The surface tension of a soap solution is 3.5 × 10^ -2 N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is α × 10^ -6 J. The value of α is _____. ( π = 22/7 )

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The work done to increase the radius of a soap bubble is given by

W = T \times \Delta A

Since a soap bubble has two free surfaces, the change in surface area is

\Delta A = 2 \times 4\pi (r_2^2 - r_1^2)

Substituting the given values:

W = 3.5 \times 10^{-2} \times 8\pi \times ((2 \times 10^{-2})^2 - (1 \times 10^{-2})^2)

W = 3.5 \times 10^{-2} \times 8\pi \times (4 \times 10^{-4} - 1 \times 10^{-4})

W = 3.5 \times 10^{-2} \times 8\pi \times 3 \times 10^{-4}

W = 84\pi \times 10^{-6}

Using

\pi = \dfrac{22}{7}

:

W = 84 \times \dfrac{22}{7} \times 10^{-6} = 264 \times 10^{-6}

J

Comparing with

\alpha \times 10^{-6}

J, we get

\alpha = 264

Answer:

264

The surface tension of a soap solution is 3.5 × 10^ -2 N/m. The...

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