The ring of radius 1 m is lying on the surface of liquid. It is lifted from the liquid surface by a force of 4 Newton in such a way that the liquid film in it remains intact. The surface tension of liquid will be

The force on the ring against the lift is $=2\times \mathrm{T}l$

where l is the length of the ring in contact with the surface of the water, and a factor of 2 is added due to the ring's inner and outer surfaces.

$4=2\times \mathrm{T}\times \left(2\mathrm{\pi r}\right) ⟹\mathrm{T}=\frac{1}{\mathrm{\pi }}\mathrm{N}/\mathrm{m}$

Hence the correct answer is $\frac{1}{\mathrm{\pi }}\mathrm{N}/\mathrm{m}.$