Also, \(\mathrm{P}=\mathrm{P}_{0}+\frac{4 \mathrm{T}}{\mathrm{R}}\ldots\) (ii)

From \((i) \;and\; (ii)\)

\(\rho g Z_{0}=\frac{4 \mathrm{T}}{\mathrm{R}}\)

\(\mathrm{z}_{0}=\frac{4 \mathrm{T}}{\mathrm{pgR}}\)

\(=\frac{4 \times 2.5 \times 10^{-2}}{10^{3} \times 10 \times 10^{-3}}=10^{-2} \mathrm{m}=1 \;\mathrm{cm}\)