Ncert Exercises & Solutions

10.30 Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? The surface tension of water at the temperature of the experiment is 7.3 × 10^–2 N m^–1. Take the angle of contact to be zero and density of water to be 1.0 × 10³ kg m^–3 (g = 9.8 ${ms}^{- 2}$ ).

The radius of the first bore, $r_{1} = \frac{d_{1}}{2} = 1.5 \times 10^{- 3} m$

The radius of the second bore, $r_{2} = \frac{d_{2}}{2} = 3 \times 10^{- 3} m$

The surface tension of water, s = 7.3 × 10^–2 N m^–1

The angle of contact between the bore surface and water, θ = 0

Let h₁ and h₂ be the heights to which water rises in the first and second tubes respectively.

Then,

$h_{1} = \frac{2 s cosθ}{r_{1} ρg} . . . (i)$
$h_{2} = \frac{2 s cosθ}{r_{2} pg} . . . (ii)$

The difference between the levels of water in the two limbs of the tube:

$h_{1} - h_{2} = \frac{2 s cosθ}{r_{1} ρg} - \frac{2 s cosθ}{r_{2} ρg} = \frac{2 s cosθ}{ρg} [\frac{1}{r_{1}} - \frac{1}{r_{2}}] = \frac{2 \times 7.3 \times 10^{- 2} \times 1}{1 \times 10^{3} \times 9.8} [\frac{1}{1.5 \times 10^{- 3}} - \frac{1}{3 \times 10^{- 3}}] = 4.966 \times 10^{- 3} m = 4.97 mm$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions