Class 9 Maths Chapter 11 HOTS Questions - Surface Areas and Volumes

Q1. A brick measures 30 cm × 10 cm ×  cm. How many bricks will be required for a wall 30 m long, 2 m high and (3/4)m thick?

Sol: Volume of wall (cuboid) = 30 × 2 × (3/4) m³ = 45 m³
Volume of 1 brick =

∴ Required number of bricks =

Q2. The surface area of a cube is m². Find its volume.
 Sol: Total S.A. = 6(side)² ⇒ side² =  ÷ 6

∴ Volume = (side)³  = 

Q3. A hemispherical tank is emptied by a pipe at the rate of 5 litres per minute. How long will it take to half-empty the tank if it is  a metre in diameter? [Take π = (22/7) and 1l = 1000 cm]³

Sol: Radius

Volume of the hemisphere = (1/2) × (volume of a sphere) =

Volume to be emptied 

∵ 5 litres are emptied in 1 minute

Q4. A cube and a cuboid have the same volume. The dimensions of the cuboid are 12 cm × 9 cm × 6 cm. Find the difference between the total surface areas of the cube and the cuboid.

Sol: Volume of cuboid = 12 × 9 × 6 = 648 cm³
Since the cube has the same volume:
Let side of cube = a 
⇒ a³ = 648 
⇒ a = ∛648 = 8.64 cm (approx)

Total Surface Area of cuboid = 2(lb + bh + hl)
= 2(12 × 9 + 9 × 6 + 6 × 12) = 2(108 + 54 + 72) = 2 × 234 = 468 cm²

Total Surface Area of cube = 6a² = 6 × (8.64)² = 6 × 74.65 = 447.9 cm² (approx)

Difference = 468 – 447.9 = 20.1 cm²

Q5. A cone and a hemisphere have equal radii. The curved surface area of the cone is the same as that of the hemisphere. Find the ratio of the slant height of the cone to its radius.

Sol: Let radius = r, slant height of cone = l
Curved Surface Area of cone = πrl
Curved Surface Area of a hemisphere = 2πr²

Equating:
πrl = 2πr² ⇒ l = 2r

∴ Ratio (l : r) = 2:1

Q6. A spherical ball is dropped in a cylindrical container partially filled with water. The water level rises by 2.5 cm. If the radius of the cylinder is 5 cm, find the radius of the ball.

Sol: Let the radius of the ball = r
Volume of water displaced = Volume of sphere = (4/3)πr³
Volume of water rise = πR²h = π (5²) (2.5) = 62.5 π cm³

Equating:

(4/3) πr³ = 62.5 π
⇒ r³ = (62.5 × 3)/4 = 46.875
⇒ r ≈ ∛46.875 = ~3.57 cm

Q7. The diameter of a sphere is decreased by 50%. Find the percentage decrease in its surface area.

Sol: Let original radius = r, new radius = r/2

Original Surface Area = 4πr²
New Surface Area = 4π(r/2)² = 4πr²/4 = πr²

Decrease in area = 4πr² – πr² = 3πr²

% Decrease = (3πr² / 4πr²) × 100 = 75%

Q8. A wooden toy is made by mounting a hemisphere on a cone. The height of the cone is 9 cm, and the radius of both hemispheres and the cone is 3.5 cm. Find the total surface area of the toy (excluding base).

Sol: Slant height of cone = √(9² + 3.5²) = √(81 + 12.25) = √93.25 ≈ 9.65 cm

Curved Surface Area of cone = πrl = π × 3.5 × 9.65 ≈ 106.01 cm²

Curved Surface Area of hemisphere = 2πr² = 2π × (3.5)² = 2π × 12.25 ≈ 76.96 cm²

Total surface area = 106.01 + 76.96 = 182.97 cm²