4. Total final surface area: $A_f=n\times4\pi r^2=10^6\times4\pi(10^{-5})^2=4\pi\times10^{-4}\ m^2$. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A capillary tube of inner radius 1.5 mm is dipped vertically in water. If the surface tension of water is $7\times10^{-2}\ Nm^{-1}$, then the volume of the water that rises in the capillary tube is (Acceleration due to gravity = $10\ ms^{-2}$) (A) 0.022 cc (B) 0.066 cc (C) 0.099 cc (D) 0.033 cc

   ›Reveal solutionSolution

   First find the capillary rise height from the surface-tension formula $h=2T\cos\theta/(\rho g r)$, then find the volume of the risen water column as a simple cylinder $V=\pi r^2 h$. This gives $V\approx0.066\,cc$.

   Concept and Intuition

   Capillary action pulls water up a narrow tube until the upward pull from surface tension around the tube's circumference balances the weight of the risen water column. Once we know how high the water rises, the volume risen is simply that of a cylinder of the tube's cross-section and that height — this is a standard two-step "find height, then find volume" problem, and for a wetting liquid like water on glass the contact angle is taken as $0^\circ$ so $\cos\theta=1$.

   Step-by-Step Solution

   1. Capillary rise formula: $h = \dfrac{2T\cos\theta}{\rho g r}$, with $\theta \approx 0^\circ$ for water in a clean glass capillary, so $\cos\theta=1$.
   2. Given: $T = 7\times10^{-2}\,N/m$, $\rho = 1000\,kg/m^3$ (water), $g=10\,ms^{-2}$, $r = 1.5\,mm = 1.5\times10^{-3}\,m$.
   3. $h = \dfrac{2\times7\times10^{-2}}{1000\times10\times1.5\times10^{-3}} = \dfrac{0.14}{15} = 9.33\times10^{-3}\,m \approx 9.33\,mm$.
   4. Volume of the risen column (a cylinder of radius $r$ and height $h$): $V = \pi r^2 h$. …

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A metallic wire of density $\rho$ is placed horizontally on the surface of a liquid with surface tension T. What is the maximum radius the wire can have so that it remains supported by surface tension? (A) $\sqrt{\dfrac{2T}{\pi \rho g}}$ (B) $\sqrt{\dfrac{T}{\pi \rho g}}$ (C) $\sqrt{\dfrac{3\rho g}{2T}}$ (D) $\sqrt{\dfrac{2\pi}{T \rho g}}$

   ›Reveal solutionSolution

   Tests the classic "needle floats on water" surface-tension balance, generalized to find the maximum radius a wire can have and still be supported.

   Concept and Intuition

   A dense object can float on a liquid surface without being buoyant at all — surface tension along the two contact lines (one on each side of the wire) pulls upward and can support the object's weight, up to a limit. Beyond a critical radius, the weight (which grows as $r^2$, i.e. volume) outgrows the supporting force (which is fixed by $T$ and length, independent of $r$), and the wire sinks.

   Step-by-Step Solution

   1. The wire has length $L$, radius $r$, density $\rho$; its weight is $mg = \rho\,(\text{volume})\,g = \rho(\pi r^2 L)g$.
   2. Surface tension acts along both lines of contact where the liquid surface meets the wire, each of length $L$, giving a maximum vertical support force of $2TL$ (the tension pulls essentially vertically when the wire is on the verge of breaking through).
   3. At the maximum supportable radius, the two are exactly balanced: $2TL = \rho\pi r^2 L g$ …

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4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.A long cylindrical vessel of glass having a hole of 0.5 mm radius at its bottom is slowly lowered vertically into a deep water bath. If the surface tension of water is $7\times10^{-2}\ \mathrm{Nm^{-1}}$, then the maximum depth the vessel can be submerged without water entering through the hole is (Acceleration due to gravity = $10\ \mathrm{ms^{-2}}$) (A) $4.2\ cm$ (B) $5.6\ cm$ (C) $2.8\ cm$ (D) $1.4\ cm$

   ›Reveal solutionSolution

   This tests the surface-tension pressure-barrier formula for a small hole resisting liquid entry — the maximum safe submersion depth comes out to $2.8\ cm$.

   Concept and Intuition

   As the vessel (with air trapped inside) is lowered into water, the hole at the bottom would let water in, but surface tension at the tiny hole forms a curved meniscus (like a hemispherical cap) that resists this, providing an effective excess pressure of $\dfrac{2T}{r}$ (analogous to the excess pressure inside a spherical liquid surface of radius $r$). Water starts entering only when the external hydrostatic pressure at that depth exceeds this surface-tension barrier. So the maximum depth is exactly when the two are equal.

