StreakPeaked· Practice

ExamsJEE MainPhysics

A drop of liquid of density ρ is floating half immersed in a liquid of density σ and surface tension 7.5 × 10⁻⁴ N cm⁻¹. The radius of drop in cm will be if (g = 10 m s⁻²)

  1. 15/√(2ρ−σ)
  2. 15/√(ρ−σ)
  3. 3/(2√(ρ−σ))
  4. 3/(20√(2ρ−σ))

Correct answer: 15/√(2ρ−σ)

Solution

For a half-immersed drop: (4/3)pi*r^3*rho*g = (2/3)pi*r^3*sigma*g + 2*pi*r*T, giving r^2 = 3T/(g(2rho - sigma)). With T = 7.5e-4 N/cm = 7.5e-2 N/m and g = 10, r = 0.15/sqrt(2rho - sigma) m = 15/sqrt(2rho - sigma) cm.

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →