Chapter 33.7 - Volume of Sphere - Solved examples

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.

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

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, r_c = 8 cm
• So Volume, V_c =  ¹⁄₃π(r_c)²h = ¹⁄₃×π×8²×16
• Volume of sphere, V_s = ⁴⁄₃πr³ = ⁴⁄₃×π×10³
5. Taking ratios, we get:
^V_c⁄_Vs = {¹⁄₃×π×8²×16} ÷ {⁴⁄₃×π×10³}
⟹ ^V_c⁄_Vs = {8²×16} ÷ {4×10³} 
= {8×8×16} ÷ {4×10³}
= {2×8×16} ÷ {10³}
= {256} ÷ {1000}
= ³²⁄₁₂₅
11. Thus ^V_c⁄_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

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.

Fig.33.35

• 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, h_c = r
• Radius of the cone, r_c = r
• So Volume, V_c =  ¹⁄₃π(r_c)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

• 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, V_p = ¹⁄₃ × base area × height = ¹⁄₃ × √[2]r ×√[2]r × r = ²⁄₃×r³
7. Now we can take the ratio:
^V_p⁄_Vc = {²⁄₃×r³} ÷ {¹⁄₃×π×r³} = {²⁄₃} ÷ {¹⁄₃×π} = {2} ÷ {π}
• Thus we get:
 Vp : Vc = 2 : π

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We have completed this discussion on solids. In the next chapter, we will see Geometry and Algebra.

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Chapter 33.6 - Surface area and Volume of Sphere

In the previous section we saw volume of cones. We also saw some solved examples. In this section, we will learn about spheres.

In earlier sections, we have seen cylinders and cones. 
1. Consider the cylinder in fig.33.25(a) below.

Fig.33.25

• It is cut by a plane. 
[A cutting plane can be of any shape. Triangular, square, rectangular, pentagonal etc., Whatever the shape be, it must be a plane. That is., the surface should not have any curves or undulations] 
• In fig.a, the cutting plane is parallel to the base of the cylinder. 
2. After the cut is made, if we remove the top portion of the cylinder and the cutting plane, we will be able to see the cross section of the cylinder. 
• Fig.c shows the cross section when the cutting plane is parallel to the base. It is a circle. 
3. In fig.b, the cutting plane is not parallel to the base. 
• In this case, the result is shown in fig.d. The cross section will not be a circle.  It will resemble an ellipse

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We will get a similar result for cones also:
• If the cutting plane is parallel to the base of the cone, the cross section obtained will be a circle
• If the cutting plane is not parallel to the base of the cone, the cross section obtained will not be a circle. It will resemble an ellipse.

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Now let us consider a sphere. It has no base. So how will we cut it?
The answer is:
■ In whichever way we cut a sphere using a cutting plane, the cross section will be a circle. This is shown in the following figs:

Fig.33.26

1. Fig.33.26(a) shows a sphere. 
• Fig.b show the sphere being cut by a cutting plane. 
• Fig.c shows the result: 
    ♦ The cross section is a circle
2. Now consider fig.33.27 below:

Fig.33.27

• Fig.a shows a sphere and it's cutting plane. But the cutting plane is inclined.
• Fig.b shows the result:
    ♦ The cross section is a circle even when the cutting plane is in an inclined position.
3. But the radii of the circles will be different.
• The radius of the circle in the cross section obtained in fig.33.26(c) will be different from that of the circle in fig.33.27(b)

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• We know that the distance from the center of a circle to any point on that circle will be the same. 
    ♦ Also we know that this constant distance is the radius of that circle
In a similar way:
• The distance from the center of a sphere to any point on the surface of that sphere will be the same.
    ♦ This constant distance is called the radius of the sphere
• Twice this distance is called the diameter of the sphere
1. Consider the cutting plane in fig.33.28(a) below. 

Fig.33.28

• That cutting plane can be horizontal, vertical or even inclined. 
    ♦ What ever be the orientation, that cutting plane has a special property:
■ The centre of the sphere lies on that cutting plane. 
• In such a case, the sphere will be split into two equal halves. 
    ♦ Each half is called a hemisphere. 
• The circle obtained in such a cross section also has some special properties:
    ♦ The center of that circle is same as the center of the sphere
    ♦ The radius of that circle is same as the radius of the sphere
    ♦ The diameter of that circle is same as the diameter of the sphere

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• In the case of cones, we cut it open and laid it flat on a level surface. See fig.33.22. 

• But a sphere cannot be cut open and laid flat. 

    ♦ If we want it flat, there will be some stretching and folding. 

    ♦ So the flat surface thus obtained will not represent the true surface area of the sphere.