   Step-by-Step Solution

   1. Surface-tension pressure barrier at the hole: $\Delta P = \dfrac{2T}{r}$.
   2. Hydrostatic pressure at depth $h$: $P = \rho g h$.
   3. Maximum depth (water on the verge of entering): $\rho g h = \dfrac{2T}{r}$. …

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5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A liquid drop of diameter D splits into 3375 small identical drops. If S is the surface tension of the liquid, then the change in the surface energy in the process is (A) $44D^2S$ (B) $44\pi D^2S$ (C) $56D^2S$ (D) $56\pi D^2S$

   ›Reveal solutionSolution

   The extra surface energy created is $14\pi D^2 S \approx 44 D^2 S$.

   Concept and Intuition

   Breaking a large drop into many small ones conserves volume but sharply increases total surface area. The energy needed equals surface tension multiplied by the increase in area: $\Delta U = S\,\Delta A$.

   Step-by-Step Solution

   1. Volume conservation: $\tfrac{4}{3}\pi (D/2)^3 = 3375\cdot\tfrac{4}{3}\pi r^3$. Since $3375 = 15^3$, $r = \dfrac{D/2}{15} = \dfrac{D}{30}$.
   2. Initial area: $A_i = 4\pi (D/2)^2 = \pi D^2$.
   3. Final area: $A_f = 3375\cdot 4\pi r^2 = 3375\cdot 4\pi\left(\dfrac{D}{30}\right)^2 = \dfrac{3375\cdot 4}{900}\pi D^2 = 15\pi D^2$.
   4. Change in energy: $\Delta U = S(A_f - A_i) = S(15\pi D^2 - \pi D^2) = 14\pi D^2 S$. …

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6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.A mercury drop of radius 1 cm is divided into $10^6$ droplets of equal size. If surface tension of mercury is $35 \times 10^{-3}\ \mathrm{N\,m^{-1}}$, then the change in surface energy in the process is (A) 8712 $\mu$J (B) 8712 erg (C) 4356 $\mu$J (D) 4356 erg

   ›Reveal solutionSolution

   Splitting a drop into many smaller droplets increases total surface area, and the extra surface energy comes from work done against surface tension. Answer: 4356 $\mu$J.

   Concept and Intuition

   When a liquid drop is broken into many smaller droplets, the total volume is conserved but the total surface area increases (smaller spheres have a larger surface-to-volume ratio). Since surface energy $=T\times\text{area}$, this increase in area must come from work done on the liquid, drawn from an external agent (this is why atomising a liquid, e.g., in a spray, requires energy input).

   Step-by-Step Solution

   1. Volume conservation for splitting one drop of radius $R$ into $n$ droplets of radius $r$: $\dfrac{4}{3}\pi R^3=n\cdot\dfrac{4}{3}\pi r^3 \Rightarrow r=\dfrac{R}{n^{1/3}}$.
   2. Here $R=1\ \text{cm}=0.01\ \text{m}$, $n=10^6$, so $n^{1/3}=100$ and $r=0.01/100=10^{-4}\ \text{m}$.
   3. Initial surface area: $A_i=4\pi R^2=4\pi(0.01)^2=4\pi\times10^{-4}\ \text{m}^2$.
   4. Final total surface area: $A_f=n\times4\pi r^2=10^6\times4\pi\times(10^{-4})^2=4\pi\times10^{-2}\ \text{m}^2$. …

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7. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.The work to be done to blow a soap bubble of radius $3\times10^{-3}\ \mathrm{m}$ is nearly (surface tension of soap solution $= 20\times10^{-3}\ \mathrm{N\,m^{-1}}$) (A) $4.5\times10^{-4}$ J (B) $4.5\times10^{-5}$ J (C) $4.5\times10^{-6}$ J (D) $4.5\times10^{-7}$ J

   ›Reveal solutionSolution

   Because a soap bubble has two surfaces, the work to blow it is $W = 8\pi T r^2$, which evaluates to about $4.5\times10^{-6}$ J.

   Concept and Intuition

   Surface tension does work whenever the surface area of a liquid film increases: $W = T\times\Delta A$. A soap film making a bubble is a thin liquid membrane with air on both the inside and the outside, so it presents two surfaces to the surrounding air, each of area $4\pi r^2$ for a sphere of radius $r$. Starting from essentially zero area (a tiny initial droplet/film) and blowing it up to radius $r$, the total new surface area created is $2\times4\pi r^2 = 8\pi r^2$.

