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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2026 · 22 January · Evening Shift · Physics · Mechanical Properties of Fluids

When a part of a straight capillary tube is placed vertically in a...

When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height hh. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by 1%1 \% each, then the height of the liquid in the tube will change by %\%.

Options

  1. A+1+1
  2. B-3
  3. C+3+3
  4. D-1

Correct answer

A. +1+1

Step-by-step solution

Capillary height formula: h=2σcos⁡θρgrh = \frac{2\sigma\cos\theta}{\rho g r}
When radius, density, and surface tension each decrease by 1%: r′=0.99rr' = 0.99r, ρ′=0.99ρ\rho' = 0.99\rho, σ′=0.99σ\sigma' = 0.99\sigma
New height: h′=2(0.99σ)cos⁡θ(0.99ρ)g(0.99r)=h⋅0.990.99×0.99=h⋅10.99=h⋅1.0101...h' = \frac{2(0.99\sigma)\cos\theta}{(0.99\rho) g (0.99r)} = h \cdot \frac{0.99}{0.99 \times 0.99} = h \cdot \frac{1}{0.99} = h \cdot 1.0101...
Percentage change: (1.0101−1)×100≈+1.01%(1.0101 - 1) \times 100 \approx +1.01\%, which rounds to +1%.
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