-2

A pipe can empty a tank in 40 min. A second pipe with diameter twice as much as that of the first is also attached with the tank to empty it. The two together can empty the tank in ?

I am attaching my solution also here but I want a good verification of it because in the textbook which I am following says the other way.

Solution :

Let the Diameter of the 1st pipe be $d_1$ and that of the other pipe be $d_2$, where $d_2 = 2d_1$. So, technically speaking the other pipe can transfer 4 times the amount transferred by the 1st pipe.

So, it would take 1/4 of the time taken by the 1st pipe.

Therefore, Pipe 1 takes 40 mins. Pipe 2 will take 10 mins only.

So, together they will take 8 mins to empty the tank.

Is it correct ?

Cyclone
  • 1,853
Shubham Raj
  • 93
  • How do you get to this conclusion ? –  Sep 11 '17 at 21:32
  • Pipes are assumed to be cylindrical. So, if we find out the volume of the 1st pipe to the 2nd pipe, it would come as 1/4.

    We have to use the data Diameter2 = 2x Diameter1

     – Shubham Raj Sep 11 '17 at 22:25
  • 2
    Its worth noting that laminar flow through a pipe goes as radius^4, so your larger pipe is actually able to discharge fluid at 16 times the rate of the smaller. I'll leave it to you to decide if the author of the question is aware of this, however. – Eddy Sep 11 '17 at 23:39
  • @ShubhamRaj: no, the conclusion the 8 follows from 40 and 10 ! –  Sep 12 '17 at 06:53
  • @Eddy: I'll leave it to you to decide if the tank is containing superfluid helium ;-) –  Sep 12 '17 at 06:55
  • 1
    This guy might be an engineering student, in which case he should assume pipe flow. Indeed, if this question was asked as part of a course on superfluids, he should assume its a superfluid. We don't know, so I thought I'd tell. – Eddy Sep 12 '17 at 07:48
  • yes i am an engineering student folks..!!

    And, according to me, I should spare the " SuperFluid " concept and simply assume the pipe to be cylindrical. And, that's what should be the best thing to do according to my textbook.

     – Shubham Raj Sep 12 '17 at 15:03
  • @YvesDaoust yes sir.. I have done exactly what you wrote in the comment.

    Pipe1 = 40 min Pipe2 = 10 min

    So, together 1/ ( 1/40 + 1/10 ) = 8 min

     – Shubham Raj Sep 12 '17 at 15:05
  • @Eddy Sir, I guess I should be solving this simply as a arithmetic maths question.

    Anyways, thanks for telling other insights..

     – Shubham Raj Sep 12 '17 at 15:08
  • Anyways, the answer in my textbook is given as 13.33 min.

    I am assuming that answer to be wrong then.

     – Shubham Raj Sep 12 '17 at 15:11
  • Also, the question doesn't specify that both pipes are attached to the tank at the same height. If the second pipe is not attached at the bottom of the tank, then while the emptying time will be less than 40 mins but depends on the height of the pipe above the bottom of the tank. – Cyclone Sep 12 '17 at 21:33

2 Answers2

1

The section of the two pipes together is five times larger than the first, hence $\dfrac{40}5$ minutes to empty.

0

The "formula" to use here is $W = RT$ where $W$ = work, $R$ = rate in work per minute, and $T$ = time in minutes. The tank is empty when $W=1$.

For the first pipe, $W_1 = 1$ when $T_1 = 40$. So $R_1 = \dfrac{1}{40}$.

For the second pipe, $R_2 = 4R_1 = \dfrac{4}{40}$

The combined rate is $R_{together}=R_1 + R_2 = \dfrac{5}{40} = \dfrac 18$.

So, $T_{together}=8$ minutes.

Steven Alexis Gregory
  • 26,769