   Step-by-Step Solution

   1. Total surface area of the bubble (two surfaces): $\Delta A = 2\times4\pi r^2 = 8\pi r^2$.
   2. Substitute $r = 3\times10^{-3}\ \text{m}$: $r^2 = 9\times10^{-6}\ \text{m}^2$, so $\Delta A = 8\pi\times9\times10^{-6} = 72\pi\times10^{-6}\ \text{m}^2 \approx 2.262\times10^{-4}\ \text{m}^2$. …

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8. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.A wire of length 20 cm is placed horizontally on the surface of water and is gently pulled up with a force of $1.456\times10^{-2}\ \mathrm{N}$ to keep the wire in equilibrium. The surface tension of water is (A) $0.00364\ \mathrm{N\,m^{-1}}$ (B) $0.0364\ \mathrm{N\,m^{-1}}$ (C) $0.00464\ \mathrm{N\,m^{-1}}$ (D) $0.0864\ \mathrm{N\,m^{-1}}$

   ›Reveal solutionSolution

   Because the water film clings to both faces of the wire as it is lifted off the surface, the pulling force balances TWO surface-tension lengths, not one; dividing correctly gives $T=0.0364\ \mathrm{N\,m^{-1}}$ — option (B).

   Concept and Intuition

   Surface tension is force per unit length acting along the free surface of a liquid, tangential to it. When a straight wire lying on a water surface is pulled straight up, the film does not merely pull along one edge — it wets the wire along both its front and back faces (think of a thin rectangular film bounded by the wire on one side, exactly like the classic soap-film-on-a-slider problem). So the total upward-resisting force from surface tension is $F=T\times(2L)$, twice what you'd get from naively using just the wire's length.

   Step-by-Step Solution

   1. Identify the two contact lines of length $L$ each (front and back faces of the wire) along which surface tension acts: total length $=2L$.
   2. Equilibrium condition: applied force = surface-tension force, i.e. $F=T\times 2L$.
   3. Solve for $T$: $T=\dfrac{F}{2L}$. …

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9. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.Two soap bubbles of radii $R_1$ and $R_2$ are kept in vacuum at constant temperature, the ratio of masses of air inside them are (A) $\dfrac{m_1}{m_2} = \dfrac{R_1}{R_2}$ (B) $\dfrac{m_1}{m_2} = \dfrac{R_2}{R_1}$ (C) $\dfrac{m_1}{m_2} = \dfrac{R_2^2}{R_1^2}$ (D) $\dfrac{m_1}{m_2} = \dfrac{R_1^2}{R_2^2}$

   ›Reveal solutionSolution

   This tests combining the soap-bubble excess-pressure formula ($4T/R$, two surfaces) with the ideal gas law to find how enclosed air mass scales with bubble radius.

   Concept and Intuition

   A soap bubble has two liquid-air surfaces (inner and outer film surface), so its excess pressure is $\Delta P = 4T/R$ (twice that of a single water drop). In vacuum, there's no external atmospheric pressure to add, so the absolute pressure of air trapped inside the bubble is exactly this excess pressure. Using the ideal gas law at fixed temperature, the mass of gas enclosed is proportional to $PV$, and substituting both $P\propto 1/R$ and $V\propto R^3$ shows the radius-dependence combines to give mass $\propto R^2$.

   Step-by-Step Solution

   1. Absolute pressure of air inside a soap bubble in vacuum: $P=\dfrac{4T}{R}$ (T = surface tension, same liquid film for both bubbles).
   2. Volume of the bubble: $V=\dfrac43\pi R^3$.
   3. From ideal gas law at constant temperature, $PV = \dfrac{m}{M}RT_{gas} \Rightarrow m \propto PV$.
   4. So $m \propto \dfrac{4T}{R}\times\dfrac43\pi R^3 = \dfrac{16\pi T}{3}R^2 \propto R^2$. …

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10. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
    Q.A glass capillary tube with a radius $r = 0.02$ cm is immersed into water to a depth of $d = 2$ cm. To blow an air bubble out of the lower end of the tube the pressure required is (Given surface tension $T = 7 \times 10^{-2}\ \text{Nm}^{-1}$ and density of water $\rho = 10^3\ \text{kgm}^{-3}$; acceleration due to gravity $g = 10\ \text{ms}^{-2}$). (A) $480\ \text{Nm}^{-2}$ (B) $900\ \text{Nm}^{-2}$ (C) $200\ \text{Nm}^{-2}$ (D) $700\ \text{Nm}^{-2}$

    ›Reveal solutionSolution

    The air bubble forming at the submerged tube mouth needs enough pressure to overcome both the hydrostatic pressure of the water above it and the extra (Laplace) pressure due to its single curved surface, $2T/r$.

