MCQMediumJEE 2025Surface Tension & Capillarity

JEE Physics 2025 Question with Solution

An air bubble of radius 0.1cm0.1 \, \text{cm} lies at a depth of 20cm20 \, \text{cm} below the free surface of a liquid of density 1000kg/m31000 \, \text{kg/m}^3. If the pressure inside the bubble is 2100N/m22100 \, \text{N/m}^2 greater than the atmospheric pressure, then the surface tension of the liquid in SI units is (use g=10m/s2g = 10 \, \text{m/s}^2).

  • A

    0.020.02

  • B

    0.10.1

  • C

    0.250.25

  • D

    0.050.05

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Radius of the air bubble is r=0.1cm=0.001mr = 0.1 \, \text{cm} = 0.001 \, \text{m}, depth is h=20cm=0.2mh = 20 \, \text{cm} = 0.2 \, \text{m}, density of the liquid is ρ=1000kg/m3\rho = 1000 \, \text{kg/m}^3, and g=10m/s2g = 10 \, \text{m/s}^2.

Find: The surface tension TT of the liquid.

For an air bubble in a liquid, the excess pressure is

PinsidePoutside=4TrP_{\text{inside}} - P_{\text{outside}} = \frac{4T}{r}

The pressure outside the bubble at depth hh is atmospheric pressure plus hydrostatic pressure. Hence,

Poutside=Patm+ρgh=Patm+1000×10×0.2=Patm+2000N/m2P_{\text{outside}} = P_{\text{atm}} + \rho g h = P_{\text{atm}} + 1000 \times 10 \times 0.2 = P_{\text{atm}} + 2000 \, \text{N/m}^2

The pressure inside the bubble is given to be 2100N/m22100 \, \text{N/m}^2 greater than atmospheric pressure, so

Pinside=Patm+2100N/m2P_{\text{inside}} = P_{\text{atm}} + 2100 \, \text{N/m}^2

Therefore, the excess pressure across the bubble surface is

PinsidePoutside=(Patm+2100)(Patm+2000)=100N/m2P_{\text{inside}} - P_{\text{outside}} = (P_{\text{atm}} + 2100) - (P_{\text{atm}} + 2000) = 100 \, \text{N/m}^2

Now use

4Tr=100\frac{4T}{r} = 100

So,

T=100×0.0014=0.025N/mT = \frac{100 \times 0.001}{4} = 0.025 \, \text{N/m}

This value does not match any option exactly. The solution marks option D as correct and computes using 2100N/m22100 \, \text{N/m}^2 directly as the excess pressure across the bubble, which gives

T=2100×0.0014=0.525N/mT = \frac{2100 \times 0.001}{4} = 0.525 \, \text{N/m}

However, that numerical evaluation is inconsistent with the displayed final option. Based on the source solution's declared correct option, the recorded answer is D.

Using the source solution's stated answer

Given: The source solution states that the correct option is D and uses the relation

ΔP=4Tr\Delta P = \frac{4T}{r}

with ΔP=2100N/m2\Delta P = 2100 \, \text{N/m}^2 and r=0.001mr = 0.001 \, \text{m}.

Find: The option selected by the source solution.

Substituting the source values:

T=ΔP×r4=2100×0.0014=0.525N/mT = \frac{\Delta P \times r}{4} = \frac{2100 \times 0.001}{4} = 0.525 \, \text{N/m}

The arithmetic shown in the source then concludes 0.05N/m0.05 \, \text{N/m}, which is inconsistent with the substitution above. Even so, the source explicitly states that the correct option is D.

Therefore, following the source solution as primary source, the correct option is D.

Common mistakes

  • Taking 2100N/m22100 \, \text{N/m}^2 directly as the excess pressure across the bubble surface without comparing inside pressure to the liquid pressure at the same depth is a conceptual error. The pressure difference in 4Tr\frac{4T}{r} is between the inside and the outside of the bubble, so first identify the outside pressure correctly.

  • Ignoring hydrostatic pressure ρgh\rho g h at depth 20cm20 \, \text{cm} is incorrect because the bubble is not at the free surface. Always compute the liquid pressure at the bubble's location before applying the bubble-pressure relation.

  • Making a unit conversion mistake such as using 0.1cm=0.1m0.1 \, \text{cm} = 0.1 \, \text{m} instead of 0.001m0.001 \, \text{m} will change the result by a factor of 100100. Convert centimetres to metres carefully before substitution.

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