# 1994 Physics Olympiad Semi-Final Exam Question

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1. Aug 27, 2014

### HPphysics

1. The problem statement, all variables and given/known data
An experiment was conducted to investigate the time it takes water to pour out of a can through a little hole and the amount of water in the can. To find the dependence on the size of the hole, five large cylindrical containers of water of the same size were emptied through relatively small circular openings of different diameter. To find the dependence on the amount of water the same containers were filled to different heights.

Each measurement was repeated several times using a handheld stopwatch to measure the times to the nearest 1/10th of a second. The averages of the times are recorded in the following table.

Height cm​
Diameter (cm)
30 15 8 4 1 ​
1.5
97.4 sec 68.9 sec 50.3 sec 35.6 sec 17.8 sec ​
2.0
54.8 sec 38.7 sec 28.3 sec 20.0 sec 10.0 sec ​
3.0
24.3 sec 17.22 sec 12.6 sec 8.9 sec 4.4 sec ​
4.0
13.7 sec 9.7 sec 7.1 sec 5.0 sec 2.5 sec ​
5.0
8.8 sec 6.2 sec 4.5 sec 3.2 sec 1.6 sec ​

Find the relationship between the time and:

a) the height of the water in the can

b) the diameter of the hole

c) using the results of a) and b), formulate a general expression for the time y as a function of both the diameter and the height

d) Use the expression from c) to predict the time required to empty a can with a 3.5cm diameter to a height of 20cm

2. Relevant equations
Knowledge of log-log and semi-log graphs..

3. The attempt at a solution

a) We plotted a graph using a constant diameter (8cm) of time vs. height.
Time on y axis and height on x axis. The graph appeared to be logarithmic, so we linearized the graph by log"ing each side. Our equation: y=.5logx+b

b) We basically did the same thing as question a, using a constant height (8cm) and plotting time vs. diameter. Our equation: y=-2logx+b

c) Our answer is y=kh^1/2d^-2 We are not exactly sure what our reasoning is

d) We aren't sure about c) so we haven't ventured here yet

2. Aug 27, 2014

### Nathanael

Welcome to physics forums!

Do you perhaps see a relationship more simple than logarithms?

Hint:
What happens to the time if you quadruple the height?

What happens to the time if you double the diameter?

(Make sure to check these relationships in different cases, not just the easy case that I hinted at)

3. Aug 27, 2014

### Orodruin

Staff Emeritus
In general, I suggest using the variable names instead of x and y when quoting the equations you get.

So, you have deduced the following;

a) For constant diameter, you have $\log(T(h,d)) = \frac 12 \log(h) + C_1(d)$. Note that the constant in general will depend on $d$.

b) For constant height, you have $\log(T(h,d)) = - 2 \log(d) + C_2(h)$.

For (c), ask yourself what the functions $C_1$ and $C_2$ could be such that $T(d,h)$ is the same for both expressions - then solve for $T$.

Bonus problem: Can you argue for why the time behaves like this?

4. Aug 27, 2014

### HPphysics

@orodruin

We still aren't sure what to do. We're thinking of graphing both expressions and finding the intersection point of the two lines. Would that serve any purpose? Not exactly sure how to find out what C1 and C2 should be.. We are only in our first year of calculus

5. Aug 27, 2014

### Staff: Mentor

Are you allowed to use a graphics package such as Excel or Kaleidagraph? If so, make a plot of t vs h for each of the five diameters on a single plot. Change the scale on each of the axes to be logarithmic. Use the curve fitting capabilities to get the equation for each of the five lines in the form t = k hn. In Kaleidagraph, this is the "power" fit option. The parameter n should come out to 0.5. Next plot the parameter k as a function of the hole diameter d. Again, change the scales to logarithmic, and fit the data in the form k = C dm. The parameter m should come out to -2. The curve fit will give you the value of the parameter C. So, combining these results, you get t = Chndm. You know the parameter C so, you are done. The equation should fit all the data.

Chet

6. Aug 28, 2014

### Orodruin

Staff Emeritus
So the expressions for $T(h,d)$ has to be equal, this tells you that
$$\frac 12 \log(h) + C_1(d) = -2 \log(d) + C_2(h).$$
Note that $C_1$ cannot depend on $h$ and $C_2$ cannot depend on $d$. Both sides must show the same behavior with both variables in order to be equal so $C_1$ must have the same behavior with $d$ as the right-hand side does and so on. Therefore
$$C_1(d) = -2\log(d) + C, \quad C_2(h) = \frac 12 \log(h) + C.$$
So, in essence
$$\log(T(h,d)) = \frac 12 \log(h) - 2 \log(d) + C.$$
You can now solve for $T$, what do you obtain?

Plotting in the same graph really does not add much information.

7. Aug 28, 2014

### Orodruin

Staff Emeritus
On a separate note: Are you sure you quoted the units in the table correctly? I find it hard to imagine a 5 cm hole in a 1 cm high cylinder as a "small hole". The data also seems to fit a bit too well with the logarithmic behavior to be from a real experiment (this can be checked but really is out of the scope for the question), it is likely mock data.

8. Aug 28, 2014

### ehild

How would you generate data for a test so as they give the desired fitting parameters? I do not think you would perform an experiment and measure... :tongue:

ehild

9. Aug 28, 2014

### Orodruin

Staff Emeritus
I would add a random normal distributed number to each data point with a standard deviation the size of a reasonable experimental error. This way nasty physicists with a statistical mind set will not complain about my mock data on internet forums. :tongue:

10. Aug 28, 2014

### ehild

I would not bother those far-away nasty physicists.

I used to include errors in my test examples but they were linear relations, it was easy.

ehild

11. Aug 28, 2014

### Staff: Mentor

Plot C1 vs d using log scales for both and see what you get.

Chet