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

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: \(264\)