Question

Find the surface energy of water kept in a cylindrical vessel of radius 6 cm surface tension of water 0.75 jul e m square

Answer 1

A = πr²

A = π(0.06)²

A = 0.0113m²

U = SxA

U = 0.75 x 0.0113

U = 8.475x10^-3

Answer 2

The surface energy of water kept in cylindrical vessel is 84.78 x $10^{-4}$ J

• The surface energy is defined as the sum of all inter molecular forces that are on the surface of a material, the degree of attraction or repulsion force of a material surface exerts on another material.

       Given,

                       r = 6 cm = 6 x $10^{-2}$ m

                       T = 0.75 J/$m^{2}$

                        Area = π$r^{2}$ = (22/7)(6 x $10^{-2})^{2}$ = 113.04 x $10^{-4} m^{2}$

• Surface energy of water = T x A

                                                 = 0.75 x 113.04 x $10^{-4}$

                                                 = 84.78 x $10^{-4}$ J