Volume of the hemispherical bowl: \[ = \left[\frac{2}{3}\pi\left(\frac{8}{2}\right)^3 - \frac{2}{3}\pi\left(\frac{4}{2}\right)^3\right] \text{ cm}^3 = \left[\frac{2}{3}\pi(4^3 - 2^3)\right] \text{ cm}^3 = \left[\frac{2}{3}\pi(64 - 8)\right] \text{ cm}^3 = \frac{2 \times 22 \times 56}{3 \times 7} \text{ cm}^3 \] Let the height of the cone be \(h\) cm. Radius of the cone = \(\frac{8}{2} = 4\) cm Therefore, volume of the cone: \[ = \frac{1}{3}\pi \times 4^2 \times h \text{ cm}^3 = \frac{1}{3} \times \frac{22}{7} \times 16h \text{ cm}^3 = \frac{22 \times 16h}{3 \times 7} \text{ cm}^3 \] According to the question: \[ \frac{22 \times 16h}{3 \times 7} = \frac{2 \times 22 \times 56}{3 \times 7} \Rightarrow 16h = 112 \Rightarrow h = 7 \] Therefore, the height of the cone is 7 cm.