    Concept and Intuition

    An air bubble growing at the open end of a capillary tube dipped in water has just ONE curved liquid surface (unlike a soap bubble in air, which has two surfaces and needs $4T/r$). To push air out at depth $d$, the air pressure must exceed the local water pressure at that depth (atmospheric + $\rho g d$) by an extra amount $2T/r$ to overcome surface tension and form the curved bubble surface.

    Step-by-Step Solution

    1. Hydrostatic (gauge) pressure at depth $d$: $\rho g d = 1000\times10\times0.02 = 200\ \text{N/m}^2$.
    2. Convert the radius: $r = 0.02\ \text{cm} = 2\times10^{-4}\ \text{m}$. …

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11. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.If $S_1$, $S_2$ and $S_3$ are the tensions at liquid-air, solid-air and solid-liquid interfaces respectively, and $\theta$ is the angle of contact at the solid-liquid interface, then (A) $S_1\cos\theta + S_2\sin\theta = S_3$ (B) $S_1\cos\theta + S_3 = S_2$ (C) $S_2\cos\theta + S_3 = S_1$ (D) $S_3\cos\theta + S_1 = S_2$

    ›Reveal solutionSolution

    This is Young's equation relating the three interfacial tensions and the contact angle: $S_1\cos\theta+S_3=S_2$.

    Concept and Intuition

    At the line where solid, liquid and air (or vapour) meet, the surface tensions acting along their respective interfaces must balance for mechanical equilibrium along the solid surface direction. The solid–air tension $S_2$ pulls the contact line one way; the solid–liquid tension $S_3$ and the horizontal component of the liquid–air tension $S_1\cos\theta$ pull the other way (where $\theta$ is measured through the liquid).

    Step-by-Step Solution

    1. At the three-phase contact line, resolve tensions along the solid surface.
    2. $S_2$ (solid–air) acts to pull the contact line outward along the solid surface. …

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12. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
    Q.Two soap bubbles of radii $r_1$ and $r_2$ in vacuum coalesce isothermally to form a new bubble. The radius of the new soap bubble is (A) $\sqrt{r_1^2 + r_2^2}$ (B) $\sqrt{2 r_1 r_2}$ (C) $\sqrt{\dfrac{r_1 r_2}{2}}$ (D) $\dfrac{r_1 r_2}{r_1 + r_2}$

    ›Reveal solutionSolution

    Conserving $pV$ for two soap bubbles merging in vacuum (where excess pressure is purely from surface tension) gives the new radius as $r=\sqrt{r_1^2+r_2^2}$.

    Concept and Intuition

    A soap bubble has two liquid surfaces (inner and outer film), so its excess internal pressure due to surface tension is $\Delta p = \dfrac{4T}{r}$. In vacuum, there's no atmospheric pressure term to add, so the absolute pressure inside each bubble is purely $p = 4T/r$. When the process is isothermal and we treat the enclosed air in each bubble as an ideal gas, the product $pV$ (proportional to the number of moles times $RT$, with $T$ constant) is additive when the two amounts of gas combine into one bubble: $p_1V_1 + p_2V_2 = pV$.

    Step-by-Step Solution

    1. Pressure inside bubble 1 (in vacuum): $p_1 = \dfrac{4T}{r_1}$; volume $V_1 = \dfrac{4}{3}\pi r_1^3$.
    2. Similarly for bubble 2: $p_2 = \dfrac{4T}{r_2}$, $V_2 = \dfrac43\pi r_2^3$.
    3. New bubble: $p = \dfrac{4T}{r}$, $V = \dfrac43\pi r^3$.
    4. Isothermal conservation: $p_1V_1 + p_2V_2 = pV$.
    5. Substituting: $\dfrac{4T}{r_1}\cdot\dfrac43\pi r_1^3 + \dfrac{4T}{r_2}\cdot\dfrac43\pi r_2^3 = \dfrac{4T}{r}\cdot\dfrac43\pi r^3$. …

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