" The surface energy of liquid film on a ring of area "0.15m^(2)" is (Surface tension of liquid "=5Nm^(-1)" ) "

The surface energy of liquid film on a ring of area 0.15 m^(2) is (surface tension of liquid 5Nm^(-1) )

The surface energy of soap liquid film on a ring of area 0.15m^2 is (Surface tension of liquid = 5 N//m)

Determine the surface energy of a liquid film formed on a ring of area 0.15 m^2 . The surface tension of the liquid =5N. m^-1 .

The surface tension of liquid is 0.5 N / m . If a film is held on a ring of area 0.02 m 2, its surface energy is

If the surface tension of a liquid is 5 N/m then the surface energy of the liquid film on a ring of area 0.15 m^(2) is