A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 5.6 cm and the total height of the toy is 11.6 cm, then what is the volume of the wooden toy (in cm3 ) (rounded off to the closest integral value)?

Given:

Radius of hemisphere (r) = 5.6 cm

Total height of toy = 11.6 cm

Height of cone (h) = Total height - Radius of hemisphere = 11.6 - 5.6 = 6 cm

Formula used:

Volume of hemisphere = \(\dfrac{2}{3}\pi r^3\)

Volume of cone = \(\dfrac{1}{3}\pi r^2 h\)

Total volume of the toy = Volume of hemisphere + Volume of cone

Calculation:

Volume of hemisphere = \(\dfrac{2}{3} \times \pi \times (5.6)^3\)

⇒ Volume of hemisphere = \(\dfrac{2}{3} \times 3.1416 \times 175.616\)

⇒ Volume of hemisphere = 367.62 cm³

Volume of cone = \(\dfrac{1}{3} \times \pi \times (5.6)^2 \times 6\)

⇒ Volume of cone = \(\dfrac{1}{3} \times 3.1416 \times 31.36 \times 6\)

⇒ Volume of cone = 196.94 cm³

Total volume of toy = Volume of hemisphere + Volume of cone

⇒ Total volume = 367.62 + 196.94 = 564.56

⇒ Total volume ≈ 565 cm³

∴ The correct answer is option (2).