Page 248 - ICSE Math 8

Solution: Radius of bucket (r) = 14 cm
Let the rise in water level be h cm, then
Volume of cylindrical column of height h = Volume of rectangular solid

22
fi × 14 × 14 × h = 28 × 11 × 10
7
28 11 10 7× × ×
fi h = = 5
22 14 14× ×
Hence, the level of water rises by 5 cm.
Example 26: An iron cylindrical pipe has thickness of 0.5 cm and external diameter of 4.5 cm. If the mass
3
of 1 cm of the iron is 8 g, find the mass of a 77 cm long pipe.
Solution: External diameter = 4.5 cm
45.
\ External radius (R) = cm
2
Thickness = 0.5 cm

Ê 45 . ˆ Ê 45 1. - ˆ 35.
\ Internal radius (r) = Á Ë 2 - 05. ˜ cm = Á Ë 2 ˜ ¯ cm = 2 cm
¯
Height (h) = 77 cm
Volume of pipe = External volume – Internal volume
2
2
= ph(R – r ) = ph(R + r)(R – r)
22 Ê 45. 35. ˆ Ê 45. 35. ˆ 22 8 1
3
= × 77 × Á + ˜ Á - ˜ cm = × 77 × × cm 3
7 Ë 2 2 ¯ Ë 2 2 ¯ 7 2 2
1
3
3
= 22 × 11 × 4 × cm = 11 × 11 × 4 cm = 484 cm 3
2
3
Mass of 1 cm = 8 g
3
\ Mass of 484 cm = (8 × 484) g = 3,872 g = 3.872 kg
Hence, the mass of the iron cylinder is 3.872 kg.
Example 27: A 11 cm × 4 cm rectangular sheet of paper is folded and taped without overlapping to make a
cylinder of height 4 cm. Find the volume of the cylinder so formed.

Solution: 11 cm
11 cm

4 cm 4 cm

Let the radius of the cylinder be r cm.
Height of cylinder (h) = 4 cm
Circumference of base = 2pr = 11 cm

22 711¥ 7
fi 2 × × r = 11 cm fi r = = cm
7 222¥ 4

7
22 Ê ˆ 2 22 7 7
3
2
3
\ Volume of cylinder = pr h = ¥ Á ˜ ¥ 4 cm = × × × 4 cm = 38.5 cm 3
7 Ë 4¯ 7 4 4
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