In the previous section we saw surface area and volume of spheres and hemispheres. We also saw some solved examples. In this section, we will see a few more solved examples.
Solved example 33.29
The surface area of a solid sphere is 120 sq.cm. If it is cut into two halves, what would be the surface area of each hemisphere?
Solution:
1. Let r cm be the radius of the total sphere. Then it's surface area would be 4πr2 cm2
2. This surface area is given as 120 cm2. So we can equate the two:
4πr2 = 120 ⟹ πr2 = 30 ⟹ r2 = 30⁄π
3. The curved surface area of each hemisphere will be 120⁄2 = 60 cm2.
4. When the sphere is cut, each hemisphere will have a base also. This base is a circular area.
• The radius of the base circle will be the same 'r'. So we can write:
Base area of each hemisphere = πr2 = π × 30⁄π = 30 cm2
5. So total surface area of each hemisphere = (60 + 30) = 90 cm2
An easy method:
• Curved surface area of a hemisphere = 1⁄2 × 4πr2 = 2πr2.
• Base area = πr2
• So total surface area of a hemisphere = 2πr2 + πr2 = 3πr2
• From (2) we have: r2 = 30⁄π
• So total surface area of each hemisphere = 3π × 30⁄π = 90 cm2
Solved example 33.30
The volume of two spheres are in the ratio 27 : 64.
(i) What is the ratio of their radii?
(ii) What is the ratio of their surface areas?
Solution:
Part (i):
1. Let the volume and radius of the first sphere be V1 and r1 respectively
• Let the volume and radius of the second sphere be V2 and r2 respectively
2. (i) Then volume of first sphere = V1 = 4⁄3 × [π(r1)3]
(ii) volume of second sphere = V2 = 4⁄3 × [π(r2)3]
3. Given that V1⁄V2 = 27⁄64
⟹ 4⁄3 × [π(r1)3] ÷ 4⁄3 × [π(r2)3] = 27⁄64
⟹ [π(r1)3] ÷ [π(r2)3] = 27⁄64
⟹ [(r1)3] ÷ [(r2)3] = 27⁄64
⟹ (r1⁄r2)3 = 27⁄64
⟹ (r1⁄r2)3 = (3⁄4)3
⟹ (r1⁄r2) = (3⁄4)
Part (ii):
1. Let S1 and S2 be the surface areas. Then we get:
S1⁄S2 = [4π(r1)2] ÷ [4π(r2)2]
⟹ S1⁄S2 = [(r1)2] ÷ [(r2)2]
⟹ S1⁄S2 = (r1⁄r2)2
2. But from (3) in part (i), we have: (r1⁄r2) = (3⁄4)
• So (r1⁄r2)2 = (3⁄4)2 = (9⁄16)
• Thus we get: S1⁄S2 = (r1⁄r2)2 = (9⁄16)
Solved example 33.31
The base radius and length of a metal cylinder are 4 cm and 10 cm. If it is melted and recast into spheres of radius 2 cm, how many spheres can be made?
Solution:
1. Total volume available for melting = Volume of the cylinder
= πr2h = π×42×10 = 160π cm3
2. Volume of one sphere = 4⁄3πr3 = 4⁄3 × π × 23 = 32⁄3 × π cm3.
3. Number of spheres = [Total volume] ÷ [Volume of one sphere]
= [160π] ÷ [32⁄3 × π]
= [160] ÷ [32⁄3]
= [160] × [3⁄32]
= [16×10] × [3⁄16×2] = 15 Nos.
Solved example 33.32
A metal sphere of radius 12 cm is melted and recast into 27 small spheres. What is the radius of each sphere?
Solution:
1. Total volume available for melting = Volume of the sphere
= 4⁄3πr3 = 4⁄3×π×123 = 2304π cm3
2. Let 'r' be the radius of one small sphere.
Then volume of one small sphere = 4⁄3πr3
3. Number of spheres = [Total volume] ÷ [Volume of one sphere]
= [2304π] ÷ [4⁄3πr3]
= [2304] ÷ [4⁄3×r3]
= [2304] × [3⁄4×(1⁄r3)]
= [1728×(1⁄r3)]
4. But number of spheres is given as 27. So we can write:
[1728×(1⁄r3)] = 27
⟹ (1⁄r3) = 27⁄1728
⟹ (1⁄r3) = 1⁄64
⟹ r3 = 64 = 43
⟹ r = 4 cm
Solved example 33.33
From a solid sphere of radius 10 cm, a cone of height 16 cm is carved out. What fraction of the volume of the sphere is the volume of the cone?
