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

Asked by Nivedha(student) , on 12/2/14


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Pls I want the answer 4 this question fast....!!!

Posted by Nivedha(student)

Given, internal diameter of cylindrical pipe = 2 cm

So, radius of cylindrical pipe (r) = 2/ 2 = 1 cm

And,

Radius of cylindrical tank (R) = 40 cm

Let level of water rise to the height of h cm.

Thus,

Now,

Volume of cylindrical tank = Volume of water flown out in half an hour

Hence, level of water rise to the height of 45 cm.

Posted by Cindrella Kumar...(student)

Given, internal diameter of cylindrical pipe = 2 cm

So, radius of cylindrical pipe (r) = 2/ 2 = 1 cm

And,

Radius of cylindrical tank (R) = 40 cm

Let level of water rise to the height of h cm.

Thus,

Now,

Volume of cylindrical tank = Volume of water flown out in half an hour

Hence, level of water rise to the height of 45 cm.

Posted by Cindrella Kumar...(student)

this is the correct answer

Given, internal diameter of cylindrical pipe = 2 cm

So, radius of cylindrical pipe (r) = 2/ 2 = 1 cm

Area of cross section of pipe= r2= (1)2= cm2

Speed of water= 0.4 m/s =0.4 x 100 cm/s = 40 cm/s

Thus,

Volume of water flown out in half an hour = x 40 x 30 x 60 = 72000 cm2

And,

Radius of cylindrical tank (R) = 40 cm

Let level of water rise to the height of h cm.

Thus,

Volume of cylindrical tank = R2h = (40)2h = 1600h cm2.

Now,

Volume of cylindrical tank = Volume of water flown out in half an hour

1600h = 7200

h = 45cm

Hence, level of water rise to the height of 45 cm.

Hope it helps

Thankyou

Posted by Cindrella Kumar...(student)

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