Rising and Falling Water Levels

When a solid sinks in a tank of water, the level rises by just enough to make room for it. Use this to find the rise, the volume of an irregular solid, the overflow, the fall when a solid is lifted out, and the depth round a block that stands out of the water.

Before you start

Volume of water in a tank = inside base area × depth of the water, and when water is poured or a solid is melted and recast its volume stays the same (Pouring and Recasting).
The volume of a cuboid is length × breadth × height; of any solid with a uniform cross-section, area of cross-section × length (Volume of Prisms); of a cylinder, $\pi r^{2}h$ (Volume of Cylinders).
To find an unknown length, substitute the numbers you know and solve the equation: $432x = 216$ gives $x = 216 \div 432 = 0.5$ .

The Rise in Water Level

Lower a heavy solid into a tank of water. It sinks, and it pushes the water out of the space it now fills. That water has nowhere to go but up, so the level rises.

When the solid ends completely under water, the water that has been pushed up is a layer as wide as the tank and as thick as the rise, and its volume is exactly the volume of the solid. So the volume of the solid = inside base area × rise.

Water running in from a tap raises the level too; how fast it does so is the next lesson, Rate of Flow.

Rise in water level

When a solid is put into the water in a tank, sinks and is completely covered, and no water overflows:

volume of the solid = inside base area of the tank × rise in water level

new depth of water = depth before + rise

Use it only when all three of these are true. (1) The tank has a uniform cross-section and stands on its base (a cuboid tank or an upright cylinder), so every layer of water has the same area. (2) You use the tank's inside base area. (3) The solid sinks and ends completely under water: a block of wood floats, and a tall block may stand out of the water (the last section shows what to do then).

Example 1: How far does the water rise?

A rectangular tank whose inside is 24 cm long, 18 cm wide and 15 cm high holds water 10 cm deep. A solid metal cube of edge 6 cm is lowered into the tank and sinks to the bottom.

Find (a) the rise in the water level, (b) the new depth of the water.

Solution

The cube is 6 cm high and the water is already 10 cm deep, so the cube is completely under water: the rule holds.

Volume of the cube $= 6^{3} = 216 \text{ cm}^{3}$

Inside base area of the tank $= 24 \times 18 = 432 \text{ cm}^{2}$

(a) Let the rise be $x$ cm. Volume of the cube = base area × rise:

$432x = 216$

$x = 216 \div 432 = 0.5$

(b) New depth $= 10 + 0.5 = 10.5$ cm, which is below the top of the tank (15 cm), so none overflows.

(a) The water rises by 0.5 cm. (b) The water is now 10.5 cm deep.

Example 2: The volume of an irregular stone

A stone has no formula for its volume, but a rise in water level measures it. Ravi puts a stone into a cylindrical container of internal radius 8 cm holding some water. The stone sinks and is completely covered, and the water level rises from 12 cm to 14.5 cm. Find the volume of the stone. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

Solution

Rise in water level $= 14.5 - 12 = 2.5$ cm

Volume of the stone = inside base area × rise

$= \pi \times 8^{2} \times 2.5$

$= 160\pi$

$= 502.65\ldots \text{ cm}^{3} \approx 503 \text{ cm}^{3}$

The volume of the stone is about $503 \text{ cm}^{3}$ .

Example 3: When a solid makes the water overflow

The inside of a fish tank is 30 cm long, 20 cm wide and 25 cm high, and it holds water 23 cm deep. Hui Min puts in a brick 20 cm long, 10 cm wide and 8 cm high, which sinks to the bottom. How much water overflows?

Solution

The water can rise only $25 - 23 = 2$ cm before it reaches the top.

Space above the water $= 30 \times 20 \times 2 = 1200 \text{ cm}^{3}$

Volume of the brick $= 20 \times 10 \times 8 = 1600 \text{ cm}^{3}$

The brick pushes up $1600 \text{ cm}^{3}$ of water, but only $1200 \text{ cm}^{3}$ fits above the old level. The water rises to the brim (25 cm, so the 8 cm brick is still completely covered) and the rest spills.

Volume that overflows $= 1600 - 1200 = 400 \text{ cm}^{3}$ , which is 400 mL.

400 mL of water overflows, and the tank is left full to the brim.

Taking a Solid Out

Taking a solid out of the water is the rise in reverse: the space it filled is filled by water from above, so the level falls. The same three conditions apply: a tank of uniform cross-section, its inside base area, and a solid that was completely under water.

