A soap film of thickness h = 1.0 µm is formed in wire loop of radius r = 5.0 cm. Surface tension and density of the soap solution are σ = 0.025 N/m and ρ = 1000 kg/m3 respectively. The soap film is punctured in the center. How long will it take to disappear ? Assume that the kinetic energy of moving liquid is equal to the surface energy of film disappeared till then.

Engineering

Physics

Conservation of Energy

Newtons Second Law of Motion

Surface Tension and Surface Energy

Question

A soap film of thickness h = 1.0 µm is formed in wire loop of radius r = 5.0 cm. Surface tension and density of the soap solution are σ = 0.025 N/m and ρ = 1000 kg/m³ respectively. The soap film is punctured in the center. How long will it take to disappear ? Assume that the kinetic energy of moving liquid is equal to the surface energy of film disappeared till then.

5 × 10^–3 sec

5 × 10^–4 sec

$\frac{5}{\sqrt{2}} \times 10^{- 3} \sec$

$5 \sqrt{2} \times 10^{- 3} \sec$

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$2 {πr}^{2} \times s = \frac{1}{2} {mv}^{2}$

$= \frac{1}{2} ρ \times {πr}^{2} t^{2}$

$\Rightarrow v = \sqrt{\frac{4 σ}{ρt}}$

$t = \frac{r}{v} = \frac{5 \times 10^{- 2}}{\sqrt{\frac{4 \times 0 .025}{10^{3} \times 10^{- 6}}}} = 5 \times 10^{- 3} \sec$