Question

Water is flowing at the rate of 15 km/ hr through a pipe of diamerer 14cm into a cuboidal pond which is 50m long and 44m wide . In what time will the level of water in pond rise by 21 cm ?


Answer should be well explained please...and no irrelevant answers...

Answer 1

volume of water to be filled in pond=50*44*21= 46200
area of cross section of water=π*r²=22/7*49 =154
volume of water coming through pipe in 1 sec =15*154=2310
time required to fill the pond =46200/2310 =20sec

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Answer 2

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Given:

Speed of water (h) = 15km/hr

= 15 x 1000m

= 15000 m.

Diameter of pipe = 14 cm

So,we know that,

radius = d/2

radius = 14/2

radius = 7 cm.

convert into meter,

radius = (7/100) m because we know that

1 m = 100cm :

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Length of the pond = 50m

breadth of the pond = 44m

height of the pond = 21 cm
convert into meter,

height of the pond = (21/100) m


We have to find the time in which the water wil rise by 21 cm.


Suppose that,water in the pond rises by 21 cm in ' x ' hours.


We know that pipe is form of cylinder and we also know that,

$volume \: of \: cylinder \: = \pi \: r {}^{2} h$

First we will calculate the Volume of water flowing out of the pipe in 1 hour.

Now put the given values we get,

$volume \: = \frac{22}{7} \times ( \frac{7}{100} ) \times ( \frac{7}{100} ) \times 15000 \\ \\ volume \: of \: pipe = 231 \: m {}^{3}$
And the volume of water which is flowing out in the pipe in x hours wil be = 231x m^3


Now we will find the volume of water in the cuboidal pond,

We know that,

Volume of cuboid = l x b x h

Volume of cuboidal pond =
$55 \times 44 \times \frac{21}{100} \\ \\ = 462 \: m {}^{3}$

By the given statements we can understand that,

volume of the water which is flowing out of the pipe in ' x ' hours = Volume of water which is flowing in the cuboidal pond

So,by applying this statement we get,

231x = 462

x = 462/231

x = 2 hours.

Therefore, the level of the water in the pond will rise by 21 cm in 2 hours.


I hope this will helps you......☺☺☺⭐

Thanks....


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