Surface Areas & Volumes — Chapter Test | Class 9 IMO

Q1

A heap of wheat is in the form of a cone of diameter 10.5 m and height 3 m. Find the area of the canvas required to just cover the heap. (Use π = 22/7)

Radius r = 10.5/2 = 5.25 m and height h = 3 m. Slant height l = √(r² + h²) = √(27.5625 + 9) = √36.5625 ≈ 6.05 m. Canvas needed = curved surface area = πrl = (22/7) × 5.25 × 6.05 ≈ 99.83 m².

Q2

If the length of the diagonal of a cube is 8√3 cm, then its surface area is:

Diagonal of a cube = a√3. So, a√3 = 8√3 → a = 8 cm. TSA = 6a² = 6 × 8² = 6 × 64 = 384 cm².

Q3

If a right circular cone has radius 5 cm and slant height 13 cm, what is its volume?

Radius r = 5, slant height l = 13. Height h = √(l² - r²) = √(169 - 25) = √144 = 12 cm. Volume = (1/3)πr²h = (1/3)π(25)(12) = 100π cm³.

Q4

Three metal cubes of edges 3 cm, 4 cm and 5 cm are melted and cast into a single cube. The edge of the new cube is:

Volume of new cube = 3³ + 4³ + 5³ = 27 + 64 + 125 = 216 cm³. Since V = a³, a = ∛216 = 6 cm.

Q5

A capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Radius r = 2.5 mm. Length of cylindrical part = Total length - 2(radius) = 14 - 2(2.5) = 14 - 5 = 9 mm. Total Surface Area = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 2(2πr²) = 2πr(h + 2r) = 2 × (22/7) × 2.5 × (9 + 5) = 2 × (22/7) × 2.5 × 14 = 220 mm².

Q6

If a right circular cylinder and a right circular cone have the same radius and same volume, then the ratio of the height of the cylinder to the height of the cone is:

V_cylinder = πr²h₁. V_cone = (1/3)πr²h₂. Since they are equal: πr²h₁ = (1/3)πr²h₂ → h₁ = (1/3)h₂ → h₁/h₂ = 1/3. Ratio is 1:3.

Q7

The lateral surface area of a cylinder of radius 7 cm and height 10 cm is (take π = 22/7):

CSA of cylinder = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm².

Q8

The length, breadth, and height of a cuboid are 15 cm, 10 cm, and 20 cm respectively. The total surface area of the cuboid is:

TSA of cuboid = 2(lb + bh + hl) = 2(15×10 + 10×20 + 20×15) = 2(150 + 200 + 300) = 2(650) = 1300 cm².

Q9

The ratio of the volumes of two cones is 4:5 and the ratio of the radii of their bases is 2:3. Find the ratio of their vertical heights.

V₁/V₂ = [(1/3)πr₁²h₁] / [(1/3)πr₂²h₂] = (r₁/r₂)²(h₁/h₂). 4/5 = (2/3)² × (h₁/h₂) = 4/9 × (h₁/h₂). Therefore, h₁/h₂ = (4/5) × (9/4) = 9/5. Ratio is 9:5.

Q10

The total surface area of a solid cylinder of radius 14 cm and height 20 cm is:

TSA = 2πr(h + r) = 2 × (22/7) × 14 × (20 + 14) = 88 × 34 = 2992 cm².

Q11

The cost of painting the total outside surface of a closed cylindrical oil tank at ₹60 per sq. m is ₹237.60. The height of the tank is 6 times the radius of the base of the tank. Find its volume.

Total surface area = cost ÷ rate = 237.60 ÷ 60 = 3.96 m². With h = 6r, TSA = 2πr(r + h) = 2πr(7r) = 14πr² = 44r², so 44r² = 3.96, r² = 0.09 and r = 0.3 m (h = 1.8 m). Volume = πr²h = (22/7) × 0.09 × 1.8 ≈ 0.509 m³.

Q12

Anita is making a conical tent for a village fair in Gujarat. The tent is 10 m high and the radius of its base is 24 m. Find the cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹70.

h = 10 m, r = 24 m. l = √(24² + 10²) = √(576 + 100) = √676 = 26 m. CSA = πrl = (22/7) × 24 × 26. Area ≈ 1961.14 m². Cost = (22/7) × 24 × 26 × 70 = 22 × 24 × 26 × 10 = ₹1,37,280.

Q13

If a sphere and a cube have the same surface area, then the ratio of the volume of the sphere to the volume of the cube is:

Surface areas equal: 4πr² = 6a² → a² = (4π/6)r² = (2π/3)r² → a = r√(2π/3). Ratio of volumes = (4/3)πr³ / a³ = (4/3)πr³ / (r³(2π/3)√(2π/3)) = (4/3)π / ((2π/3)√(2π/3)) = 2 / √(2π/3) = √6 / √π = √6 : √π.

Q14

Two cubes each of volume 64 cm³ are joined end to end. The surface area of the resulting cuboid is:

Volume of one cube = 64 cm³ → side a = 4 cm. When joined, length of cuboid l = 4+4 = 8 cm, breadth b = 4 cm, height h = 4 cm. TSA = 2(lb + bh + hl) = 2(32 + 16 + 32) = 2(80) = 160 cm².

Q15

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. The outer curved surface area of the bowl is:

Inner radius r = 5 cm, thickness = 0.25 cm. Outer radius R = 5 + 0.25 = 5.25 cm. Outer CSA = 2πR² = 2 × (22/7) × 5.25 × 5.25 = 2 × 22 × 0.75 × 5.25 = 173.25 cm².