Math, asked by reply2gcarnav2004, 1 year ago

Water is flowing through a cylindrical pipe of Intemal diameter 2cm, into a cylindrical tank of base radius
40cm, at the rate of 0.4m/sec Determine the rise in level of water in the tank in half an hour.

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Answered by Anonymous
7

your answer:-

radius of cylindrical pipe , r = 1cm

radius of cylindrical tank , R = 40cm

water is flowing through a cylinderical pipe into cylindrical tank at the rate of 0.4m/s or 40cm/s.

length of water through pipe , l= rate × time taken

= 40cm/s × 1/2 hr

= 40cm/s × 30 × 60

= 40 × 1800 cm

= 72000 cm

now, volume of water flows through pipe = volume of water rise in cylindrical tank

or, πr² × l = πR² × h

or, (1cm)² × 72000 cm = (40cm)² × h

or, 72000cm³ = 1600 cm² × h

or, h = 720/16 cm = 45cm

hence, 45cm rise in the level of water in the rank in half an hour.

Answered by ʙʀᴀɪɴʟʏᴡɪᴛᴄh
4

Answer:

Diameter of circular end of pipe = 2 cm

of circular end of pipe =

Area of cross-section =

Speed of water = 0.4 m/s = 0.4 60 = 24 metre/min

Volume of water that flows in 1 minute from pipe =

Volume of water that flows in 30 minutes from pipe = 30

Radius (r2) of base of cylindrical tank = 40 cm = 0.4 m

Let the cylindrical tank be filled up to h m in 30 minutes.

Volume of water filled in tank in 30 minutes is equal to the volume of water flowed out in 30 minutes from the pipe.

Therefore, the rise in level of water in the tank in half an hour is 45 cm.

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