StreakPeaked· Practice

ExamsJEE AdvancedPhysics

Water is contained in a cylindrical vessel of radius 6.0 cm. Taking the surface tension of water to be 0.075 J/m², determine the surface energy of the exposed top water surface.

  1. 4.25 * 10⁻⁴ J
  2. 8.5 * 10⁻⁴ J
  3. 12.75 * 10⁻⁴ J
  4. 1.7 * 10⁻³ J

Correct answer: 8.5 * 10⁻⁴ J

Solution

Surface energy equals surface tension multiplied by the free surface area. The top surface is circular, area = pi*r².

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →