A circular pipe is to be designed in such a way that water flowing through it at a velocity of 7 m per min. is collected at its open end at the rate of 11 cubic metre per min., what should be the inner radius of the pipe

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13 Mar 2015 Volume and surface area A circular pipe is to be designed in such a way that water flowing through it at a velocity of 7 m per min. is collected at its open end at the rate of 11 cubic metre per min., what should be the inner radius of the pipe: OPtion
1) Sqrt(2) m
2) 2 m
3) 1/Sqrt(2) m
4) 1/2 m
5) 1/4 m
6) 1/Sqrt(3)
7) 1/3
8) 1/Sqrt(5)
9) 5 m
10)1/5 m; Correct Option: 3 Best Solution (41)

Surit Datta 9 years AGO

We know
pi*r*r*h is the area volume of
circular pipe
so,
pi*r*r*h = 11
here h = 7
so,
pi*r*r*7 = 11
solving this we get,
r = 1/sqrt(2)
so option 3
that's it.

Like? Yes (29) | No (5)

Sarabeshwar 9 years AGO

Formula to be used, pi* r^2 *h

(22/7)*r^2 * 7 = 11

r^2 = 1/2

r =1/sqrt(2)

Like? Yes (3) | No

Supriya Mondal 9 years AGO

Let, inner radius of the pipe is r metre
Then, according the question we can write,
π×r^2×7=11
So, r=1/√2
So, option 3)1/√2 is the right answer

Like? Yes (3) | No

Jaspalsingh Parmar 9 years AGO

Formula : pi * radius * radius * height
R^2 * 22/7*7 = 11
R^2 = 11/22 = 1/2
R = 1/ (sqrt(2)) answer.

Like? Yes (3) | No

sheikh shahrukh 9 years AGO

just see the units....
11 m^3/minute=7m/minute * m^2 ( to balance m^3/minute on both sides we multiplied m^2 and m^2 is area so...)
11/7=area of pipe
11/7=22/7 * radius^2
on solving
radius^2=1/2
so radius= 1/sqrt(2 meter)........
answer is option 3 thats it

Like? Yes (3) | No

JANARTHANAN RAM 9 years AGO

volume of cylinder==pi *(r^2) * h
r^2 * 22/7*7 = 11
r^2 = 11/22 = 1/2
r = 1/ (sqrt(2))

Like? Yes (3) | No

nitin kishore 9 years AGO

Volume of water flowing through the pipe per minute= pi*(radius)^2 * velocity = 11
so,
radius = 1/sqrt(2) m

Like? Yes (2) | No

Jay parmar 9 years AGO

Formula : pi * radius * radius * height
R^2 * 22/7*7 = 11
R^2 = 11/22 = 1/2
R = 1/ (sqrt(2)) answer.

Like? Yes (3) | No (1)

Samyak 9 years AGO

surfacre * velocity = rate

pi*r2 *7 = 11
therefor, r = 0.707

So 1/sqrt(2) = 0.707

Like? Yes (2) | No

ABHINAV PUSHPAM 9 years AGO

22/7.r^2.h=11
R^2*22/7*7=11
R^2=11/22
R= 1/sqrt(2)
Option 3)

Like? Yes (2) | No

Meenakshi 9 years AGO

volume=pi*r*r*h
given,h=7 m per min
volume=11 cubic metre per min
11=22/7*r^2*7
11=22*r^2
11/22=r^2
1/2=r^2
1/sqrt(2)=r

Like? Yes (2) | No

tasho k 9 years AGO

Pi*r^2*h=11
22/7*r*r*7=11
So r=1/sqrt (2

Like? Yes (2) | No

NANGINEDI RAVITEJA 9 years AGO

area*7=11 so opt 3

Like? Yes (2) | No

J.DHANDAPANI 9 years AGO

(pi) r^2 (7) = 11 =>r=1/Sqrt(2) m.

Like? Yes (2) | No

MUKKA ARUN KUMAR 9 years AGO

flow rate=(1/4)*(22/7)*(pipe diameter)*(pipe diameter)*(velocity)

Like? Yes (2) | No

mandeep 9 years AGO

Area*7=11

Like? Yes (1) | No

RAKESH 9 years AGO

22/7 * r^2 * 7 = 11

r = 1/Sqrt(2) m

Like? Yes (1) | No

Anshul Garg 9 years AGO

7*pi*r^2 = 11

Solving for r we get r =1/sqrt(2)

Like? Yes (1) | No

KALYAN KUMAR REDDY VALLURU 9 years AGO

1/sqrt(2)

Like? Yes (1) | No

arush gupta 9 years AGO

ans 1/Sqrt(2)

Like? Yes (1) | No

Sachit Kwatra 9 years AGO

PIE*r^2=VOlume/speed=Area
22/7*r^2=11/7
r=1/sqrt(2)

Like? Yes (1) | No

KUMAR HARSH 9 years AGO

h=7,r=?
22/7.r^2.h=11
R^2*22/7*7=11
R^2=11/22
R= 1/sqrt(2)

Like? Yes (1) | No

darshana kumrawat 9 years AGO

let length of pipe = l
radius of pipe = r
volume of pipe = pi*r^2*l
water flowing out in 1 min = (pi*r^2*l)*(7/l) = pi*r^2*7
pi*r^2*7=11
therefore, r = 1/sqrt(2)

Like? Yes (1) | No

Math Lovers 9 years AGO

Circular pipe is a Cylinder.
Outflow/minute = Volume of cylinder/minute
Volume of cylinder = pi * radius * radius * height

R^2 * 22/7*7 = 11

R^2 = 11/22 = 1/2

R = 1/ (sqrt(2)) answer.

Like? Yes (1) | No

praveen kumar 9 years AGO