A dripping tap: measuring a rate and scaling it to litres per day and per year

The idea

A rate compares two quantities with different units, so a volume collected in a measured time gives millilitres per minute, which is the gradient of a straight volume-time graph and scales by unit conversion to litres per day and per year.

What you need

• A classroom or science-room sink tap that can be set to a steady drip
• 250 mL plastic measuring cylinder with 2 mL graduations, stood in the sink under the tap (it holds 10 min of any drip up to 25 mL per minute)
• 10 mL plastic measuring cylinder with 0.2 mL graduations
• Stopwatch or phone timer
• Calculator or spreadsheet

How to do it

1. Set the tap to a slow, steady drip. Count the drips in 60 s three times and take the median.
2. Stand the 250 mL cylinder under the tap, start the timer, and read the volume every 2 min for 10 min without moving the cylinder (if it nears 250 mL, stop and note the time).
3. Tabulate volume against time, plot the points and find the gradient in mL per minute.
4. Compute the volume of one drip: mL per minute divided by drips per minute. Check it by counting the drips that fill the 10 mL cylinder to its 5.0 mL mark and dividing 5.0 mL by that count.
5. Convert the rate to litres per day (× 1440 min, ÷ 1000) and litres per year (× 365), and compute how many days the drip takes to fill a 9 L bucket.
6. Set a faster drip and repeat; compare the two gradients. Compare both daily figures with Riverina Water's statement that dripping taps can waste 30 to 200 litres of water a day, and compute the rate in mL per minute that 30 litres a day requires.
7. Financial extension: multiply the yearly volume in kilolitres by the usage price per kilolitre printed on a household water bill.

What you should see

The volume-time points lie on a straight line through the origin because the drip is steady, and its gradient is the rate. As a worked check, a tap dripping 45 times a minute that fills 32 mL in 10 min runs at 3.2 mL/min, so one drip is 0.071 mL and about 70 drips fill the 10 mL cylinder to its 5.0 mL mark (5.0 ÷ 0.071); that is 4.61 L a day and 1682 L (1.68 kL) a year, and a 9 L bucket fills in 1.95 days. Reaching 30 L a day needs 20.8 mL/min, about 6.5 times that drip, which is a fast drip or a thin trickle. Drips vary in size from tap to tap: the USGS drip calculator takes a drip as 0.25 mL, which at 45 drips a minute gives 11.25 mL/min, 112.5 mL in 10 min (still inside the 250 mL cylinder) and 20 drips to the 5.0 mL mark. The learner knows it worked when the gradient of the graph and the total volume divided by the total time agree within one graduation of the cylinder, and the yearly figure comes out the same by two routes (from mL per minute and from L per day).

What changes

What you change

time (min)

What you measure

volume collected (mL)

What you keep the same

• tap setting left untouched during a run
• the same collecting cylinder
• timer started when the cylinder is placed

Common misconceptions

Each of these ideas is wrong, and the activity is a chance to test it.

• A small drip wastes an insignificant amount over a year.
• A rate is the same thing as an amount.
• Collecting for twice as long doubles the rate.

Safety card

Curriculum references

The NSW syllabus outcomes and Australian Curriculum v9 codes this activity supports. They are references, not a verified or complete curriculum alignment.

• Mathematics K–10 Syllabus (2022), Stage 4 Ratios and rates; page read 2026-09-22MA4-RAT-C-01
• Australian Curriculum v9AC9M8M05AC9M8M07

Sources

The pages the author read to write this activity.

1. curriculum.nsw.edu.au/learning-areas/mathematics/mathematics-k-10-2022/content/stage-4/faf41ae0d9
2. rwcc.nsw.gov.au/your-water/savewater/at-home/taps
3. water.usgs.gov/edu/activity-drip.html