Surface Areas and Volumes — Practice

Question 1

Find the volume of a sphere of radius 3 cm (π = 22/7).

Accepted Answer

Step 1 — V = (4/3)πr³ = (4/3)(22/7)(27) = (4/3)(594/7) = 792/7 ≈ 113.1 cm³.

✦ Answer: ≈ 113.1 cm³.

Question 2

A cone has radius 5 cm and height 12 cm. Find its slant height.

Accepted Answer

Step 1 — l = √(5² + 12²) = √169 = 13 cm.

✦ Answer: 13 cm.

Question 3

Find the total surface area of a hemisphere of radius 7 cm (π = 22/7).

Accepted Answer

Step 1 — TSA = 3πr² = 3 × 22/7 × 49 = 462 cm².

✦ Answer: 462 cm².

Question 4

A toy is a cone of radius 3.5 cm mounted on a hemisphere of the same radius. If the cone's slant height is 6 cm, find the total surface area (π = 22/7).

Accepted Answer

Step 1 — Exposed surfaces = cone CSA + hemisphere CSA.

Step 2 — πrl + 2πr² = 22/7 × 3.5 × 6 + 2 × 22/7 × 3.5² = 66 + 77 = 143 cm².

✦ Answer: 143 cm².

Question 5

A capsule is a cylinder of length 5 cm (radius 0.5 cm) with a hemisphere on each end. Find its total length and volume (π = 22/7).

Accepted Answer

Step 1 — Total length = 5 + 0.5 + 0.5 = 6 cm.

Step 2 — V = πr²h + 2·(2/3)πr³ = 22/7×0.25×5 + (4/3)(22/7)(0.125) ≈ 3.93 + 0.52 = 4.45 cm³.

✦ Answer: length 6 cm, volume ≈ 4.45 cm³.

Question 6

A metallic sphere of radius 6 cm is melted and drawn into a wire of radius 0.2 cm. Find the length of the wire (π = 22/7).

Accepted Answer

Step 1 — Conserve volume: (4/3)π(6³) = π(0.2)²·L.

Step 2 — (4/3)(216) = 0.04·L ⇒ 288 = 0.04L.

Step 3 — L = 7200 cm = 72 m.

✦ Answer: 72 m.

Question 7

How many spherical lead shots of radius 1 cm can be made from a cuboid of lead 8 cm × 6 cm × 4 cm? (π = 22/7)

Accepted Answer

Step 1 — Cuboid volume = 8×6×4 = 192 cm³.

Step 2 — One shot = (4/3)π(1³) = 88/21 ≈ 4.19 cm³.

Step 3 — Number = 192 ÷ (88/21) = 192 × 21/88 = 45.8 ⇒ 45 whole shots.

✦ Answer: 45 shots.

Question 8

A bucket (frustum) has radii 20 cm and 10 cm and height 24 cm. Find its volume (π = 3.14).

Accepted Answer

Step 1 — V = ⅓πh(R² + r² + Rr) = ⅓ × 3.14 × 24 × (400 + 100 + 200).

Step 2 — = ⅓ × 3.14 × 24 × 700 = 3.14 × 8 × 700 = 17584 cm³.

✦ Answer: 17584 cm³ ≈ 17.58 litres.

Question 9

A well of diameter 3 m is dug 14 m deep. The earth is spread evenly to form a 4 m wide embankment around it. Find the embankment's height (π = 22/7).

Accepted Answer

Step 1 — Well (cylinder) volume: r = 1.5, V = 22/7 × 1.5² × 14 = 99 m³.

Step 2 — Embankment is a ring: inner r = 1.5, outer R = 1.5+4 = 5.5; area = π(R²−r²) = 22/7(30.25−2.25) = 88 m².

Step 3 — Height = Volume/Area = 99/88 = 1.125 m.

✦ Answer: 1.125 m.

Question 10

A solid cylinder of radius 6 cm and height 15 cm is melted and recast into cones of radius 2 cm and height 3 cm. How many cones are formed? (π cancels)

Accepted Answer

Step 1 — Cylinder V = πr²h = π×36×15 = 540π.

Step 2 — Cone V = ⅓πr²h = ⅓π×4×3 = 4π.

Step 3 — Number = 540π / 4π = 135.

✦ Answer: 135 cones.

Practice — Surface Areas and Volumes

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