Ex 14.2 Class 9 Maths Ganita Manjari Solutions | Sid Classes

Class 9 Maths Ganita Manjari Ex 14.2 solutions on volume of solid shapes, step by step. 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 14.2 Surface Areas and Volumes of Cylinders, Recasting Solids, and Dimensional Ratios. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough covers cylinder curved surface areas, volume ratios, maximum cylindrical cutouts from cubes, percentage changes in dimensions, and melting/recasting metallic solids entirely clear.

🔑 Key Formulas & Principles:

    Cylinder Formulas: Curved Surface Area (\(\text{CSA}\)) = \(2\pi rh\), Total Surface Area (\(\text{TSA}\)) = \(2\pi r(r + h)\), Volume (\(V\)) = \(\pi r^2 h\).
    Conservation of Volume: When a solid shape is melted and recast into another shape, the total volume remains constant (\(V_{\text{initial}} = V_{\text{final}}\)).

NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.2 Solutions

Question 1

Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A, and the height of cylinder B is half that of cylinder A. Find the ratio of the curved surface area of A to the curved surface area of B. Also find the ratio of the volume of A to the volume of B.

Solution:

1. Let the radius of cylinder A be \(r_A = r\) and height be \(h_A = h\).
2. Then, radius of cylinder B is \(r_B = 2r\) and height is \(h_B = \frac{1}{2}h\).

(i) Ratio of Curved Surface Areas (\(\text{CSA}\)):
* \(\text{CSA of A} = 2\pi r_A h_A = 2\pi rh\)
* \(\text{CSA of B} = 2\pi r_B h_B = 2\pi (2r)\left(\frac{1}{2}h\right) = 2\pi rh\)
* Ratio = \(\frac{2\pi rh}{2\pi rh} = \mathbf{1 : 1}\).

(ii) Ratio of Volumes (\(V\)):
* \(\text{Volume of A} = \pi r_A^2 h_A = \pi r^2 h\)
* \(\text{Volume of B} = \pi r_B^2 h_B = \pi (2r)^2 \left(\frac{1}{2}h\right) = \pi (4r^2) \left(\frac{1}{2}h\right) = 2\pi r^2 h\)
* Ratio = \(\frac{\pi r^2 h}{2\pi r^2 h} = \mathbf{1 : 2}\).

Question 2

The radii of two cylinders are in the ratio 2:3, and their heights are in the ratio 3:2. Find (a) the ratio of their volumes, and (b) the ratio of their curved surface areas.

Solution:

1. Let the radii be \(r_1 = 2x\) and \(r_2 = 3x\), and their heights be \(h_1 = 3y\) and \(h_2 = 2y\).

(a) Ratio of their volumes:
* \(\frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} = \frac{(2x)^2 (3y)}{(3x)^2 (2y)} = \frac{4x^2 \times 3y}{9x^2 \times 2y} = \frac{12}{18} = \frac{2}{3} = \mathbf{2 : 3}\).

(b) Ratio of their curved surface areas:
* \(\frac{\text{CSA}_1}{\text{CSA}_2} = \frac{2\pi r_1 h_1}{2\pi r_2 h_2} = \frac{(2x)(3y)}{(3x)(2y)} = \frac{6xy}{6xy} = \mathbf{1 : 1}\).

Question 3

The edge of a cube measures \(r\) cm. The largest possible right circular cylinder is cut out of the cube. What do you think is the volume of the cylinder (in cm\(^3\))?

Solution:

1. For the largest possible right circular cylinder cut out from a cube of edge \(r\):
- The diameter of the cylinder's base equals the edge of the cube, so radius \(R = \frac{r}{2}\).
- The height of the cylinder equals the edge of the cube, so height \(h = r\).
2. Volume of the cylinder = \(\pi R^2 h = \pi \left(\frac{r}{2}\right)^2 (r) = \pi \left(\frac{r^2}{4}\right)r = \mathbf{\frac{\pi r^3}{4}\text{ cm}^3}\).

Question 4

The radius of the base of a cylinder is increased by 10%. At the same time, the height of the cylinder is decreased by \(x\%\). Given that the volume of the cylinder remains unchanged, find the value of \(x\).

Solution:

1. Let initial radius be \(r\) and height be \(h\). Initial volume \(V = \pi r^2 h\).
2. New radius \(r' = r + 0.10r = 1.1r\).
3. New height \(h' = h \left(1 - \frac{x}{100}\right)\).
4. New volume \(V' = \pi (1.1r)^2 h' = \pi (1.21 r^2) h \left(1 - \frac{x}{100}\right)\).
5. Since volume remains unchanged (\(V = V'\)):
\(\pi r^2 h = \pi (1.21 r^2) h \left(1 - \frac{x}{100}\right)\)
\(1 = 1.21 \left(1 - \frac{x}{100}\right)\)
\(\frac{100}{121} = 1 - \frac{x}{100} \implies \frac{x}{100} = 1 - \frac{100}{121} = \frac{21}{121}\)
\(x = \frac{2100}{121} \approx \mathbf{17.36}\).

Question 5

A solid metallic cube of side \(12\text{ cm}\) is melted and recast into solid cylindrical rods, each having radius \(2\text{ cm}\) and height \(12\text{ cm}\). Find:
(i) the volume of the cube,
(ii) the volume of one cylindrical rod,
(iii) the approximate number of complete cylindrical rods that can be formed. (\(\pi \approx \frac{22}{7}\))

Solution:

* (i) Volume of the cube:
\(\text{Volume} = \text{side}^3 = 12^3 = \mathbf{1728\text{ cm}^3}\).

* (ii) Volume of one cylindrical rod:
\(\text{Volume} = \pi r^2 h = \frac{22}{7} \times (2)^2 \times 12 = \frac{22}{7} \times 4 \times 12 = \frac{1056}{7} \approx \mathbf{150.86\text{ cm}^3}\).

* (iii) Number of complete cylindrical rods:
\(\text{Number of rods} = \frac{\text{Volume of the cube}}{\text{Volume of one rod}} = \frac{1728}{\frac{1056}{7}} = \frac{1728 \times 7}{1056} = \frac{12096}{1056} \approx 11.45\).
Since only complete rods can be formed, the number of complete cylindrical rods is \(\mathbf{11}\).

Frequently Asked Questions (FAQs)

Q1: What happens to the volume of a cylinder if its radius is doubled?
A: Because volume depends on the square of the radius (\(\pi r^2 h\)), doubling the radius quadruples the volume of the cylinder.

Q2: How do you find the number of solid items formed by melting another solid?
A: Divide the total volume of the original melted solid by the volume of a single recast item, then round down to the nearest whole number for complete items.

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