Height of liquid in capillary \(=\frac{2 T}{r \rho g}=h\)

Pressure we need to apply \(=\rho g h+P_0\)   \(\left\{\begin{array}{l}\text { Where, } \\ T=\text { Surface tension } \\ r=\text { Radius of capillary } \\ \rho=\text { Density of liquid } \\ P_0=\text { Atmospheric pressure }\end{array}\right.\)

Substitute value of \(h\)

\(P=\rho g \times \frac{2 T}{r \rho g}+P_0=\frac{2 T}{r}+P_0=\frac{4 T}{d}+P_0\)

\(\Rightarrow P=\frac{4 \times 0.07}{\left(0.28 \times 10^{-3}\right)}+P_0=1000 \,Nm ^{-2}+10^5 \,Nm ^{-2}\)    \(\left\{\begin{array}{l}\text { Given, } \\ T=0.07 \,N / m \\ d=0.28 \,mm \end{array}\right.\)

\(\Rightarrow P=\left(10^3+10^5\right) \,Nm ^{-2}=101 \times 10^3 \,Nm ^{-2}\)