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From a solid cylinder of height 24 cm and radius 5 cm, two cones of height 12 cm and radius 5 cm are hollowed out. Find the volume and surface area of the remaining solid.
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Volume of remaining solid = Volume of cylinder - Volume of two cones
$= \frac{22}{7} \times 5 \times 5 \times 24 - 2 \times \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 = \frac{8800}{7}$ or 1257.14 cm$^3$ approx.
$l = \sqrt{(12)^2 + (5)^2} = 13$ cm
Surface Area of remaining solid = Curved Surface Area of cylinder + Curved Surface Area of two cones
$= 2 \times \frac{22}{7} \times 5 \times 24 + 2 \times \frac{22}{7} \times 5 \times 13 = \frac{8140}{7}$ or 1162.85 cm$^2$ approx.
$= \frac{22}{7} \times 5 \times 5 \times 24 - 2 \times \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 = \frac{8800}{7}$ or 1257.14 cm$^3$ approx.
$l = \sqrt{(12)^2 + (5)^2} = 13$ cm
Surface Area of remaining solid = Curved Surface Area of cylinder + Curved Surface Area of two cones
$= 2 \times \frac{22}{7} \times 5 \times 24 + 2 \times \frac{22}{7} \times 5 \times 13 = \frac{8140}{7}$ or 1162.85 cm$^2$ approx.