Page 296 - Start Up Mathematics_8 (Non CCE)

Example 32: A milk tank is in the form of a cylinder whose radius is 1.5 m and height is 7 m. Find the
quantity of milk in litres that can be stored in the tank. (NCERT)
Solution: Radius of milk tank (r) = 1.5 m, Height of milk tank (h) = 7 m
}
2
3
15 ¥)
Volume of milk tank = pr h = { 22 ¥ (. 2 7 m = Ê 22 ¥ . ¥ . ¥ ˆ 3 3
15 15 7 m = 49.5 m
Á
˜
¯
Ë
3
Now 1 m = 1,000 L 7 7
3
fi 49.5 m = (49.5 × 1,000) L = 49,500 L
Hence, 49,500 litres of milk can be stored in the milk tank.
Example 33: The circumference of the base of a cylinder is 264 cm and its height is 24 cm. Find the volume
of the cylinder.
Solution: Circumference of base = 264 cm
22
fi 2pr = 264 fi 2 × × r = 264
7
264 7¥
fi r = = 42 cm
222¥
Height of cylinder (h) = 24 cm

3
2
\ Volume of cylinder = pr h = 22 × 42 × 42 × 24 cm = 1,33,056 cm 3
7
Example 34: Two cylindrical cans have equal height. If the radius of one is two-third of the other and it
holds 20 litres of liquid, find the capacity of the second can. (Height and radius are in metres.)

Solution: Let the radius of the second can (r ) be (r) m.
2
Ê 2 ˆ
r m
Then, the radius of the first can (r ) = Á Ë 3 ¯ ˜
1
Let the height of the two cans be (h) m.

20
3
Volume of the first can (V ) = 20 litres = 1 000, m 3 (  1 m = 1,000 L)
1
20 Ê 2 ˆ 2 20
2
3
fi pr h = 1 000, m fi p × Á Ë 3 ¯ ˜ × h = 1 000, m 3
r
1
4 20 20 ¥ 9
2
2
3
fi p × r h = m fi p × r h = m 3
9 1 000, 1 000 ¥, 4
2
2
3
Volume of the second can (V ) = pr h = pr h = 1 000 ¥, 20 ¥ 9 4 m = 45 L
2
2
\ The capacity of the second tank = 45 L
3
This question can also be done without changing 20 L into m .
Example 35: The radius and height of a cylinder are in the ratio 7 : 2. If the volume of the cylinder is
3
8,316 cm , find its total surface area.
Solution: Let the radius of the cylinder (r) be 7x cm and the height of the cylinder (h) be 2x cm.
2
Volume of cylinder = pr h
22 22
2
2
3
8,316 = × (7x) × (2x) cm fi × 49x × 2x = 8,316
7 7
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