Math, asked by napajayashree, 1 year ago

Water flows through a cylindrical pipe whose inner diameter is 1cm at the rate of 80cm per second in an cylindrical tank , the radius of whose base is 40cm. What is the rise of water level in tank in half an hour??.

Answers

Answered by pranavawasarmol123
9

Answer:

Step-by-step explanation:

Given, radius of tank, r1 = 40 cm

Let height of water level in tank in half an hour = 1 cm.

Also, given internal radius of cylindrical pipe, r2 = 1 cm

and speed of water = 80 cm/s i.e., in 1 water flow = 80 cm

In 30 (min) water flow = 80x 60 x 30 = 144000 cm According to the question,

Hence, the level of water in cylindrical tank rises 90 cm in half an hour.

Answered by rahulthakur1722006
4

Answer:

Inner radius of cylindrical pipe=1cm

Speed of water = 80cm/s

It means,in one sec water flow for a distance of 80 cm

Therefore,water flow in 30min

30×60×80=144000cm

Now, volume of water in cylindrical tank= volume of water flow in the pipe in half an hour.

πr^2h= π^2h2

r^2h1=r^2h2

40×40×h1=1×1×144000

h1= 144000/1600

= 90 cm

Hence, the level of water in the cylindrical tank rises 90 cm in half an hour.

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