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Naman Academy For Class 9
Chapter 14: Math of Space: Surface Area and VolumePage 140–141All 8 Questions on This Page

Class 9 Maths Chapter 14 Exercise Set 14.4 Solutions

Step-by-step solutions for Class 9 Mathematics Ganita Manjari Part 1I Chapter 14 Exercise Set 14.4 (Questions 1–8, Page 140–141). Covers surface area and volume of spheres and hemispheres, radius determination from mass/weight ratio of similar metal spheres, ratio of Moon to Earth surface areas, curved and total surface areas of a hemisphere, volume of preservative liquid around packed spheres in a cuboidal box, percentage increase relationships between sphere radius and volume, and painting cost calculation for a hemispherical dome.

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Key Concepts for Exercise Set 14.4

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1

Surface Area of a Sphere

A sphere of radius r has a single continuous curved surface whose total area is equal to four times the area of its great circular cross-section.

S = 4πr²
2

Volume of a Sphere

The measure of three-dimensional space enclosed by a sphere of radius r is four-thirds pi times the cube of its radius.

V = (4/3)πr³
3

Curved and Total Surface Area of a Hemisphere

When a solid sphere of radius r is cut into two equal halves by a plane through its center, each half is a hemisphere. Its curved surface area is half of the sphere's surface area, and its total surface area includes the flat circular base.

CSA = 2πr², TSA = CSA + πr² = 3πr²
4

Volume of a Hemisphere

The volume of a hemisphere of radius r is exactly half the volume of the complete sphere.

V = (2/3)πr³
5

Scaling Laws and Percentage Changes for Spheres

For spheres of different sizes or undergoing uniform dilation: surface areas scale as the ratio of squared radii (r₁/r₂)², while volumes (and masses for constant density) scale as the ratio of cubed radii (r₁/r₂)³. An increase in radius by p% yields a volume multiplied by (1 + p/100)³.

S₁/S₂ = (r₁/r₂)², V₁/V₂ = (r₁/r₂)³, V_new = (1 + p/100)³ × V_old
Question 1
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A ball bearing has a radius of 0.7 cm. Find its surface area.

Solution
Given:
Radius, r = 0.7 cm

The surface area of a sphere is

S = 4πr²
Taking π ≈ 22/7,
S = 4 × (227) × (0.7)²
= 4 × (227) × 0.49
= 6.16 cm²
Answer:Surface area = 6.16 cm²
Question 2
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Two solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.

Solution
Given:

Weight of the larger sphere = 5920 g

Weight of the smaller sphere = 740 g

Diameter of the smaller sphere = 5 cm

Since both spheres are made of the same metal, their masses are proportional to their volumes. Therefore,

Vlarge / Vsmall = 5920740
= 8

The volume of a sphere is

V = (43)πr³

Since the factor (43)π is the same for both spheres, their volume ratio is the ratio of the cubes of their radii.

Let the radii of the larger and smaller spheres be R and r, respectively.

R³ / r³ = 8
Taking cube roots,
R / r = ∛8 = 2

The radius of the smaller sphere is

r = 52 = 2.5 cm

Therefore,
R = 2 × 2.5
= 5 cm
Answer:Radius of the larger sphere = 5 cm
Question 3
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The diameter of the moon is approximately one fourth the diameter of the earth. Given that the moon and earth are both roughly spherical, find the ratio of their surface areas.

Solution

Let the radii of the Moon and Earth be r and R, respectively.

Since the Moon's diameter is approximately one fourth of the Earth's diameter,

r / R = 14

The surface area of a sphere is

S = 4πr²
Therefore,

Surface area of Moon / Surface area of Earth = 4πr²4πR²

= (r / R)²
= (14)²
= 116
Hence,
Answer:Moon : Earth surface areas ≈ 1 : 16
Question 4
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Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.

Solution
Given:
Radius, r = 21 cm
(i) Curved surface area

The curved surface area of a hemisphere is

CSA = 2πr²
Taking π ≈ 22/7,
CSA = 2 × (227) × 21²
= 2 × 22 × 3 × 21
= 2772 cm²
Curved surface area = 2772 cm²
(ii) Total surface area

The total surface area includes the curved surface and the circular base.

