Question

From a solid sphere of radius 15cm, a right circular cylinder hole of radius 9 cm whose axis passing through the centre is removed. The total surface area of the remaining solid is: A)1188pi b)108pi c)1170pi d)144pi

Answer 1

Answer = 1152 π cm³      (the given options are wrong)

Pls see the diagrams for understanding the solution.

A sphere with a cylindrical hole through its center is called spherical ring (like spherical beads in a chain).

Radius of cylindrical hole = r
Radius of Sphere = R
Height of the hole = Height of Spherical Ring = H = 2 h
                      R² = h² + r²
Height of spherical cap = R - h = h'

Curved Surface Area of Spherical cap (at top/bottom) removed
                 = 2πR (R - h) = 2 π R h'

Curved surface area of Spherical Ring (of height H = 2h)
                 = 2 π R H = 4 π R h

Curved cylindrical surface (hole) inside spherical Ring (height H=2h)
                 = 2 π r H = 4 π r h

Total Surface Area of Spherical Ring = 2 π (R + r) H

Answer:  r = 9 cm,  R = 15 cm,   h = √(R²-r²) = 12 cm,   H = 24 cm
               T.S.A. = 2 π (15 + 9) 24 = 1152π cm²

Total surface area of two caps removed = 2π R (2R - H)

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h' = height of small spherical cap removed

Volume of a Spherical Cap (top/bottom) removed
           = π/3 * (R-h)² (2R+h)  = π/3 h' ² (3R - h')

Volume of Spherical Ring = π/6 * H³ = 4π/3 * (R²-r²)³/²