Ex 14.4 Class 9 Maths Ganita Manjari Solutions | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.4 Surface Areas and Volumes of Spheres, Hemispheres, Density Ratios, and Scaling Principles. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough covers ball bearing surface areas, mass-volume relationships, moon-to-earth surface area ratios, hemispherical areas, liquid displacement packing, percentage volume expansions, and dome painting costs entirely clear.
🔑 Key Formulas & Principles:
- Sphere Formulas: Surface Area \(S = 4\pi r^2\), Volume \(V = \frac{4}{3}\pi r^3\).
- Hemisphere Formulas: Curved Surface Area (\(\text{CSA}\)) = \(2\pi r^2\), Total Surface Area (\(\text{TSA}\)) = \(3\pi r^2\), Volume = \(\frac{2}{3}\pi r^3\).
NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.4 Solutions
Question 1
A ball bearing has a radius of \(0.7\text{ cm}\). Find its surface area.
Solution:
1. Given: Radius of the sphere (\(r\)) = \(0.7\text{ cm}\).
2. Surface area of a sphere = \(4\pi r^2\).
3. Substituting the values (\(\pi = \frac{22}{7}\)):
\(\text{Surface Area} = 4 \times \frac{22}{7} \times (0.7)^2 = 4 \times \frac{22}{7} \times 0.49 = 4 \times 22 \times 0.07 = \mathbf{6.16\text{ cm}^2}\).
Question 2
Two solid spheres made of the same metal have weights \(5920\text{ g}\) and \(740\text{ g}\). Determine the radius of the larger sphere, if the diameter of the smaller one is \(5\text{ cm}\).
Solution:
1. Since both spheres are made of the same metal, their density is constant. Therefore, weight is directly proportional to volume (\(\text{Weight} \propto V\)).
2. Ratio of volumes = Ratio of weights:
\(\frac{V_1}{V_2} = \frac{5920}{740} = 8\).
3. Since volume of a sphere \(V = \frac{4}{3}\pi r^3\), the ratio of their volumes is equal to the cube of the ratio of their radii:
\(\left(\frac{r_1}{r_2}\right)^3 = 8 \implies \frac{r_1}{r_2} = 2\).
4. Diameter of the smaller sphere = \(5\text{ cm} \implies \text{Radius } r_2 = \frac{5}{2} = 2.5\text{ cm}\).
5. Radius of the larger sphere (\(r_1\)) = \(2 \times r_2 = 2 \times 2.5 = \mathbf{5\text{ cm}}\).
Question 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.
Solution:
1. Let the diameter of the earth be \(D\) and its radius be \(R = \frac{D}{2}\).
2. The diameter of the moon is \(\frac{D}{4}\) and its radius is \(r = \frac{D}{8}\).
3. Ratio of surface areas = \(\frac{\text{Surface area of moon}}{\text{Surface area of earth}} = \frac{4\pi r^2}{4\pi R^2} = \left(\frac{r}{R}\right)^2\).
4. \(\left(\frac{D/8}{D/2}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} = \mathbf{1 : 16}\).
Question 4
Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius \(21\text{ cm}\).
Solution:
Given: Radius (\(r\)) = \(21\text{ cm}\).
* (i) Curved Surface Area (\(\text{CSA}\)) of hemisphere:
\(\text{CSA} = 2\pi r^2 = 2 \times \frac{22}{7} \times (21)^2 = 2 \times \frac{22}{7} \times 441 = 2 \times 22 \times 63 = \mathbf{2772\text{ cm}^2}\).
* (ii) Total Surface Area (\(\text{TSA}\)) of hemisphere:
\(\text{TSA} = 3\pi r^2 = 3 \times \frac{22}{7} \times 441 = 3 \times 22 \times 63 = \mathbf{4158\text{ cm}^2}\).
Question 5
Metal spheres, each of radius \(2\text{ cm}\), are packed into a rectangular box of internal dimensions \(16\text{ cm} \times 8\text{ cm} \times 8\text{ cm}\). 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.
Solution:
1. Volume of the rectangular box = \(16 \times 8 \times 8 = 1024\text{ cm}^3\).
2. Volume of one metal sphere of radius \(2\text{ cm}\):
\(V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (2)^3 = \frac{4}{3} \times \frac{22}{7} \times 8 = \frac{704}{21} \approx 33.52\text{ cm}^3\).
3. Total volume of 16 spheres = \(16 \times 33.5238 \approx 536.38\text{ cm}^3\).
4. Volume of preservative liquid = Volume of box - Volume of 16 spheres
\(= 1024 - 536.38 = 487.62\text{ cm}^3 \approx \mathbf{488\text{ cm}^3}\).
Question 6
The radius of a sphere is increased by \(10\%\). Show that the volume increases by approximately \(33.1\%\).
Solution:
1. Let the original radius be \(r\). Original volume \(V = \frac{4}{3}\pi r^3\).
2. New radius \(r' = r + 0.10r = 1.1r\).
3. New volume \(V' = \frac{4}{3}\pi (1.1r)^3 = \frac{4}{3}\pi (1.331 r^3) = 1.331 V\).
4. Percentage increase in volume = \(\frac{V' - V}{V} \times 100 = \frac{1.331 V - V}{V} \times 100 = 0.331 \times 100 = \mathbf{33.1\%}\).
Question 7
The radius of a sphere is increased by \(x\%\). The volume of the sphere increases by \(72.8\%\). Find the value of \(x\).
Solution:
1. New volume \(V' = V + 0.728 V = 1.728 V\).
2. Let new radius be \(r'\). Then \(\left(\frac{r'}{r}\right)^3 = 1.728\).
3. Taking cube root on both sides: \(\frac{r'}{r} = \sqrt[3]{1.728} = 1.2\).
4. Since \(r' = 1.2r = r + 0.2r\), the radius is increased by \(0.2 \times 100\% = \mathbf{20\%}\) (so \(x = 20\)).
Question 8
The hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is \(35.2\text{ m}\), find the cost of painting it, given the cost of painting is ₹10 per \(100\text{ cm}^2\).
Solution:
1. Given base circumference \(2\pi r = 35.2\text{ m}\):
\(2 \times \frac{22}{7} \times r = 35.2 \implies \frac{44}{7}r = 35.2 \implies r = \frac{35.2 \times 7}{44} = 5.6\text{ m}\).
2. Curved Surface Area of dome = \(2\pi r^2 = 2 \times \frac{22}{7} \times (5.6)^2 = 2 \times \frac{22}{7} \times 31.36 = 197.12\text{ m}^2\).
3. Converting area to square centimetres (\(1\text{ m}^2 = 10,000\text{ cm}^2\)):
\(\text{Area} = 197.12 \times 10,000 = 1,971,200\text{ cm}^2\).
4. Cost of painting at ₹10 per \(100\text{ cm}^2\):
\(\text{Cost} = \frac{1,971,200}{100} \times 10 = 19,712 \times 10 = \mathbf{₹1,97,120}\).
Frequently Asked Questions (FAQs)
Q1: How does the volume of a sphere scale when its radius is increased by a percentage?
A: Because volume depends on the cube of the radius (\(r^3\)), scaling the radius by a factor \(k\) scales the volume by \(k^3\).
Q2: What is the difference between curved surface area and total surface area of a hemisphere?
A: Curved surface area accounts only for the dome part (\(2\pi r^2\)), whereas total surface area also includes the flat circular base (\(\pi r^2\)), totaling \(3\pi r^2\).
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