Solution:
1. Consider the red sphere in fig.33.32(a) below.
• Two ellipses are drawn inside it: A dotted ellipse and a dashed ellipse
[The dotted ellipse is shown just to give an emphasis to the 'spherical shape'. It does not come in any of our calculations]
• 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 (shown in cyan colour) 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.33.33(b) below. In that fig. we can see that the, base of the new cone is below the dashed ellipse
3. In fig.33.33(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:
AB2 = OB2 - OA2 ⟹ AB2 = 102 - 62 ⟹ AB2 = 100 - 36 ⟹ AB2 = 64 ⟹ AB = 8 cm
4. Thus we have:
• Height of the cone, h = 16 cm
• Radius of the cone, rc = 8 cm
• So Volume, Vc = 1⁄3π(rc)2h = 1⁄3×π×82×16
• Volume of sphere, Vs = 4⁄3πr3 = 4⁄3×π×103
5. Taking ratios, we get:
Vc⁄Vs = {1⁄3×π×82×16} ÷ {4⁄3×π×103}
⟹ Vc⁄Vs = {82×16} ÷ {4×103}
= {8×8×16} ÷ {4×103}
= {2×8×16} ÷ {103}
= {256} ÷ {1000}
= 32⁄125
11. Thus Vc⁄Vs = 32⁄125
⟹ Vc = 32⁄125 × Vs
• So 'volume of the cone' is 32⁄125 of the 'volume of the sphere'
Solved example 33.34
The picture shows the dimensions of a petrol tank. How many litres of petrol can it hold?
Solution:
1. The tank has two hemispherical parts and one cylindrical part
• The yellow dashed line indicates the axis of the tank
• From the fig., it is clear that radius of the hemisphere is 1 m.
• So it's volume = Vh = 2⁄3πr3 = 2⁄3×π×13= 2⁄3×π m3
• Thus volume of two hemispheres = 2 × 2⁄3×π = 4⁄3×π m3
2. Height of a hemisphere will be equal to it's radius.
So length of the cylindrical part = [6 - (2×1)] = 4 m
3. Volume of cylinder = Vc = πr2h = π×12×4 = 4π m3
4. Thus total volume = 4⁄3×π + 4π = 16⁄3×π m3.
5. We know that 1 liter is the volume of a cube of edge 10 cm (Details here)
• So 1 liter = 103 cm3 = 1000 cm3
• Now, (16⁄3×π) m3 = [(16⁄3×π) × 1000000] cm3 = 16746666.67 cm3. (∵ 1 m = 100 cm)
• Thus the no. of liters = 16746666.67⁄1000 = 16746.67 liters
Solved example 33.35
A solid sphere is cut into two hemispheres. From one, a square pyramid and from the other, a cone, each of maximum possible size are carved out. What is the ratio of their volumes?
Solution:
1. Consider the red hemisphere in fig.33.35(a) below.
• A dotted ellipse and a dashed curve are drawn inside it
[The dotted curve is shown just to give an emphasis to the 'hemispherical shape'. It does not come in any of our calculations]
• The dashed ellipse represents the base of the hemisphere
• For maximum possible volume, this base is taken as the base of the cone (shown in cyan colour) in fig.33.35(b)
• We can see that, the cone fits perfectly in the hemisphere.
• This is shown more clearly in fig.33.35(c)
2. From fig.c, we have:
• Height of the cone, hc = r
• Radius of the cone, rc = r
• So Volume, Vc = 1⁄3π(rc)2h = 1⁄3×π×r2×r = 1⁄3×π×r3
3. Consider the red hemisphere in fig.33.36(a) below. It is the same hemisphere of radius r, that we saw for the cone above
• A square (seen as a rhombus in view) is drawn in the base of the hemisphere
• This square is the base of the pyramid
4. For maximum possible volume, the diagonal of the square must be equal to the diameter of the circle
• So in fig.c, we can write:
OP = OQ = half of diameter = radius = r
5. OPQ is a right triangle. We can apply Pythagoras theorem
• Then base edge = PQ = √[OP2 + OQ2] = √[r2 + r2] = √[2r2] = √[2]r
6. So volume of the pyramid, Vp = 1⁄3 × base area × height = 1⁄3 × √[2]r ×√[2]r × r = 2⁄3×r3
7. Now we can take the ratio:
Vp⁄Vc = {2⁄3×r3} ÷ {1⁄3×π×r3} = {2⁄3} ÷ {1⁄3×π} = {2} ÷ {π}
• Thus we get:
Vp : Vc = 2 : π
We have completed this discussion on solids. In the next chapter, we will see Geometry and Algebra.
