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Home » 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? 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 m s–2).
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? 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 m s–2).
CBSE | Class 11 | Mechanical Properties of Fluids | NCERT | Physics
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? 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 m s–2).

Answer :

According to the question, the diameter of the first bore is

d1 = 3.0 mm = 3 × 10-3 m
The radius of the first bore is

r1 = 3/2 = 1.5 x 10-3 m.
The diameter of the second bore is

d2 =6mm
The radius of the second bore is

r2 = 6/2 = 3 x 10-3 mm
The surface tension of water is

s = 7.3 × 10-2 N /m
The angle of contact between the bore surface and water is 

θ= 0
And the density of water is

ρ =1.0 × 103 kg/m-3
Acceleration due to gravity, g = 9.8 m/s2
Let h1 and h2 be the corresponding heights to which water rises in the first and second tubes.
As a result, the height differences can be writte using the expression of height

h=\frac{2s\cos \theta }{r\rho g}

{{h}_{1}}-{{h}_{2}}=\frac{2s\cos \theta }{{{r}_{1}}\rho g}-\frac{2s\cos \theta }{{{r}_{2}}\rho g}

{{h}_{1}}-{{h}_{2}}=\frac{2s\cos \theta }{\rho g}\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]

{{h}_{1}}-{{h}_{2}}=\frac{2\times 7.3\times {{10}^{-2}}\times 1}{{{10}^{3}}\times 9.8}\left[ \frac{1}{1.5\times {{10}^{-3}}}-\frac{1}{3\times {{10}^{-3}}} \right]

{{h}_{1}}-{{h}_{2}}=4.97mm

Therefore, 2.482 mm is the difference in the water levels of the two arms.

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