Question

Calculate the surface energy of ring floating on the liquid surface? (surface tension of liquid is 75 N/m and are
pf ring is 0.04 m2

Answer 1

The surface energy of ring floating on the liquid surface is 3 N-m.

Explanation:

Surface energy 'S.E' is the enrgy which is associated with intermolecular level and is developed with the formation of intermolecular bond between liquid and a surface.

Given,

            Surface tension of liquid 'T' = 75 N/m

            Surface area of ring interaction with the liquid surface 'S.A'  = 0.04 $m^{2}$

We know that  

                          Surface energy = Surface tension x Surface area

                                   ⇒  S.E = T x S.A

                                   ⇒  S.E = 75 x 0.04

                                   ⇒  S.E = 3 N-m

Surface energy of the ring floating on liquid is 3 N-m.