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Mechanical Properties Of Fluids

Question
CBSEENPH11020019

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).

Solution
Diameter of the first bore, d1 = 3.0 mm = 3 × 10–3 m 
So, the radius of the first bore, r1 = d1/ 2  =  1.5 × 10-3 m
Diameter of the second 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 hbe the heights to which water rises in the first and second tubes respectively.

The space heights space are space given space by space the space relation comma space

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So comma space difference space between space the space levels space of space water space in space the space two space
limbs space of space the space tube space can space be space calculated space as colon

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space space space space space space equals space 4.97 space mm space

Therefore comma space the space difference space between space levels space of
water space in space the space two space bores space is space 4.97 space mm.