Class nine NCERT Sol Maths Surface Areas Volumes
Chapter 13: Surface Areas and Volumes
TEXTBOOK’S EXERCISE – 13.6
1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm3 = 1l)
Sol. Here, h = 25 cm, 2πr = 132 cm.
2πr = 132 ⇒ 2 × 22/7 × r = 132
= 21 cm
Volume of the cylinder = πr2h
= 22/7 × 21 × 21 × 25 cm3 = 34650 cm3
= 34.65 litres
2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.
Sol. Here, inner radius (r) = 24/2 cm = 12 cm
Outer radius (R) = 28/2 cm = 14 cm, h = 35 cm
Volume of the wood used in the pipe
= π(R2 – r2) h
= 22/7 [(14)2 – (12)2] × 35 cm3
= 22/7 × 26 × 2 × 35 cm3 = 5720 cm3
Mass of 1 cm3 of wood = 0.6 g
∴ Mass of 5720 cm3 of wood
= 0.6 × 5720 g = 3432 g
= 3.432 kg.
3. A soft drink is available in two packs — (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?
Sol. For tin can with rectangular base
l = 5 cm, b = 4 cm, h = 15 cm
Volume of the tin can
= lbh = 5 × 4 × 15 cm3 = 300 cm3
For plastic cylinder with circular base
r = 7/2 cm = 3.5 cm, h = 10 cm
Volume of the plastic cylinder = πr2h
= 22/7 × 3.5 × 3.5 × 10 cm3 = 385 cm3
Difference in the capacities of the two containers = (385 – 300) cm3 = 85 cm3
Hence, the plastic cylinder with circular base has greater capacity by 85 cm3.