Solved example 33.29
The surface area of a solid sphere is 120 sq.cm. If it is cut into two halves, what would be the surface area of each hemisphere?
Solution:
1. Let r cm be the radius of the total sphere. Then it's surface area would be 4πr2 cm2
2. This surface area is given as 120 cm2. So we can equate the two:
4πr2 = 120 ⟹ πr2 = 30 ⟹ r2 = 30⁄π
3. The curved surface area of each hemisphere will be 120⁄2 = 60 cm2.
4. When the sphere is cut, each hemisphere will have a base also. This base is a circular area.
• The radius of the base circle will be the same 'r'. So we can write:
Base area of each hemisphere = πr2 = π × 30⁄π = 30 cm2
5. So total surface area of each hemisphere = (60 + 30) = 90 cm2
An easy method:
• Curved surface area of a hemisphere = 1⁄2 × 4πr2 = 2πr2.
• Base area = πr2
• So total surface area of a hemisphere = 2πr2 + πr2 = 3πr2
• From (2) we have: r2 = 30⁄π
• So total surface area of each hemisphere = 3π × 30⁄π = 90 cm2
Solved example 33.30
The volume of two spheres are in the ratio 27 : 64.
(i) What is the ratio of their radii?
(ii) What is the ratio of their surface areas?
Solution:
Part (i):
1. Let the volume and radius of the first sphere be V1 and r1 respectively
• Let the volume and radius of the second sphere be V2 and r2 respectively
2. (i) Then volume of first sphere = V1 = 4⁄3 × [π(r1)3]
(ii) volume of second sphere = V2 = 4⁄3 × [π(r2)3]
3. Given that V1⁄V2 = 27⁄64
⟹ 4⁄3 × [π(r1)3] ÷ 4⁄3 × [π(r2)3] = 27⁄64
⟹ [π(r1)3] ÷ [π(r2)3] = 27⁄64
⟹ [(r1)3] ÷ [(r2)3] = 27⁄64
⟹ (r1⁄r2)3 = 27⁄64
⟹ (r1⁄r2)3 = (3⁄4)3
⟹ (r1⁄r2) = (3⁄4)
Part (ii):
1. Let S1 and S2 be the surface areas. Then we get:
S1⁄S2 = [4π(r1)2] ÷ [4π(r2)2]
⟹ S1⁄S2 = [(r1)2] ÷ [(r2)2]
⟹ S1⁄S2 = (r1⁄r2)2
2. But from (3) in part (i), we have: (r1⁄r2) = (3⁄4)
• So (r1⁄r2)2 = (3⁄4)2 = (9⁄16)
• Thus we get: S1⁄S2 = (r1⁄r2)2 = (9⁄16)
Solved example 33.31
The base radius and length of a metal cylinder are 4 cm and 10 cm. If it is melted and recast into spheres of radius 2 cm, how many spheres can be made?
Solution:
1. Total volume available for melting = Volume of the cylinder
= πr2h = π×42×10 = 160π cm3
2. Volume of one sphere = 4⁄3πr3 = 4⁄3 × π × 23 = 32⁄3 × π cm3.
3. Number of spheres = [Total volume] ÷ [Volume of one sphere]
= [160π] ÷ [32⁄3 × π]
= [160] ÷ [32⁄3]
= [160] × [3⁄32]
= [16×10] × [3⁄16×2] = 15 Nos.
Solved example 33.32
A metal sphere of radius 12 cm is melted and recast into 27 small spheres. What is the radius of each sphere?
Solution:
1. Total volume available for melting = Volume of the sphere
= 4⁄3πr3 = 4⁄3×π×123 = 2304π cm3
2. Let 'r' be the radius of one small sphere.
Then volume of one small sphere = 4⁄3πr3
3. Number of spheres = [Total volume] ÷ [Volume of one sphere]
= [2304π] ÷ [4⁄3πr3]
= [2304] ÷ [4⁄3×r3]
= [2304] × [3⁄4×(1⁄r3)]
= [1728×(1⁄r3)]
4. But number of spheres is given as 27. So we can write:
[1728×(1⁄r3)] = 27
⟹ (1⁄r3) = 27⁄1728
⟹ (1⁄r3) = 1⁄64
⟹ r3 = 64 = 43
⟹ r = 4 cm
Solved example 33.33
From a solid sphere of radius 10 cm, a cone of height 16 cm is carved out. What fraction of the volume of the sphere is the volume of the cone?
