The problem of shallow water tank

[kan]

A cylindrical water tank, [MATH]5m[/MATH] radius, [MATH]20m[/MATH] height is being drained at the bottom of the tank. The amount of water released with average speed [MATH]0.5\sqrt{h}[/MATH] (h is the height of the water tank). Ask after how long the tank will run out?

Some answers are given:

a) 20 hours

b) 620 minutes

c) 400.862 minutes

d) 1404.962 minutes

My method is as follows: [MATH](\pi∗5^{2}∗20)/(0.5\sqrt{20})=702.481[/MATH] minutes

I do not know where was wrong. Looking forward to your help. Thanks!

 

[lev888]

A cylindrical water tank, [MATH]5m[/MATH] radius, [MATH]20m[/MATH] height is being drained at the bottom of the tank. The amount of water released with average speed [MATH]0.5\sqrt{h}[/MATH] (h is the height of the water tank). Ask after how long the tank will run out?

Some answers are given:

a) 20 hours

b) 620 minutes

c) 400.862 minutes

d) 1404.962 minutes

My method is as follows: [MATH](\pi∗5^{2}∗20)/(0.5\sqrt{20})=702.481[/MATH] minutes

I do not know where was wrong. Looking forward to your help. Thanks!

Is h height of the tank or current water level?
What are the units of speed?

 

[Steven G]

A cylindrical water tank, [MATH]5m[/MATH] radius, [MATH]20m[/MATH] height is being drained at the bottom of the tank. The amount of water released with average speed [MATH]0.5\sqrt{h}[/MATH] (h is the height of the water tank). Ask after how long the tank will run out?

Some answers are given:

a) 20 hours

b) 620 minutes

c) 400.862 minutes

d) 1404.962 minutes

My method is as follows: [MATH](\pi∗5^{2}∗20)/(0.5\sqrt{20})=702.481[/MATH] minutes

I do not know where was wrong. Looking forward to your help. Thanks!

I have one question for you? Why minutes? Why not hours or seconds?

If this was a test problem I would be happy to see it since it can't be done with the information given.

 

[Deleted member 4993]

A cylindrical water tank, [MATH]5m[/MATH] radius, [MATH]20m[/MATH] height is being drained at the bottom of the tank. The amount of water released with average speed [MATH]0.5\sqrt{h}[/MATH] (h is the height of the water tank). Ask after how long the tank will run out?

Some answers are given:

a) 20 hours

b) 620 minutes

c) 400.862 minutes

d) 1404.962 minutes

My method is as follows: [MATH](\pi∗5^{2}∗20)/(0.5\sqrt{20})=702.481[/MATH] minutes

I do not know where was wrong. Looking forward to your help. Thanks!

Aside from being ill-defined problem (as explained above), you have to remember that 'h' is a function of time. Hence the "The amount of water released with average speed", is a function of time. It is not constant. You have to use calculus to solve it.

 

[Steven G]

Aside from being ill-defined problem (as explained above), you have to remember that 'h' is a function of time. Hence the "The amount of water released with average speed", is a function of time. It is not constant. You have to use calculus to solve it.

I thought the same thing but it clearly said that wrt sqrt(h) that h is the height of the tank. So who is correct? Also note (not that it means anything) but this problem was posted in the algebra section, not calculus.

 

[Deleted member 4993]

I thought the same thing but it clearly said that wrt sqrt(h) that h is the height of the tank. So who is correct? Also note (not that it means anything) but this problem was posted in the algebra section, not calculus.

I think this is a translated problem.

I think, if we "understood" the original problem, it would have said "h is the height of the water-level inside the tank".