NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 14 Math of Space Surface Area and Volume Exercise 14.2

Exercise Set 14.2

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.
Answer:

Let cylinder A have radius r and height h. Then cylinder B has radius 2r and height \( \frac{h}{2} \).
Curved surface areas:
CSA of A = 2πrh
CSA of B = 2π(2r) × \( \frac{h}{2} \) = 2πrh
CSA of A : CSA of B = 2πrh : 2πrh = 1 : 1
Volumes:
Volume of A = πr²h
Volume of B = π(2r)² × \( \frac{h}{2} \) = π × 4r² × \( \frac{h}{2} \) = 2πr²h
Volume of A : Volume of B = πr²h : 2πr²h = 1 : 2
CSA of A : CSA of B = 1 : 1, and volume of A : volume of B = 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.
Answer:

Take the radii as 2x and 3x, and the heights as 3y and 2y.
(a) \( \frac{V_1}{V_2} = \frac{\pi (2x)^2 (3y)}{\pi (3x)^2 (2y)} = \frac{12x^2y}{18x^2y} = \frac{2}{3} \)
So V₁ : V₂ = 2 : 3.
(b) \( \frac{\text{CSA}_1}{\text{CSA}_2} = \frac{2\pi (2x)(3y)}{2\pi (3x)(2y)} = \frac{6xy}{6xy} = 1 \)
So CSA₁ : CSA₂ = 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 cm³)?
Answer:

For the largest cylinder, its circular end must fit exactly inside a square face of the cube (touching all four sides), and its height must be the whole edge of the cube.
Diameter of the base = edge = r, so the radius is \( \frac{r}{2} \) cm.
Height of the cylinder = r cm.
V = π × (radius)² × height = π × \( \left(\frac{r}{2}\right)^2 \) × r = π × \( \frac{r^2}{4} \) × r = \( \frac{\pi r^3}{4} \)
With π = \( \frac{22}{7} \), V = \( \frac{11r^3}{14} \) ≈ 0.79 r³, i.e. the cylinder uses about 78.5% of the cube (\( \frac{\pi}{4} \) ≈ 0.785).
The volume of the cylinder is \( \frac{\pi r^3}{4} \) cm³.

Teacher's Note:
1. The radius is half the edge, not the edge itself; students often write πr³ by mistake.
2. The fraction π/4 ≈ 0.785 does not depend on the size of the cube.

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.
Answer:
Let the original radius be r and height h, so the original volume is πr²h.
New radius = r + 10% of r = 1.1r.
New height = h − x% of h = \( \left(1 - \frac{x}{100}\right)h \).
The volume does not change, so
\( \pi (1.1r)^2 \left(1 - \frac{x}{100}\right) h = \pi r^2 h \)
\( 1.21\left(1 - \frac{x}{100}\right) = 1 \;\Rightarrow\; 1 - \frac{x}{100} = \frac{1}{1.21} = \frac{100}{121} \)
\( \frac{x}{100} = 1 - \frac{100}{121} = \frac{21}{121} \;\Rightarrow\; x = \frac{2100}{121} = 17\frac{43}{121} \)
x = \( \frac{2100}{121} \) ≈ 17.36, so the height must be reduced by about 17.36%.

Teacher's Note:
1. The answer is not 10% (or 21%): the radius is squared, so its effect is 1.21 times, and a percentage decrease works on the new total.
2. Cancelling πr²h at the start keeps the working short.

Question 5. 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 ≈ \( \frac{22}{7} \))
Answer:

(i) Volume of the cube = 12³ = 12 × 12 × 12 = 1728 cm³
(ii) Volume of one rod (r = 2 cm, h = 12 cm) = πr²h = \( \frac{22}{7} \) × 2² × 12 = \( \frac{22 \times 48}{7} \) = \( \frac{1056}{7} \) ≈ 150.86 cm³
(iii) Melting and recasting does not change the volume of metal, so
Number of rods = 1728 ÷ \( \frac{1056}{7} \) = \( \frac{1728 \times 7}{1056} \) = \( \frac{12096}{1056} \) ≈ 11.45
Only complete rods count, so we round down.
11 complete rods can be made; about 1728 − 11 × 150.86 ≈ 68.6 cm³ of metal is left over.

Teacher's Note:
1. For 'complete' objects always round down, even though 11.45 is closer to 11 anyway.
2. The leftover metal is less than one rod's volume, which confirms the answer.