StreakPeaked· Practice

ExamsJEE MainPhysics

A dilute solution of oleic acid contains 0.01 cm³ of oleic acid per cm³ of solution. Using 100 spherical drops each of radius (3/(4*pi))^(1/3) * 10⁻³ cm, a monomolecular film of this solution is spread over an area of 4 cm². The resulting thickness of the oleic acid layer is x * 10⁻¹⁴ m. Find x.

  1. 150
  2. 200
  3. 250
  4. 300

Correct answer: 250

Solution

The special radius makes each drop's volume exactly 10⁻⁹ cm³. Multiply by 100 drops, take 1% for the oleic acid, and divide by the film area to get the thickness.

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →