Question

Q. WATER FLOWS OUT THROUGH A CIRCULAR PIPE WHOSE INTERNAL
radius is 1 cm at the rate of 80 cm/sec into an empty cylindrical tank , the radius of whose base is 40 cm . By how much will the level of water rise in the tank in half an hour ?

Answer 1

aloha!

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let the radius of pipe be represented by R 
and the height be represented by H 

and radius of cylindrical tank be r
height be h

now, 

we know that area of circle is $\pi r^{2}$

now,

$\frac{22}{7} X 1 X 1 = \frac{22}{7} cm^{2}$ ---------(1)

it is given that water flows through the pipe at the rate of 80 cm/sec.

now we should multiply (1) by 80.


$\frac{22}{7} X 80 = \frac{1760}{7}$$cm^{3}$ { it would become the volume }

 in 1 sec this much water flows through the pipe =$\frac{1760}{7}$  cm3
 
in one minute it would be= $\frac{1760}{7}$  cm3 X 60
                                           = 15084 cm3 { approx. }

now, for 30 minutes it would be = 15084 X 30 cm3


Radius of the base of the cylindrical tank = 40 cm
Let the rise in level of the water in 30 min = h cm


Volume of the water filled in 30 min = πr2h x 30  
                                                             =  (22/7) x (40)2 x h x 30 cm3

(22/7) x (40)2 x h x 30 = 15084 x 30
 ⇒ h = 3cm  { approx. }


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hope it helps :)