**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 cm ^{2}, and length 1.26 cm.**

**Solution:** Volume of a rail of uniform cross section = area x length

= (12.8 x 1,26) cm^{3}

= 16.128 cm^{3}

**(2) Find the area of the cross section, assuming it to be uniform, of a solid, given that its volume is 92.8 cm ^{3}, 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 cm^{2}

**(5) The cross section area of a pipe is 42 cm ^{2}, 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) m^{3}

= 0.315 m^{3}

**Solution: **

∴ Volume of the solid = (area of cross section) x length

Area of cross section = (6 x 2) + (4 x 2) cm^{2}

= 12 + 8 cm^{2}

= 20 cm^{2}

∴ Volume = 20 x 4 cm^{3}

= 80 cm^{3}

**(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 cm ^{2} 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 m^{3}

= 243 litre.

9 question is incorrect I have done it if you want send me feedback on my email

Send us Varun at comment box