TSA = 2πr² + πr² = 3πr²
Therefore,
TSA = 3 × (227) × 21²
= 3 × 22 × 3 × 21
= 4158 cm²
Answer:Curved surface area = 2772 cm² | Total surface area = 4158 cm²
Question 5
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Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 cm. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.

16 spheres of radius 2 cm (diameter 4 cm) pack tightly inside the 16 cm × 8 cm × 8 cm box in a 4 × 2 × 2 arrangement, with preservative liquid filling the unoccupied volume.

Solution
Given:

Radius of each sphere, r = 2 cm

Number of spheres = 16

Dimensions of the box = 16 cm × 8 cm × 8 cm

The volume of a cuboid is

Vbox = lwh
Therefore,
Vbox = 16 × 8 × 8
= 1024 cm³

The volume of a sphere is

Vsphere = (43)πr³

Taking π ≈ 227, the volume of one sphere is

Vsphere = (43) × (227) × 2³
= 70421 cm³

The total volume of 16 spheres is

Vspheres = 16 × (70421)
= 1126421 cm³

The liquid fills the space remaining inside the box.

Vliquid = Vbox - Vspheres
= 1024 - (1126421)
= 1024021
≈ 487.62 cm³

Rounding to the nearest integer,

Answer:Volume of preservative liquid = 488 cm³
Question 6
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The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.

Solution

Let the original radius of the sphere be r.

The original volume is

V1 = (43)πr³

After increasing the radius by 10%, the new radius becomes

r + (10100)r = 1.1r

The new volume is

V2 = (43)π(1.1r)³
Since
(1.1)³ = 1.331
we get
V2 = 1.331 × ((43)πr³)
= 1.331V1
Thus, the new volume is 133.1% of the original volume.

The percentage increase is

Percentage increase = 133.1% - 100%
= 33.1%
Answer:Increase in volume = 33.1%
Question 7
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The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.

Solution

Let the original radius of the sphere be r.

The original volume is

V1 = (43)πr³

Since the radius increases by x%, the new radius is

r + (x/100)r = (1 + x/100)r

The new volume is

V2 = (43)π[(1 + x/100)r]³

The volume increases by 72.8%. Therefore, the new volume is

V2 = V1 + (72.8100)V1
= 1.728V1

Equating the two expressions for the new volume,

(43)π(1 + x/100)³r³ = 1.728 × ((43)πr³)

Cancelling the common factors,

(1 + x/100)³ = 1.728
(1 + x/100)³ = 17281000
(1 + x/100)³ = (1.2)³
Taking cube roots on both sides,
1 + x/100 = 1.2
x/100 = 1.2 - 1
x/100 = 0.2
x = 20
Hence,
Answer:x = 20%
Question 8
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The hemispherical dome of a building needs to be painted (see Fig. 14.16). If the circumference of the base of the dome is 35.2 m, find the cost of painting it, given the cost of painting is ₹10 per 100 cm².

Fig. 14.16: The hemispherical dome has a base circumference of 35.2 m (radius r = 5.6 m). Only the curved surface of area 197.12 m² (1,971,200 cm²) is painted.

Solution
Given:
Circumference of the base = 35.2 m
Painting cost = ₹10 per 100 cm²
Finding the radius of the dome

The circumference of a circle is

C = 2πr
Taking π = 22/7,
35.2 = 2 × (227) × r
r = 35.2 × 744
r = 5.6 m
Finding the area to be painted

Only the curved surface of the hemispherical dome needs to be painted.

The curved surface area of a hemisphere is

CSA = 2πr²
Substituting r = 5.6 m,
CSA = 2 × (227) × 5.6²
= 2 × (227) × 31.36
= 197.12 m²
Convert the area into square centimetres because the painting rate is given per 100 cm².

Since 1 m² = 10,000 cm²,

197.12 m² = 197.12 × 10,000 cm²
= 1,971,200 cm²
Finding the cost of painting

Number of 100 cm² units:

1,971,200100 = 19,712

Each unit costs ₹10. Therefore,

Total cost = 19,712 × 10
= ₹1,97,120
Answer:Cost of painting = ₹1,97,120

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