A spherical soap bubble is blown such that its radius increases at rate of \frac{1}{2 \pi} cm./s. Ignoring the pressure outside the bubble, the power delivered, at the moment when the radius is 5 cm will be : (Surface tension of soap solution is 40 × 10–3 Nm

Engineering

Physics

Surface Tension

Question

A spherical soap bubble is blown such that its radius increases at rate of $\frac{1}{2 π}$ cm./s. Ignoring the pressure outside the bubble, the power delivered, at the moment when the radius is 5 cm will be : (Surface tension of soap solution is 40 × 10^–3 Nm^–1).

3.2 × 10^–3 watt

8 × 10^–3 watt

1.6 × 10^–4 watt

4 × 10^–3 watt

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$U = 8 {πr}^{2} T$

$\frac{dU}{dt} = 16 πr (\frac{dr}{dt}) \cdot T$

$= 16 π \times \frac{5}{100} \times \frac{1}{2 π} \times \frac{1}{100} \times 40 \times 10^{- 3}$

= 1600 × 10^–7 watt

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