Page 297 - Start Up Mathematics_8 (Non CCE)

8 316,
3
3
fi 308x = 8,316 fi x = = 27
308
fi x = 27 = 3
3
\ Radius (r) = 7 × 3 = 21 cm, Height (h) = 2 × 3 = 6 cm
22
\ Total surface area = 2pr(h + r) = 2 × × 21 × (21 + 6) cm 2
7
22
2
= 2 × × 21 × 27 cm = 3,564 cm 2
7
Example 36: A cylindrical bucket, 14 cm in radius, is filled with water to some height. If a rectangular
solid of size 28 cm × 11 cm × 10 cm is immersed in water, find the height by which water
rises in the bucket.

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 37: 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 metal 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 - 05. ˜ cm = Á Ë ˜ ¯ cm = 2 cm
¯
Height (h) = 77 cm 2 2
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
= × 77 × Á + ˜ Á - ˜ cm 3
7 Ë 2 2 ¯ Ë 2 2 ¯
22 8 1
= × 77 × × cm 3
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.

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