Table of Contents
Exercise 10.2 Solutions
Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad
Problem 1
A toy is in the form of a cone mounted on a hemisphere of the same diameter. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. [use π = 3.14]
Solution:
Given: Diameter = 6 cm ⇒ Radius (r) = 3 cm
Cone height (h) = 4 cm
Slant height of cone (l) = √(r² + h²) = √(9 + 16) = 5 cm
Curved surface area of cone = πrl = 3.14 × 3 × 5 = 47.1 cm²
Curved surface area of hemisphere = 2πr² = 2 × 3.14 × 9 = 56.52 cm²
Total surface area = Cone CSA + Hemisphere CSA = 47.1 + 56.52 = 103.62 cm²
Problem 2
A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm and the heights of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the total surface area of the solid. [use π = 3.14]
Solution:
Given: Radius (r) = 8 cm
Cylinder height (h₁) = 10 cm, Cone height (h₂) = 6 cm
Cylinder CSA = 2πrh₁ = 2 × 3.14 × 8 × 10 = 502.4 cm²
Cone slant height (l) = √(r² + h₂²) = √(64 + 36) = 10 cm
Cone CSA = πrl = 3.14 × 8 × 10 = 251.2 cm²
Hemisphere CSA = 2πr² = 2 × 3.14 × 64 = 401.92 cm²
Total surface area = Cylinder CSA + Cone CSA + Hemisphere CSA = 502.4 + 251.2 + 401.92 = 1155.52 cm²
Problem 3
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm and the thickness is 5 mm. Find its surface area.
Solution:
Given: Total length = 14 mm, Diameter = 5 mm ⇒ Radius (r) = 2.5 mm
Height of cylinder = Total length – 2 × radius = 14 – 5 = 9 mm
Cylinder CSA = 2πrh = 2 × 3.14 × 2.5 × 9 ≈ 141.3 mm²
Two hemispheres = 1 full sphere surface area = 4πr² = 4 × 3.14 × 6.25 ≈ 78.5 mm²
Total surface area = 141.3 + 78.5 ≈ 219.8 mm²
Problem 4
Two cubes each of volume 64 cm³ are joined end to end together. Find the surface area of the resulting cuboid.
Solution:
Volume of each cube = 64 cm³ ⇒ Side length (a) = ∛64 = 4 cm
When joined, cuboid dimensions become: Length = 8 cm, Breadth = 4 cm, Height = 4 cm
Total surface area = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32 + 16 + 32) = 160 cm²
Problem 5
A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at rate of ₹20 per m².
Solution:
Given: Diameter = 1.4 m ⇒ Radius (r) = 0.7 m, Cylinder height (h) = 8 m
Cylinder CSA = 2πrh = 2 × (22/7) × 0.7 × 8 = 35.2 m²
Two hemispheres = 1 full sphere surface area = 4πr² = 4 × (22/7) × 0.49 ≈ 6.16 m²
Total surface area = 35.2 + 6.16 = 41.36 m²
Cost of painting = 41.36 × 20 = ₹827.20
Problem 6
A sphere, a cylinder and a cone have the same radius and same height. Find the ratio of their volumes.
Solution:
Given: Same radius (r) and height (h), and for sphere: diameter = height ⇒ h = 2r
Volume of sphere = (4/3)πr³
Volume of cylinder = πr²h = πr²(2r) = 2πr³
Volume of cone = (1/3)πr²h = (1/3)πr²(2r) = (2/3)πr³
Ratio = Sphere : Cylinder : Cone = (4/3) : 2 : (2/3) = 4 : 6 : 2 = 2 : 3 : 1
Problem 7
A hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the side of the cube. Determine the total surface area of the remaining solid.
Solution:
Let side of cube = a ⇒ Radius of hemisphere = a/2
Total surface area of cube = 6a²
Area removed (circle) = π(a/2)² = πa²/4
Curved surface area added by hemisphere = 2π(a/2)² = πa²/2
Net change = -πa²/4 + πa²/2 = +πa²/4
Total surface area = 6a² + πa²/4 = a²(6 + π/4)
Problem 8
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its radius of the base is 3.5 cm, find the total surface area of the article.
Solution:
Given: Radius (r) = 3.5 cm, Cylinder height = 10 cm
Cylinder CSA = 2πrh = 2 × (22/7) × 3.5 × 10 = 220 cm²
Two hemispheres = 1 full sphere surface area = 4πr² = 4 × (22/7) × 12.25 = 154 cm²
Area of two circular tops removed = 2 × πr² = 2 × (22/7) × 12.25 = 77 cm²
Net surface area = Cylinder CSA + Sphere CSA – Removed circular areas = 220 + 154 – 77 = 297 cm²