A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If surface tension of soap solution is 40 dyne/cm then the new radius is ______ cm. (Take $$\pi = \frac{22}{7}$$)

Work done in blowing a soap bubble from radius $$r_1$$ to $$r_2$$:

$$W = 8\pi T(r_2^2 - r_1^2)$$ (factor 8 because soap bubble has two surfaces)

Initial diameter = 7 cm, so $$r_1 = 3.5$$ cm. $$T = 40$$ dyne/cm. $$W = 36960$$ erg.

$$36960 = 8 \times \frac{22}{7} \times 40 \times (r_2^2 - 12.25)$$

$$36960 = \frac{7040}{7}(r_2^2 - 12.25)$$

$$36960 \times 7 = 7040(r_2^2 - 12.25)$$

$$r_2^2 - 12.25 = \frac{258720}{7040} = 36.75$$

$$r_2^2 = 49$$, $$r_2 = 7$$ cm

The answer is 7.