Question

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-shaped 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 $1.0\times {10}^{3}kg{m}^3(g=9.8m{s}^{-2})$

• A. 3 mm
• B. 2 mm
• C. 4 mm
• D. 5.0 mm

Answer 1

The correct option is A 3 mm

Given that $r_1=\frac{3.0}{2}=1.5mm=1.5\times {10}^{-3}m,r_2=\frac{6.0}{2}=3.0mm=3.0\times {10}^{-3}m$

$T=7.3\times {10}^{-2}N{m}^{-1},\theta =0^{\circ }\rho =1.0\times {10}^{3}kg{m}^{-1},g=9.8m{s}^{-2}$
When angle of contact is zero degree the radius of the meniscus equals radius of bore
Excess pressure in the first bore $P_1=\frac{2T}{r_2}=\frac{2\times 7.3\times {10}^{-2}}{1.5\times {10}^{-3}}=97.3$ Pascal
Excess pressure in the second bore,$P_2=\frac{2T}{r_2}=\frac{2\times 7.3\times {10}^{-2}}{3\times {10}^{-3}}=48.7$ Pascal
Hence, pressure difference in the two limbs of the tube
$\Delta P=P_1-P_2=h\rho g$
or $h=\frac{P_1-P_2}{\rho g}=\frac{97.3-48.7}{1.0\times {10}^3\times 9.8}=5.0mm$