Question

from a solid sphere of radius 15 cm , a right circular cylindrical hole of radius 9cm whose axis passing through the centre is removed . the total surface area of the remaining solid is.............

Answer 1

Radius of sphere=R
Radius of cylindrical hole=r
Height of cylindrical hole =H=2h

Curved surface area of spherical ring =2πRH =4πRh

Curved surface area of spherical cap of height (R-h ) at the top or bottom =2π(R-h) (Not needed in this question just information)

curved surface area of cylindrical hole inside sphere of height H and radius of cross section r =2πrH = 4πrh

Total surface area of spherical ring= CSA of spherical ring+ CSA of cylindrical hole= 2πRH +2πrH= 2π(R+r)H

Now,
Given,
R=15cm
r=9cm
h=√(R²-r²)=√(225-81)=√144=12cm
H=2h=24cm
Total surface area of spherical ring=2π(15+9)24=1152πcm²=3620.57cm²