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Naman Academy For Class 9
Chapter 14: Math of Space: Surface Area and VolumePage 129–130All 5 Questions on This Page

Class 9 Maths Chapter 14 Exercise Set 14.2 Solutions

Step-by-step solutions for Class 9 Mathematics Ganita Manjari Part 1I Chapter 14 Exercise Set 14.2 (Questions 1–5, Page 129–130). Covers curved surface area and volume of cylinders, scaling ratios of dimensions and volumes, volume of the largest cylinder cut from a cube, percentage decrease in cylinder height to maintain constant volume, and recasting a metallic cube into cylindrical rods.

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Key Concepts for Exercise Set 14.2

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1

Curved Surface Area of a Cylinder

The curved surface area (CSA) of a right circular cylinder with base radius r and height h is the area of the unrolled rectangular lateral surface.

CSA = 2πrh
2

Volume of a Cylinder

The volume V of a cylinder measures the three-dimensional space it occupies, calculated as base area times height.

V = πr²h
3

Ratios of Surface Areas and Volumes for Cylinders

For two cylinders with radii in ratio r<sub>1</sub>/r<sub>2</sub> and heights in ratio h<sub>1</sub>/h<sub>2</sub>, the curved surface areas scale as (r<sub>1</sub>/r<sub>2</sub>)(h<sub>1</sub>/h<sub>2</sub>) and volumes scale as (r<sub>1</sub>/r<sub>2</sub>)²(h<sub>1</sub>/h<sub>2</sub>).

CSA<sub>1</sub> / CSA<sub>2</sub> = (r<sub>1</sub>/r<sub>2</sub>)(h<sub>1</sub>/h<sub>2</sub>), V<sub>1</sub> / V<sub>2</sub> = (r<sub>1</sub>/r<sub>2</sub>)²(h<sub>1</sub>/h<sub>2</sub>)
4

Largest Cylinder Inscribed in a Cube

The largest right circular cylinder that can be cut from a cube of edge r has its circular base inscribed in a square face (diameter = r, radius R = r/2) and its height equal to the edge length (h = r).

R = r/2, h = r ⇒ V = π(r/2)²r = (πr³) / 4
5

Conservation of Volume in Melting and Recasting

When a solid metallic object is melted and recast into other solids without loss of material, the total volume remains unchanged. The number of complete recast items is the floor of the ratio of the original volume to the volume of one recast unit.

Volume of initial solid = n × Volume of one recast unit
Question 1
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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

Let the radius and height of cylinder A be r and h, respectively.

For cylinder B,

Radius = 2r, Height = h/2
(i) Ratio of curved surface areas

The curved surface area of a cylinder is

CSA = 2πrh

For cylinder A,

CSAA = 2πrh

For cylinder B,

CSAB = 2π(2r)(h/2)
= 2πrh
Therefore,
CSAA : CSAB = 2πrh : 2πrh
= 1 : 1
(ii) Ratio of volumes

The volume of a cylinder is

V = πr²h

For cylinder A,

VA = πr²h

For cylinder B,

VB = π(2r)²(h/2)
= 2πr²h
Therefore,
VA : VB = πr²h : 2πr²h
= 1 : 2
Answer:CSAA : CSAB = 1 : 1 | VA : VB = 1 : 2
Question 2
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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
Given,

r1 / r2 = 23, h1 / h2 = 32

(a) Ratio of their volumes

The volume of a cylinder is

V = πr²h
Therefore,
V1 / V2 = πr1²h1πr2²h2
= (r1 / r2)² × (h1 / h2)
= (23)² × (32)
= 49 × 32
= 23
Hence,
V1 : V2 = 2 : 3
(b) Ratio of their curved surface areas

The curved surface area of a cylinder is

CSA = 2πrh
Therefore,
CSA1 / CSA2 = 2πr1h12πr2h2
= (r1 / r2) × (h1 / h2)
= 23 × 32
= 1
Hence,
CSA1 : CSA2 = 1 : 1
Answer:Ratio of volumes = 2 : 3 | Ratio of curved surface areas = 1 : 1
Question 3
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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 cm³)?

The largest right circular cylinder cut out of a cube of edge r has diameter equal to r (radius R = r/2) and height h = r.

Solution

The edge of the cube is r cm.

For the largest possible right circular cylinder that fits inside the cube, its circular base fits exactly inside a square face of the cube.

Therefore, the diameter of the cylinder equals the edge of the cube:
Diameter = r cm

Let the radius of the cylinder be R. Then,

R = r/2 cm

The height of the cylinder equals the edge of the cube:

h = r cm

The volume of a cylinder is

V = πR²h
Substituting R = r/2 and h = r,
V = π(r/2)²r
= π(r²/4) × r
= πr³4
Therefore,
V = πr³4 cm³
Answer:V = πr³4 cm³
Question 4
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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

Let the original radius and height of the cylinder be r and h, respectively.

The volume of a cylinder is

V = πr²h

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

r + (10100)r = 1.1r

After decreasing the height by x%, the new height becomes

h − (x/100)h = (1 − x/100)h

Since the volume remains unchanged,

πr²h = π(1.1r)²(1 − x/100)h

Cancelling the common factors πr²h,

1 = 1.21(1 − x/100)
1 − x/100 = 11.21
1 − x/100 = 100121
x / 100 = 1 − 100121
x / 100 = 21121
x = 2100121
x = 17 43121 ≈ 17.36
Therefore,
x ≈ 17.36%
Answer:x = 17 43121% ≈ 17.36%
Question 5
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A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each having radius 2 cm and height 12 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. (π ≈ 22/7)

Solution
(i) Find the volume of the cube
Side of the cube = 12 cm.

The volume of a cube is

V = a³
Therefore,
V = 12³
= 12 × 12 × 12
= 1728 cm³
Volume of the cube = 1728 cm³
(ii) Find the volume of one cylindrical rod
Given:

Radius r = 2 cm, Height h = 12 cm

The volume of a cylinder is

V = πr²h
Taking π = 22/7,
V = (227) × 2² × 12
= (227) × 4 × 12
= 10567
= 150 67
≈ 150.86 cm³
Volume of one rod ≈ 150.86 cm³
(iii) Find the approximate number of complete cylindrical rods that can be formed

Since the cube is melted and recast without loss of metal, its volume equals the total volume of the cylindrical rods formed.

Number of rods = Volume of the cubeVolume of one rod

Substituting the values,
Number of rods = 1728 / (10567)
= 1728 × 71056
= 12611
= 11 511
≈ 11.45

Since only complete rods can be formed, 11 complete rods can be made. Some metal will remain, but there will not be enough to form another complete rod.

Number of complete cylindrical rods = 11
Answer:(i) Volume of the cube = 1728 cm³ | (ii) Volume of one rod ≈ 150.86 cm³ (150 67 cm³) | (iii) Number of complete cylindrical rods = 11

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