Answer: C
Let the length of each edge of the cube be \( a \) cm. ∴ The total surface area of the cube = \( 6a^2 \) square cm If the edge length is increased by 10%, the new edge length = \( a + a \times \frac{10}{100} \) cm = \( a + \frac{a}{10} \) cm = \( \frac{11a}{10} \) cm ∴ New surface area of the cube = \[ 6\left(\frac{11a}{10}\right)^2 = \frac{726a^2}{100} \text{ square cm} \] ∴ Percentage increase in surface area = \[ \frac{\frac{726a^2}{100} - 6a^2}{6a^2} \times 100\% = \frac{\frac{126a^2}{100} \times 100}{6a^2}\% = \frac{126a^2}{6a^2}\% = 21\% \]
Let the length of each edge of the cube be \( a \) cm. ∴ The total surface area of the cube = \( 6a^2 \) square cm If the edge length is increased by 10%, the new edge length = \( a + a \times \frac{10}{100} \) cm = \( a + \frac{a}{10} \) cm = \( \frac{11a}{10} \) cm ∴ New surface area of the cube = \[ 6\left(\frac{11a}{10}\right)^2 = \frac{726a^2}{100} \text{ square cm} \] ∴ Percentage increase in surface area = \[ \frac{\frac{726a^2}{100} - 6a^2}{6a^2} \times 100\% = \frac{\frac{126a^2}{100} \times 100}{6a^2}\% = \frac{126a^2}{6a^2}\% = 21\% \]