Solution:
1. Consider the red sphere in fig.33.32(a) below.
 |
| Fig.33.32 |
• Two ellipses are drawn inside it: A dotted ellipse and a dashed ellipse
[The dotted ellipse is shown just to give an emphasis to the 'spherical shape'. It does not come in any of our calculations]
• 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 (shown in cyan colour) 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.33.33(b) below. In that fig. we can see that the, base of the new cone is below the dashed ellipse
 |
| Fig.33.33 |
• 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:
AB2 = OB2 - OA2 ⟹ AB2 = 102 - 62 ⟹ AB2 = 100 - 36 ⟹ AB2 = 64 ⟹ AB = 8 cm
4. Thus we have:
• Height of the cone, h = 16 cm
• Radius of the cone, rc = 8 cm
• So Volume, Vc = 1⁄3π(rc)2h = 1⁄3×π×82×16
• Volume of sphere, Vs = 4⁄3πr3 = 4⁄3×π×103
5. Taking ratios, we get:
Vc⁄Vs = {1⁄3×π×82×16} ÷ {4⁄3×π×103}
⟹ Vc⁄Vs = {82×16} ÷ {4×103}
= {8×8×16} ÷ {4×103}
= {2×8×16} ÷ {103}
= {256} ÷ {1000}
= 32⁄125
11. Thus Vc⁄Vs = 32⁄125
⟹ Vc = 32⁄125 × Vs
• So 'volume of the cone' is 32⁄125 of the 'volume of the sphere'
Solved example 33.34
The picture shows the dimensions of a petrol tank. How many litres of petrol can it hold?
 |
| Fig.33.34 |
1. The tank has two hemispherical parts and one cylindrical part
• The yellow dashed line indicates the axis of the tank
• From the fig., it is clear that radius of the hemisphere is 1 m.
• So it's volume = Vh = 2⁄3πr3 = 2⁄3×π×13= 2⁄3×π m3
• Thus volume of two hemispheres = 2 × 2⁄3×π = 4⁄3×π m3
2. Height of a hemisphere will be equal to it's radius.
So length of the cylindrical part = [6 - (2×1)] = 4 m
3. Volume of cylinder = Vc = πr2h = π×12×4 = 4π m3
4. Thus total volume = 4⁄3×π + 4π = 16⁄3×π m3.
5. We know that 1 liter is the volume of a cube of edge 10 cm (Details here)
• So 1 liter = 103 cm3 = 1000 cm3
• Now, (16⁄3×π) m3 = [(16⁄3×π) × 1000000] cm3 = 16746666.67 cm3. (∵ 1 m = 100 cm)
• Thus the no. of liters = 16746666.67⁄1000 = 16746.67 liters
Solved example 33.35
A solid sphere is cut into two hemispheres. From one, a square pyramid and from the other, a cone, each of maximum possible size are carved out. What is the ratio of their volumes?
Solution:
1. Consider the red hemisphere in fig.33.35(a) below.
 |
| Fig.33.35 |
[The dotted curve is shown just to give an emphasis to the 'hemispherical shape'. It does not come in any of our calculations]
• The dashed ellipse represents the base of the hemisphere
• For maximum possible volume, this base is taken as the base of the cone (shown in cyan colour) in fig.33.35(b)
• We can see that, the cone fits perfectly in the hemisphere.
• This is shown more clearly in fig.33.35(c)
2. From fig.c, we have:
• Height of the cone, hc = r
• Radius of the cone, rc = r
• So Volume, Vc = 1⁄3π(rc)2h = 1⁄3×π×r2×r = 1⁄3×π×r3
3. Consider the red hemisphere in fig.33.36(a) below. It is the same hemisphere of radius r, that we saw for the cone above
 |
| Fig.33.36 |
• This square is the base of the pyramid
4. For maximum possible volume, the diagonal of the square must be equal to the diameter of the circle
• So in fig.c, we can write:
OP = OQ = half of diameter = radius = r
5. OPQ is a right triangle. We can apply Pythagoras theorem
• Then base edge = PQ = √[OP2 + OQ2] = √[r2 + r2] = √[2r2] = √[2]r
6. So volume of the pyramid, Vp = 1⁄3 × base area × height = 1⁄3 × √[2]r ×√[2]r × r = 2⁄3×r3
7. Now we can take the ratio:
Vp⁄Vc = {2⁄3×r3} ÷ {1⁄3×π×r3} = {2⁄3} ÷ {1⁄3×π} = {2} ÷ {π}
• Thus we get:
Vp : Vc = 2 : π
