NEET2025Physics-Fluid Mechanics
NEET 2025 Physics Surface Tension MCQ Question
Type: MCQ-conceptual-Hard-Class 11
Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θ₀ (θ₀ << 1) with the x-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is:
(take θ(x) = sin θ(x) = tan θ(x) = dy/dx, g is the acceleration due to gravity)
A
d²y/dx² = ρg/S x
B
d²y/dx² = ρg/S y
C
d²y/dx² = √(ρg/S)
D
dy/dx = √(ρg/S x)
Correct Answer
Option B
Detailed Explanation
Answer: (B)
Solution:
At any point A on the liquid surface, let the height of liquid be y.
Radius of Curvature (ROC) at point A = ρ
Curvature = 1 / ROC
= (d²y/dx²) / [1 + (dy/dx)²]^(3/2)
Given θ is very small, therefore:
tanθ = dy/dx ≈ 0
Hence,
Curvature = d²y/dx²
Excess pressure due to surface tension:
ΔP = S × Curvature
= S(d²y/dx²)
Hydrostatic pressure at height y:
ΔP = ρgy
Therefore,
ρgy = S(d²y/dx²)
So,
d²y/dx² = (ρg/S) y
Hence, correct option is B .
Found an issue with this question?