Question

Two narrow bores of diameter 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 kg 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 1

Diameter of the first tube = 3 mm
Diameter of the second tube = 6 mm
Thus, Radius of the first tube, $r_1$ = 1.5 mm = 1.5 × 10^-3 m
Radius of the first tube, $r_2$= 3 mm = 3 × 10^-3 m
Surface tension of water, S = 7.3 × 10^-2 N/m
Density of water, ρ = 1000 kg/m³
Angle of contact between bore surface and water, θ = 0
Acceleration due to gravity, g = 9.8 m/s

Height of water rise, $\frac{2Scos\theta}{rho gr$
Where,
S = Surface tension
θ = Contact angle
ρ = Density
g = acceleration due to gravity
r = radius

Let $h_1$ and $h_2$ be the height rise of water in tube 1 and tube 2 respectively. Then the height 
$h_1=\frac{2Scos\theta}{\rho r_1g}$ and $h_2=\frac{2Scos\theta}{\rho r_2g}$

thus, the difference between heights is given by ∆h = $h_1-h_2$

=$\frac{2Scos\theta}{\rho g}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

= 2 × 7.3 × 10^-2 cos0°/(1000 × 9.8) × (1/1.5 × 10^-3 - 1/3 × 10^-3)

= 4.97 mm

Answer 2

Refer to the attached image.