Form a solid sphere of radius 10 centimetres, a cone of height 16 centimetres is carved out. What fraction of the volume of the sphere is the volume of the cone?

1. Consider the red sphere in fig (a) below.

• Two ellipses are drawn inside it: A dotted ellipse and a dashed ellipse

• The dashed ellipse represents a circle whose centre is same as the centre of the sphere

♦ Also this circle is horizontal

• So this circle divides the sphere into an upper hemisphere and a lower hemisphere

• This circle is taken as the base of the cone in fig.b.

• We can see that, the cone fits perfectly in the upper hemisphere.

• This is shown more clearly in fig.c

2. From fig.c we can see that, the height of the cone will be the height of the hemisphere, which is 10 cm

• But cone given in the question has a height of 16 cm.

• So the given cone does not fit inside the upper hemisphere alone.

♦ It will occupy some portion of the lower hemisphere also

• This is shown in fig (b) below. In that fig. we can see that the, base of the new cone is below the dashed ellipse

3. In fig (c), the measurements are given

• One half of the cone is represented by the right triangle ABC

• The distance of the apex C from the centre O will be the radius of the sphere, which is 10 cm

• So the remaining distance OA will be (16 – 10) = 6 cm

• Distance OB will also be the radius 10 cm

• Applying Pythagoras theorem to the right triangle OAB, we get:

–

⟹ = 10² – 6²

⟹ = 100 – 36

⟹ = 64

⟹ AB = 8 cm

4. Thus we have:

• Height of the cone, h = 16 cm

• Radius of the cone, r_c = 8 cm

• So Volume,

• Volume of sphere,

5. Taking ratios, we get:

• So 'volume of the cone' is of the 'volume of the sphere'