Question

Two narrow bores of diameters $3.0mm$ and $6.0mm$ 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 $7.3\times {10}^{-2}N{m}^{-1}$. Take the angle of contact to be zero and density of water to be $1.0\times {10}^{3}kg{m}^{-3}$ ($g=9.8m{s}^{-2}$).

Answer 1

Diameter of the first bore, $d_1=3.0mm=3\times {10}^{-3}m$

Hence, the radius of the first bore, $r_1=d_1/2=.5\times {10}^{-3}m$

Diameter of the first bore, $d_2=6.0mm=6\times {10}^{-3}mm$

Hence, the radius of the first bore, $r_2=d_2/2=3\times {10}^{-3}m$

Surface tension of water, $s=7.3\times {10}^{-2}N{m}^{-1}$

Angle of contact between the bore surface and water, $\theta =0$

Density of water, $\rho =1.0\times {10}^{3}kg/m^{-3}$
Acceleration due to gravity, $g=9.8m/s^2$
Let $h_1$ and $h_2$ be the heights to which water rises in the first and second tubes respectively. These heights are given by the relations:
$h_1=2scos\theta /r_1\rho g$ ..... (i)

$h_2=2scos\theta /r_2\rho g$ .... (ii)
The difference between the levels of water in the two limbs of the tube can be calculated as:

$=\frac{2scos\theta }{r_1\rho g}-\frac{2scos\theta }{r_2\rho g}$

$=\frac{2cos\theta }{\rho g}[\frac{1}{r_1}-\frac{1}{r_2}]$

$=\frac{2\times 7.3\times {10}^{-2}\times 1}{1\times {10}^3\times 9.8}[\frac{1}{1.5\times {10}^{-3}}-\frac{1}{3\times {10}^{-3}}]$

$=4.966\times {10}^{-3}m$

$=4.97mm$

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