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Ganita Manjari · Class 9 · Part II · Chapter 14

Ganita Manjari Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume Exercise 14.2 Solutions

Five questions on right circular cylinders: ratios of volume and curved surface area, the largest cylinder in a cube, a 10% change, and recasting a cube.

Last updated 6 October 2026

  • 5Questions
  • 0Diagrams
  • Ch 14Math of Space: Surface Area and Volume

How to answer every question

CSA = 2πrh, V = πr²h
  1. Write the formula CSA = 2πrh, TSA closed = 2πr(h + r), V = πr²h.
  2. Use symbols for ratios Call the dimensions r and h and the other cylinder 2r and h over 2.
  3. Cancel common factors π and many other terms cancel in a ratio.
  4. Round down for whole objects You cannot make part of a rod.

Curved surface area: the area of the side only, 2πrh.

Right circular cylinder: the axis is perpendicular to the circular base.

Recasting: melting a solid and making another with the same volume.

Ratios and change

Chapter 14 Exercise 14.2 Question 1 Solution Cylinder B has twice the radius and half the height of cylinder A. Find the ratio of the curved surface areas of A to B, and of the volumes of A to B.

Answer: 1 : 1 and 1 : 2
Let A have radius r and height h

Then B has radius 2r and height h2.

Cylinder ACylinder B
CSA = 2πrh2πrh2π(2r)(h2) = 2πrh
V = πr²hπr²hπ(2r)²(h2) = 2πr²h
Ratios
CSA of A : CSA of B = 2πrh : 2πrh = 1 : 1
V of A : V of B = πr²h : 2πr²h = 1 : 2

CSA ratio 1 : 1, volume ratio 1 : 2.

Chapter 14 Exercise 14.2 Question 2 Solution The radii of two cylinders are in the ratio 2 : 3 and their heights in the ratio 3 : 2. Find (a) the ratio of their volumes and (b) the ratio of their curved surface areas.

Answer: 2 : 3 and 1 : 1
Use r = 2k, 3k and h = 3m, 2m
(a) V₁V₂ = π(2k)²(3m)π(3k)²(2m) = 1218 = 23
(b) CSA₁CSA₂ = 2π(2k)(3m)2π(3k)(2m) = 1212 = 1

(a) 2 : 3. (b) 1 : 1.

Chapter 14 Exercise 14.2 Question 3 Solution The edge of a cube is r cm. The largest possible right circular cylinder is cut out of it. Find the volume of the cylinder.

Answer: πr³/4
Largest cylinder

The height is the full edge, h = r. The circular base must fit inside a square of side r, so its diameter is r and its radius is r2.

V = π(r2)² × r = πr³4 cm³

The volume is πr³4 cm³.

Chapter 14 Exercise 14.2 Question 4 Solution The radius of a cylinder is increased by 10% and the height is decreased by x% so that the volume stays the same. Find x.

StarredAnswer: x ≈ 17.36
Equal volumes

New radius 1.1r, new height h(1 − x100).

π(1.1r)²h(1 − x100) = πr²h ⟹ 1.21(1 − x100) = 1
1 − x100 = 11.21 ⟹ x100 = 0.211.21 ⟹ x = 2100121 ≈ 17.36
Check

The height becomes about 0.8264h and 1.21 × 0.8264 = 1.000 ✓.

x = 2100121 ≈ 17.36, so the height decreases by about 17.36%.

Melting and recasting

Chapter 14 Exercise 14.2 Question 5 Solution A solid metal cube of side 12 cm is melted and recast into rods of radius 2 cm and height 12 cm. Find (i) the volume of the cube, (ii) the volume of one rod, (iii) the number of complete rods. (π ≈ 227)

Answer: 1728 cm³; 150.86 cm³; 11 rods
Part (i)
12³ = 1728 cm³
Part (ii)
πr²h = 227 × 4 × 12 = 10567 ≈ 150.86 cm³
Part (iii)
1728 × 71056 = 12,0961056 ≈ 11.45

Only whole rods count, so the answer is 11 (the rest of the metal is not enough for a 12th).

(i) 1728 cm³, (ii) about 150.86 cm³, (iii) 11 complete rods.

Answers at a glance

In every ratio question, π cancels.

QuestionWhat is askedKey ideaAnswer
Q1 Radius doubled, height halvedUse r and hCSA 1 : 1, V 1 : 2
Q2 Radii 2 : 3, heights 3 : 2V ∝ r²h, CSA ∝ rhV 2 : 3, CSA 1 : 1
Q3 Cylinder in a cubeRadius r over 2, height rπr³ over 4
Q4 Radius +10%1.21(1 − x/100) = 1x ≈ 17.36
Q5 Cube to rodsSame volume1728; 150.86; 11 rods

Other chapters of Ganita Manjari (Class 9)

Source of the questions: NCERT, Ganita Manjari, Grade 9, Part II, Chapter 14, Exercise Set 14.2. The solutions, explanations and diagrams on this page are our own working.

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