011-40705070  or
Select Board & Class
• Select Board
• Select Class
from SBOA SCHOOL & JR COLLEGE, asked a question
Subject: Math , asked on 12/2/14

# 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.

This conversation is already closed by Expert

• -1
View More

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.

• 1

• 0

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.

• 0

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

• -1