Question

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 x 10⁻² N m⁻¹. Take the angle of contact to be zero and density of water to be 1.0 x 10³ kg m⁻³ (g = 9.8 m s⁻²).

Answer 1

Hey mate ^_^

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Answer:
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d1 = 3 × 10^–3 m

r1 = d1/ 2  =  1.5 × 10^-3 m

d2 = 6 × 10^–3 mm

r2 = 3 × 10^-3 m

Surface tension of water = 7.3 × 10^–2 N m^–1

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

ρ =1.0 × 10^3 kg/m^–3


These heights are given by the relations:

h1 = 2s Cosθ / r1ρg    - - - - - - -> (1)
h2 = 2s Cosθ / r2ρg   - - - - - - > (2)

Check this attachment :-)

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Final answer: 4.97 mm
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#Be Brainly❤️

Answer 2

Diameter of the first bore, d1 = 3.0 mm = 3 × 10–3 m
Hence, the radius of the first bore, r1 = d1/ 2  =  1.5 × 10-3 m
Diameter of the first bore, d2 = 6.0 mm = 6 × 10–3 mm
Hence, the radius of the first bore, r2 = d2/ 2  =  3 × 10-3 m
Surface tension of water, s = 7.3 × 10–2 N m–1
Angle of contact between the bore surface and water, θ= 0
Density of water, ρ =1.0 × 103 kg/m–3
Acceleration due to gravity, g = 9.8 m/s2
Let h1 and h2 be the heights to which water rises in the first and second tubes respectively. These heights are given by the relations:
h1 = 2s Cosθ / r1ρg      …..(i)
h2 = 2s Cosθ / r2ρg      …..(ii)
The difference between the levels of water in the two limbs of the tube can be calculated as:
= 4.966 × 10-3 m
= 4.97 mm

Hence, the difference between levels of water in the two bores is 4.97 mm.