Let the radius of the right circular cone be \( r \) cm So, the base diameter of the cone = \( 2r \) cm ∴ Height of the cone \( h = 2r \) cm ∴ Current volume of the cone = \[ \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 \times 2r = \frac{2}{3}\pi r^3 \text{ cubic cm} \] Now, if the height were 7 times the base diameter, then height = \( 2r \times 7 = 14r \) cm ∴ New volume = \[ \frac{1}{3}\pi r^2 \times 14r = \frac{14}{3}\pi r^3 \text{ cubic cm} \] According to the question: \[ \frac{14}{3}\pi r^3 - \frac{2}{3}\pi r^3 = 539 \Rightarrow \frac{12}{3}\pi r^3 = 539 \Rightarrow \frac{12}{3} \times \frac{22}{7} r^3 = 539 \Rightarrow r^3 = \frac{49 \times 7}{2 \times 4} = \frac{343}{8} \Rightarrow r = \frac{7}{2} \] ∴ Height of the cone = \( 2r = 2 \times \frac{7}{2} = 7 \) cm