Chapter 14 Exercise 14.4 Question 1 Solution : A ball bearing has radius 0.7 cm. Find its surface area.
Answer: 6.16 cm²
Sphere: 4πr²
4 × 227 × 0.7 × 0.7 = 887 × 0.49 = 6.16 cm²
The surface area is 6.16 cm².
Ganita Manjari · Class 9 · Part II · Chapter 14
Ganita Manjari Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume Exercise 14.4 Solutions
Eight questions on spheres and hemispheres: surface areas, spheres of one metal, packing, percentage changes and the cost of painting a dome.
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- Ch 14Math of Space: Surface Area and Volume
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Brilliant! You have finished every question in this exercise.
How to answer every question
A = 4πr², V = ⁴⁄₃πr³
- Pick the right formula Sphere: 4πr² and ⁴⁄₃πr³. Hemisphere: CSA 2πr², TSA 3πr², V ⅔πr³.
- Use ratios Surface areas go as r², volumes as r³.
- Convert units Use consistent units before multiplying by costs.
- Check Estimate the size of the answer.
Sphere: all points at a given distance r from the centre.
Hemisphere: half a sphere; its flat circle is the base.
Same metal: equal density, so weight is proportional to volume.
Spheres
Chapter 14 Exercise 14.4 Question 2 Solution : Two solid spheres of the same metal weigh 5920 g and 740 g. Find the radius of the larger one, if the diameter of the smaller is 5 cm.
StarredAnswer: 5 cm
Same metal: weight is proportional to volume
5920740 = 8 = VlargeVsmall = (Rr)³ ⟹ Rr = 2
The radii
The smaller sphere has radius 52 = 2.5 cm, so the larger has R = 2 × 2.5 = 5 cm.
The larger sphere has radius 5 cm.
Chapter 14 Exercise 14.4 Question 3 Solution : The diameter of the moon is about one fourth of the diameter of the earth. Find the ratio of their surface areas.
Answer: 1 : 16
Surface area is proportional to the square of the radius
Moon : Earth = 1² : 4² = 1 : 16
If the earth's radius is R, the moon's is R4, so the area ratio is 4π × R² × 14π × R² × 16 = 116.
The surface areas are in the ratio 1 : 16.
Chapter 14 Exercise 14.4 Question 4 Solution : Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
Answer: CSA 2772 cm², TSA 4158 cm²
Formulas
CSA = 2πr² = 2 × 227 × 21 × 21 = 2772 cm²
TSA = 3πr² = 3 × 227 × 441 = 4158 cm²
(i) 2772 cm², (ii) 4158 cm².
Chapter 14 Exercise 14.4 Question 5 Solution : Metal spheres of radius 2 cm are packed in a box 16 cm × 8 cm × 8 cm. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of the liquid, to the nearest integer.
Answer: 488 cm³
Volumes
box = 16 × 8 × 8 = 1024 cm³
one sphere = 43π × 2³ = 32π3 ≈ 33.51 cm³
16 spheres = 512π3 ≈ 536.17 cm³
Liquid
1024 − 536.17 = 487.83 ≈ 488 cm³
The 16 spheres fit exactly: 4 along the length (4 × 4 = 16 cm) and 2 across and 2 up (2 × 4 = 8 cm).
The liquid has volume about 488 cm³.
Chapter 14 Exercise 14.4 Question 6 Solution : The radius of a sphere is increased by 10%. Show that the volume increases by about 33.1%.
Answer: volume up by about 33.1%
Volume is proportional to r³
The new radius is 1.1r, so the new volume is 43π(1.1r)³ = 1.331 × 43πr³.
increase = 1.331 − 1 = 0.331 = 33.1%
The volume increases by 33.1%.
Chapter 14 Exercise 14.4 Question 7 Solution : The radius of a sphere is increased by x%. The volume increases by 72.8%. Find x.
StarredAnswer: x = 20
Set up
(1 + x100)³ = 1.728
Cube root
1.2³ = 1.728, so 1 + x100 = 1.2, and x = 20.
Check: 1.2 × 1.2 = 1.44 and 1.44 × 1.2 = 1.728 ✓.
x = 20: the radius increases by 20%.
Chapter 14 Exercise 14.4 Question 8 Solution : The hemispherical dome of a building is to be painted. The circumference of its base is 35.2 m. Find the cost at ₹10 per 100 cm².
Answer: ₹1,97,120
The dome
Radius
2πr = 35.2 ⟹ r = 35.2 × 72 × 22 = 246.444 = 5.6 m
Area to paint (curved surface)
CSA = 2πr² = 2 × 227 × 5.6 × 5.6 = 197.12 m²
In cm²: 197.12 × 10,000 = 1,971,200 cm².
Cost
1,971,200100 × 10 = 19,712 × 10 = ₹1,97,120
The painting costs ₹1,97,120.
Answers at a glance
Surface areas scale with r squared and volumes with r cubed.
| Question | What is asked | Key idea | Answer |
|---|---|---|---|
| Q1 | Ball bearing | 4πr² | 6.16 cm² |
| Q2 | Spheres of one metal | Weight ∝ volume | 5 cm |
| Q3 | Moon and Earth | Area ∝ r² | 1 : 16 |
| Q4 | Hemisphere r = 21 | 2πr², 3πr² | 2772; 4158 cm² |
| Q5 | Spheres in a box | 1024 − 536.17 | 488 cm³ |
| Q6 | Radius +10% | 1.1³ | 33.1% |
| Q7 | Volume +72.8% | 1.2³ = 1.728 | x = 20 |
| Q8 | Dome painting | 2πr², r = 5.6 | ₹1,97,120 |
Source of the questions: NCERT, Ganita Manjari, Grade 9, Part II, Chapter 14, Exercise Set 14.4. The solutions, explanations and diagrams on this page are our own working.