■ But it can be shown that the surface area of a sphere of radius r is 4πr²
■ Also it can be shown that the volume of a sphere of radius r is ⁴⁄₃πr³ 
We will see the derivation of these results in higher classes
Let us see an example:
In fig. 33.29(a) below, a sphere just fits inside a cube of edge 8 cm. 

Fig.33.29

What is the surface area of the sphere?
Solution:
1. Given that the sphere just fits inside the cube. Then the sides of the cube will be touching the surface of the sphere. This is shown in fig.b
• So diameter of the sphere = edge of the cube = 8 cm
• Radius of the sphere = ⁸⁄₂ = 4 cm
2. Surface area = 4πr² = 4×π×4² = 64π cm².

Another example:
A solid sphere of radius 12 cm is cut into two equal halves. What is the surface area of each hemispheres?
Solution:
■ When we say 'Surface area of a cone', there are two items:
(i) The curved surface area of the cone
(ii) The area of the base of the cone
■ When we say 'surface area of a sphere', there is only one item:
• The curved surface area of the sphere
    ♦ This is because, for a sphere, there is no base
■ But when we say 'Surface area of a hemisphere', there are two items:
(i) The curved surface area of the hemisphere
(ii) The area of the base of the hemisphere
• The curved surface area of a hemisphere will be exactly half of the surface area of the corresponding sphere. So in our present problem, we can write:
1. Curved surface area of the hemisphere = ¹⁄₂×4πr² = 2πr² = 2×π×12² = 288π cm²
2. Base area of the hemisphere = Area of the circle of radius 12 cm = πr² = π×12² = 144π cm²
3. Total surface area = 288π + 144π = 432π cm².

Another example:
In fig. 33.30(a) below, a sphere just fits inside a cylinder. 

Fig.33.30

Find the following ratios:
(i) Surface area of the given cylinder : Surface area of the given sphere
(ii) Volume of the given cylinder : Volume of the given sphere
Solution:
1. Given that the sphere just fits inside the cylinder. Then the surface of the sphere will be touching the surface of the cylinder all around. This is shown in fig.b
2. It is clear that diameter of the sphere will be equal to the diameter of the cylinder
• From this we get:
Radius of the sphere = radius of the cylinder 
• Let us put it as r
• Also, height of the cylinder will be equal to 2r
3. Let us calculate the surface areas:
(i) Surface area of the cylinder 
= Two no. base areas + Curved surface area 
= πr² + πr² + 2πrh = 2πr² + 2πr×2r = 6πr²
(ii) Surface area of sphere = 4πr²
■ So the ratio
Surface area of the given cylinder : Surface area of the given sphere
= 6πr² : 4πr² = 6 : 4 = 3 : 2
4. Let us calculate the volumes:
(i) Volume of the cylinder
= Base area × height = πr² × 2r = 2πr³ cm³.
(ii) Volume of sphere = ⁴⁄₃πr³ cm³.
■ So the ratio
Volume of the given cylinder : Volume of the given sphere
= 2πr³ : ⁴⁄₃πr³ = 2 : ⁴⁄₃ = 6 : 4 = 3 : 2
5. So we find that both ratios are 3 : 2

One more example:
A water tank is in the shape of a hemisphere attached to a cylinder. It's radius is 1.5 m and total height is 2.5 m. How many liters of water can it hold?
Solution:
1. Fig. 33.31(a) below shows the view of the tank.

Fig.33.31

• The upper part is in the shape of a cylinder
• The lower part is in the shape of a hemisphere
• In the fig., the two parts can be seen separately. 
• But in an actual tank, after the welding and painting is done, it will be difficult to visually see them separately. However, by taking accurate measurements, we will be able to mark the exact boundary between the two parts.
2. Such measurements are shown in fig.b
• The radius of both the cylinder and sphere will be 1.5 m
    ♦ This is shown by green dimension lines
• The height of the hemisphere will be equal to it's radius, which is 1.5 m
    ♦ This is shown by red dimension line
• The total height of the tank is 2.5 m
    ♦ This is shown by yellow dimension line
• From fig.b it is clear that, the height of the cylinder, h = (2.5-1.5) = 1 m
3. So we have all the required measurements. We can calculate the volumes
(a) Volume of cylinder = πr²h = π×1.5×1.5×1 = 2.25π m³.
(b) Volume of hemisphere = half of volume of sphere = ¹⁄₂ × ⁴⁄₃πr³ = ²⁄₃πr³ 
= ²⁄₃×π×1.5³ = 2.25π m³.
• Total volume = 2.25π + 2.25π = 4.5π = 14.13 m³. 
4. 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, 14.13 m³ = (14.13 × 1000000) cm³ = 14130000 cm³.
• Thus the no. of liters = ^14130000⁄₁₀₀₀ = 14130 liters

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In the next section, we will see a few more solved examples.

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