Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.4 Solutions Math of Space: Surface Area and Volume
NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.4 Math of Space: Surface Area and Volume provides step-by-step solutions to help students solve questions related to the surface area and volume of three-dimensional shapes.
These solutions are explained in simple language to make it easier for students to understand the concepts, apply the relevant formulas, and solve problems accurately. Students can access the solutions in PDF format for easy download, offline study, and revision. These solutions help improve problem-solving skills, strengthen conceptual understanding, and prepare effectively for exams.
NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.4
1. A ball bearing has a radius of 0.7 cm. Find its surface area.
Given:
Radius, r = 0.7 cm
Surface area of a sphere:
SA = 4πr²
Substituting the value:
SA = 4 × 22/7 × (0.7)²
= 4 × 22/7 × 0.49
= 6.16 cm²
Therefore, the surface area of the ball bearing is 6.16 cm².
2. 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.
Answer:
Since both spheres are made of the same metal, their weights are proportional to their volumes.
So,
Volume ratio = Weight ratio
= 5920 : 740 = 8 : 1
For spheres:
Volume ∝ r³
Therefore,
R³ : r³ = 8 : 1
So,
R : r = 2 : 1
Diameter of the smaller sphere = 5 cm
Hence, its radius:
r = 2.5 cm
Therefore,
R = 2 × 2.5 = 5 cm
Hence, the radius of the larger sphere is 5 cm.
3. 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.
Answer:
The diameter of the moon is 1/4 the diameter of the earth.
Therefore, their radii are also in the ratio:
Moon : Earth = 1 : 4
Surface area of a sphere is proportional to the square of its radius:
Surface area ∝ r²
So,
Surface area of moon : Surface area of earth
= 1² : 4²
= 1 : 16
Therefore, the ratio of their surface areas is 1 : 16.
4. Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
Answer:
Given:
Radius, r = 21 cm
(i) Curved surface area of the hemisphere
CSA = 2πr²
= 2 × 22/7 × 21²
= 2772 cm²
(ii) Total surface area of the hemisphere
TSA = 3πr²
= 3 × 22/7 × 21²
= 4158 cm²
Therefore:
(i) Curved surface area = 2772 cm²
(ii) Total surface area = 4158 cm²
5. 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.
Answer:
Volume of the rectangular box:
V = 16 × 8 × 8 = 1024 cm³
Radius of each sphere:
r = 2 cm
Volume of one sphere:
V = (4/3)πr³
= (4/3) × π × 2³
= 32π/3 cm³
Volume of 16 spheres:
16 × 32π/3 = 512π/3
Using π ≈ 3.14:
Volume of 16 spheres ≈ 535.89 cm³
Therefore, volume of preservative liquid:
= 1024 − 535.89
≈ 488.11 cm³
Rounded to the nearest integer:
488 cm³
Hence, the volume of the preservative liquid is approximately 488 cm³.
6. The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.
Answer:
Let the original radius of the sphere be r.
Original volume:
V = (4/3)πr³
If the radius is increased by 10%, the new radius is:
1.1r
New volume:
V' = (4/3)π(1.1r)³
= (4/3)π × 1.331r³
= 1.331V
Increase in volume:
V' − V = 1.331V − V = 0.331V
Percentage increase:
0.331 × 100 = 33.1%
Therefore, the volume increases by approximately 33.1%.
7. The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.
Answer:
Let the original radius of the sphere be r.
Original volume:
V = (4/3)πr³
If the radius is increased by x%, the new radius is:
r(1 + x/100)
Given that the volume increases by 72.8%, the new volume is:
1.728V
Therefore,
(1 + x/100)³ = 1.728
Taking cube roots:
1 + x/100 = 1.2
So,
x/100 = 0.2
x = 20
Therefore, the radius is increased by 20%.
8. 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.
Answer:
Given:
- Circumference of the base of the hemispherical dome = 35.2 m
- Painting cost = ₹10 per 100 cm²
First, find the radius of the dome.
Circumference = 2πr
35.2 = 2 × 22/7 × r
So,
r = 5.6 m
The area to be painted is the curved surface area of the hemisphere:
CSA = 2πr²
= 2 × 22/7 × (5.6)²
= 197.12 m²
Convert into square centimetres:
197.12 m² = 197.12 × 10,000 = 19,71,200 cm²
Cost of painting 100 cm² = ₹10.
Therefore,
Cost = (19,71,200 / 100) × 10
= ₹1,97,120
Hence, the cost of painting the hemispherical dome is ₹1,97,120.