a
Diameter of the first bore, \(d_{1}=3.0 mm =3 \times 10^{-3} m\)

Hence, the radius of the first bore, \(r_{1}=\frac{d_{1}}{2}=1.5 \times 10^{-3} m\)

Diameter of the second bore, \(d_{2}=6.0 mm\)

Hence, the radius of the second bore, \(r_{2}=\frac{d_{2}}{2}=3 \times 10^{-3} m\)

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, \(g=9.8 m / s ^{2}\) 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:

\(h_{1}=\frac{2 s \cos \theta}{r_{1} \rho g}\)

\(h_{2}=\frac{2 s \cos \theta}{r_{2} \rho g}\)

The difference between the levels of water in the two limbs of the tube can be calculated as

\(=\frac{2 s \cos \theta}{r_{i} \rho g }-\frac{2 s \cos \theta}{r_{2} \rho g }\)

\(=\frac{2 s \cos \theta}{\rho g}\left[\frac{1}{r_{1}}-\frac{1}{r_{2}}\right]\)

\(=\frac{2 \times 7.3 \times 10^{-2} \times 1}{1 \times 10^{3} \times 9.8}\left[\frac{1}{1.5 \times 10^{-3}}-\frac{1}{3 \times 10^{-3}}\right]\)

\(=4.966 \times 10^{-3} m\)

\(=4.97 mm\)

Hence, the difference between levels of water in the two bores is \(4.97\; mm\)