Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.2 Solutions Math of Space: Surface Area and Volume
NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 14 Exercise 14.2 Math of Space: Surface Area and Volume provides step-by-step solutions to help students understand the concepts of surface area and volume of three-dimensional shapes.
The solutions are explained in simple language, making it easier for students to apply the relevant formulas and solve mathematical problems accurately. Students can access the solutions in PDF format for easy download, offline study, and revision. These solutions help strengthen conceptual understanding, improve problem-solving skills, and prepare effectively for exams.
NCERT Class 9 Maths Ganita Manjari Chapter 14 Exercise Set 14.2
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 the radius and height of cylinder A be r and h.
Then for cylinder B:
Radius = 2r
Height = h/2
Curved surface area
Curved surface area of a cylinder:
CSA = 2πrh
For cylinder A:
CSAₐ = 2πrh
For cylinder B:
CSAᵦ = 2π(2r)(h/2) = 2πrh
Therefore,
CSA of A : CSA of B = 1 : 1
Volume
Volume of a cylinder:
V = πr²h
For cylinder A:
Vₐ = πr²h
For cylinder B:
Vᵦ = π(2r)²(h/2)
= 2πr²h
Therefore,
Volume of A : Volume of B = 1 : 2
Hence:
Curved surface area ratio = 1 : 1
Volume ratio = 1 : 2
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.
Let the radii of the two cylinders be 2r and 3r, and their heights be 3h and 2h.
(a) Ratio of their volumes
Volume of a cylinder:
V = πr²h
So,
V₁ : V₂ = π(2r)²(3h) : π(3r)²(2h)
= 4 × 3 : 9 × 2
= 12 : 18
= 2 : 3
Therefore, the ratio of their volumes is:
2 : 3
(b) Ratio of their curved surface areas
Curved surface area of a cylinder:
CSA = 2πrh
So,
CSA₁ : CSA₂ = 2π(2r)(3h) : 2π(3r)(2h)
= 6 : 6
= 1 : 1
Therefore:
(a) Volume ratio = 2 : 3
(b) Curved surface area ratio = 1 : 1
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:
The largest possible right circular cylinder cut from a cube of edge r cm will have:
Height = r cm
and its circular base will be inscribed in a square of side r cm.
So, the diameter of the cylinder is r, hence:
Radius = r/2
Volume of a cylinder:
V = πR²h
Substituting:
V = π(r/2)²(r)
= πr³/4
Therefore, the volume of the cylinder is:
πr³/4 cm³
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 and height of the cylinder be r and h.
Original volume:
V = πr²h
The radius is increased by 10%, so the new radius is:
1.1r
The height is decreased by x%, so the new height is:
h(1 − x/100)
Since the volume remains unchanged:
π(1.1r)²h(1 − x/100) = πr²h
Cancelling πr²h:
1.21(1 − x/100) = 1
So,
1 − x/100 = 1/1.21 = 100/121
Therefore,
x/100 = 1 − 100/121 = 21/121
x = 2100/121 ≈ 17.36
Hence, the height must be decreased by approximately 17.36%
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 ≈ 22/7)
Answer:
Given:
- Side of cube = 12 cm
- Radius of each cylindrical rod = 2 cm
- Height of each rod = 12 cm
- π ≈ 22/7
(i) Volume of the cube
Volume = side³
= 12³
= 1728 cm³
Therefore, the volume of the cube is 1728 cm³.
(ii) Volume of one cylindrical rod
Volume of a cylinder:
V = πr²h
= 22/7 × 2² × 12
= 22/7 × 48
= 1056/7
≈ 150.86 cm³
Therefore, the volume of one cylindrical rod is approximately 150.86 cm³.
(iii) Number of complete cylindrical rods
Number of rods = Volume of cube ÷ Volume of one rod
= 1728 ÷ 150.86
≈ 11.45
Only complete rods can be formed.
Therefore, approximately 11 complete cylindrical rods can be made.