Question

Sol. [Key Concept: $\Delta\ E = T \times$(surface-area increase)$= 4\pi\ T(512r^{2} - R^{2})$ with $R = 8r$.]

$$V_{1} = V_{2}$$

$$\begin{matrix} & \frac{4}{3}\pi R^{3} = 512\left( \frac{4}{3}\pi r^{3} \right) \\ & R = (512)^{1/3}r \\ & 1 = 8r \\ & r = \frac{1}{8}\text{ }mm \\ & \Delta SE = {SE}_{2} - {SE}_{1} \\ & \ = 512\left\lbrack \text{ }T \times 4\pi r^{2} \right\rbrack - \left\lbrack T \times 4\pi R^{2} \right\rbrack \\ & \ = 4\pi\text{ }T\left\lbrack 512r^{2} - R^{2} \right\rbrack \\ & \ = 4\pi(0.08)\left\lbrack 512\left( \frac{1}{8} \right)^{2} - 1^{2} \right\rbrack \times 10^{- 6} \approx 7 \times 10^{- 6} \\ & \alpha = 7 \end{matrix}$$