Mathematics — Standard (041) — CBSE Class 10 board question

Given: Cylinder: height = 24 cm, radius = 5 cm. Two cones hollowed out: height = 12 cm each, radius = 5 cm each.

Volume of remaining solid:

Volume of cylinder = $\pi r^2 h = \dfrac{22}{7} \times 5^2 \times 24 = \dfrac{22 \times 25 \times 24}{7} = \dfrac{13200}{7}$ cm³

Volume of one cone = $\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 25 \times 12 = \dfrac{2200}{7}$ cm³

Volume of two cones = $2 \times \dfrac{2200}{7} = \dfrac{4400}{7}$ cm³

Volume of remaining solid = $\dfrac{13200 - 4400}{7} = \dfrac{8800}{7} \approx \mathbf{1257.14}$ cm³

Surface Area of remaining solid:

The visible surfaces are: curved surface of cylinder + curved surfaces of two cones (the circular ends of the cylinder remain open as cone bases coincide).

CSA of cylinder = $2\pi r h = 2 \times \dfrac{22}{7} \times 5 \times 24 = \dfrac{5280}{7}$ cm²

Slant height of cone: $l = \sqrt{r^2 + h^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ cm

CSA of two cones = $2 \times \pi r l = 2 \times \dfrac{22}{7} \times 5 \times 13 = \dfrac{2860}{7}$ cm²

Total Surface Area = $\dfrac{5280 + 2860}{7} = \dfrac{8140}{7} \approx \mathbf{1162.86}$ cm²

Source: Chapter 12, Sections 12.2 and 12.3

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• Volume: Subtract both cone volumes from the cylinder volume. Since the two cones are hollowed out internally, their base circles are inside — no circular flat areas remain on the ends of the cylinder.
• Surface Area: Only the curved surfaces count. The top and bottom circular faces of the cylinder are entirely removed (each cone's base = the cylinder's circular end), so no $\pi r^2$ terms are added. The slant height $l = 13$ cm is key — calculate it via Pythagoras.
• A common mistake is adding $2\pi r^2$ (circular ends of cylinder); avoid it here since both ends are fully hollowed by the cones.