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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

Consider a narrow capillary tube dipped vertically into a large...

Consider a narrow capillary tube dipped vertically into a large container of liquid with density ρ\rho and surface tension SS. The liquid rises in the tube to a height hh above the flat surface of the liquid in the container, forming a spherical meniscus. What is the radius of curvature RR of this meniscus?

Options

  1. AR=SρghR = \dfrac{S}{\rho g h}
  2. BR=2SρghR = \dfrac{2S}{\rho g h}
  3. CR=ρgh2SR = \dfrac{\rho g h}{2S}
  4. DR=S2ρghR = \dfrac{S}{2\rho g h}

Correct answer

B. R=2SρghR = \dfrac{2S}{\rho g h}

Step-by-step solution

The pressure difference across a spherical meniscus of radius of curvature RR is given by ΔP=2SR\Delta P = \dfrac{2S}{R}.

This pressure difference is balanced by the hydrostatic pressure of the liquid column of height hh, which is ΔP=ρgh\Delta P = \rho g h.

Equating the two expressions for the pressure difference:

2SR=ρgh\dfrac{2S}{R} = \rho g h

Solving for RR:

R=2SρghR = \dfrac{2S}{\rho g h}

Answer: R=2SρghR = \dfrac{2S}{\rho g h}
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