S Chand ICSE Mathematics Class 9 Solution Eighteenth Chapter Surface Area and Volume of 3D Solids Exercise 18B
(1) Find the volume of a rail of uniform cross-section 12.8 cm2, and length 1.26 cm.
Solution: Volume of a rail of uniform cross section = area x length
= (12.8 x 1,26) cm3
= 16.128 cm3
(2) Find the area of the cross section, assuming it to be uniform, of a solid, given that its volume is 92.8 cm3, and length is 6.4 m.
Solution: ∴Volume of cross section = (Area of cross section) x length
∴ 92.8 = 640 x (area)
Or, area = 92.8 / 640
∴ Area = 0.145 cm2
(5) The cross section area of a pipe is 42 cm2, and water is pouring out of it at the rate of 1.25 m per sec . If the pipe remains full, find the number of litres discharged per minute.
Solution: Water is pouring out 1.25 m per sec
In 1 sec water is pouring 1.25 m
60 sec water is pouring = (1.25 x 60) m
= 75 m
If pipe is full water discharged = (75 x 42/10,000) m3
= 0.315 m3
∴ Volume of the solid = (area of cross section) x length
Area of cross section = (6 x 2) + (4 x 2) cm2
= 12 + 8 cm2
= 20 cm2
∴ Volume = 20 x 4 cm3
= 80 cm3
(7) a swimming pool is 50 m long and 15 m wide. Its shallow and deep ends are 1½ m and 4½ m deep respectively. If the bottom of the pool slopes uniformly, find the amount of water in litres required to fill the pool.
(8) The area of cross-section of a pipe is 5.4 cm2 and water is pumped out of it the rate of 27 km/h. Find in litres the volume of water which flows out of the pipe in one minute.
Solution: ∴ In 1 h water is pumped out = 27 km
= 27 x 1000 m
60 m water is pumped out 27000 m
1 m water is pumped out 27000 / 60 = 450 m
∴ Volume of water pumped out = 5.4/10000 x 450
= 0.243 m3
= 243 litre.