Let r be the radius of a sphere then its volume is given by V = 4/3 πr3 cu. units.
· Consider a sphere and two right circular cones of same base radius and height such that twice the radius of the sphere is equal to the height of the cones.
· Then we can observe that the contents of two cones will exactly occupy the sphere.
From the Fig.7.30, we see that
Volume of a sphere = 2 × (Volume of a cone)
where the diameters of sphere and cone are equal to the height of the cone.
Volume of a sphere = 4/3 πr3 cu. Units
Let r and R be the inner and outer radius of the hollow sphere. Volume enclosed between the outer and inner spheres
Volume of a hollow sphere = 4/3 π(R3 − r3 ) cu. Units
Let r be the radius of the solid hemisphere.
Volume of the solid hemisphere = ½ (volume of sphere)
Volume of a solid hemisphere = 2/3 πr3 cu. Units
Let r and R be the inner and outer radius of the hollow hemisphere.
Volume of a hollow hemisphere = 2/3 π(R3 − r3 ) cu. Units
Example 7.21 The volume of a solid hemisphere is 29106 cm3. Another hemisphere whose volume is two-third of the above is carved out. Find the radius of the new hemisphere.
Solution Let r be the radius of the hemisphere.
Given that, volume of the hemisphere = 29106 cm3
Now, volume of new hemisphere = 2/3 (Volume of original sphere)
= 2/3 × 29106
Volume of new hemisphere = 19404 cm3
2/3 πr 3 = 19404
Therefore, r = 21 cm
Example 7.22 Calculate the weight of a hollow brass sphere if the inner diameter is 14 cm and thickness is 1mm, and whose density is 17.3 g/ cm3.
Solution Let r and R be the inner and outer radii of the hollow sphere.
Given that, inner diameter d = 14 cm; inner radius r = 7 cm; thickness = 1 mm = 1/10 cm
Outer radius R = 7 + 1/10 = 71/10 = 7.1 cm
But, weight of brass in 1 cm3 = 17.3 gm
Total weight = 17.3×62.48 = 1080.90 gm
Therefore, total weight is 1080.90 grams.