Let take \(g=10 \,m / s ^2\)

For water to enter the sphere, pressure required is \(=\rho g h\)

\(=1 \times 10 \times \frac{40}{100} \times 1000 \quad\left(\rho=1000 \,kg / m ^3\right)\)

\(=4000 \,\frac{ N }{ m ^2}=\text { excess pressure }\)

Let the hole have radius \(=R\)

Excess pressure \(=\frac{2 T}{R}\)    [One surface air, one surface water]

\(\Rightarrow 4000=\frac{2 \times 0.07}{R}\)

\(\Rightarrow 2 R=0.07 \times 10^{-3} \,m\)

\(\Rightarrow d=0.07 \,mm\)