The excess pressure inside a liquid drop is 500 Nm⁻². If the radius of the drop is 2 mm, the surface tension of liquid is x × 10⁻³ Nm⁻¹. The value of x is _____________.

Solution

Physics sir · 3 steps

A drop has only one surface, so use Δ P = 2T/r and turn it around to find T.

1. One surface. A liquid drop has a single free surface, the outer one. Its excess pressure is Δ P = 2T/r.

   T = (Δ P r)/2

2. Put in numbers. Convert the radius first: r = 2 mm = 2 × 10⁻³ m. Then T = (500 × 2 × 10⁻³)/2 = 0.5 N m⁻¹.
3. So. T = 0.5 N m⁻¹ = 500 × 10⁻³ N m⁻¹, so x = 500.

The common slip: 250. Using the soap-bubble formula Δ P = 4T/r gives T = 0.25 N m⁻¹, that is 250, not 500. A drop has only one surface.

The idea to keep. Count the surfaces first. A drop or an air bubble inside a liquid has one surface, so Δ P = 2T/r. A soap bubble in air has two, so Δ P = 4T/r.

Answer from the NTA final answer key. Solution by PrepWiser.