The surface tension of a soap bubble is \(0.03\) N/m. The work done in increasing the diameter of bubble from \(2\) cm to \(6\) cm is \(\alpha \pi \times 10^{-4}\) J. The value of \(\alpha\) is _______. (Take \(\pi = 3.14\))

Initial radius \(r_1 = \dfrac{2}{2} = 1\) cm \(= 10^{-2}\) m

Final radius \(r_2 = \dfrac{6}{2} = 3\) cm \(= 3 \times 10^{-2}\) m

Surface tension \(T = 0.03\) N/m

A soap bubble has two free surfaces. The change in total surface area is given by:

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

\(\Delta A = 8\pi ((3 \times 10^{-2})^2 - (10^{-2})^2)\)

\(\Delta A = 8\pi (9 \times 10^{-4} - 10^{-4}) = 64\pi \times 10^{-4}\) m\(^2\)

Work done \(W = T \Delta A\)

\(W = 0.03 \times 64\pi \times 10^{-4}\)

\(W = 1.92\pi \times 10^{-4}\) J

Comparing with \(\alpha \pi \times 10^{-4}\) J, we get \(\alpha = 1.92\).

Answer: \(1.92\)