S Chand ICSE Mathematics Class 9 Solution Eighteenth Chapter Surface Area and Volume of 3D Solids Exercise 18B (O.P. Malhotra, S.K. Gupta, Anubhuti Gangal)

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

= 0.145

∴ 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 Solution:

∴ 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.

Solution: (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.    Updated: September 27, 2019 — 12:46 pm

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