The fall is worked from the solid's volume, not copied from the rise. They are the same only if no water spilt when the solid went in: if some overflowed, that water is gone, and the level ends lower than it started.

Fall in water level

volume of the solid taken out = inside base area of the tank × fall in water level (the solid was completely under water)

new depth of water = depth before − fall

Example 4: The water level falls

A cylindrical container of internal radius 8 cm holds water 18 cm deep. A solid metal cylinder of radius 4 cm and height 10 cm stands upright on the bottom, completely under water. The metal cylinder is lifted out. Find the new depth of the water.

Solution

Volume of the metal cylinder $= \pi \times 4^{2} \times 10 = 160\pi \text{ cm}^{3}$

Inside base area of the container $= \pi \times 8^{2} = 64\pi \text{ cm}^{2}$

Let the fall in water level be $x$ cm. Volume taken out = base area × fall:

$64\pi x = 160\pi$

$64x = 160$ (dividing both sides by $\pi$ )

$x = 160 \div 64 = 2.5$

New depth $= 18 - 2.5 = 15.5$ cm

The water is now 15.5 cm deep.

A Block That Is Not Covered

A block that stands on the bottom and is taller than the water will reach is not completely covered, so the rule for the rise does not hold: only the part of the block under water pushes water up, and you do not yet know how much of it that is.

Look at the water instead. Its volume has not changed, but it now stands on less floor: the base of the tank less the base of the block. So find the new depth from that smaller area, then check that the depth is lower than the top of the block.

A block standing in water, not covered

Area the water stands on = inside base area of the tank − base area of the block

volume of water = (area the water stands on) × new depth

Check: the new depth must be less than the height of the block. (If it is not, the block is covered, and the rise rule applies.) It must also be below the top of the tank, so that none overflows.

This holds for a block that is heavy enough to stay standing on the bottom, and has the same cross-section all the way up (a cuboid or an upright cylinder), so the water stands on the same area at every height.

Example 5: A tall block stood in a tank

A rectangular tank whose inside is 40 cm long, 30 cm wide and 30 cm high holds water 10 cm deep. A metal block with a base 20 cm by 15 cm and a height of 25 cm is stood upright on the bottom of the tank.

Find the new depth of the water, correct to 3 significant figures.

Solution

Volume of water $= 40 \times 30 \times 10 = 12\,000 \text{ cm}^{3}$

Suppose the block stands out of the water, so it is not covered (this is checked at the end).

Area the water stands on $= 40 \times 30 - 20 \times 15 = 1200 - 300 = 900 \text{ cm}^{2}$

Let the new depth be $d$ cm.

$900d = 12\,000$

$d = 12\,000 \div 900 = 13.33\ldots \approx 13.3$

Check: 13.3 cm is less than the block's height of 25 cm, so the block does stand out of the water; and it is below the top of the tank (30 cm), so none overflows.

Using the rise rule with the whole block would be wrong: $300 \times 25 = 7500 \text{ cm}^{3}$ , a rise of $7500 \div 1200 = 6.25$ cm and a depth of 16.25 cm. Only the part of the block under water counts: $300 \times 13.33\ldots = 4000 \text{ cm}^{3}$ , and that does give the rise, $4000 \div 1200 = 3.33\ldots$ cm, from 10 cm to 13.3 cm.

The water is now about 13.3 cm deep.

Common mistakes

Giving the rise when the question asks for the new depth, or the other way round. New depth = depth before + rise.
Using the rise rule when it does not hold: for a solid that floats, one that stands out of the water, or a container whose cross-section is not uniform.
Using the outside measurements. The water rises over the inside base of the tank: use its internal length, breadth or radius.
Forgetting the water already in the tank. Only the space above the water is free: compare the solid's volume with that space, not with the whole capacity.
Taking the fall to be the same as the rise. Work the fall from the solid's volume: after an overflow the water that spilt is gone, so the level falls further than it rose.
Not checking a block that is stood in the water. Suppose it stands out, find the depth, then check: if the depth is above the block, it is covered, and the rise rule applies instead.
Dividing by a product with $\pi$ in it without brackets. If $225\pi x = 2400$ , then $x = 2400 \div (225\pi)$ : 2400 divided by the whole of $225\pi$ . Keyed in as $2400 \div 225 \times \pi$ it gives the wrong answer; put brackets round $225\pi$ .

Practice

Try each question before you open its solution. Where π is needed, the question says which value to use; give answers that are not exact correct to 3 significant figures.

Question 1

A tank whose inside is 50 cm long, 40 cm wide and 30 cm high holds water 20 cm deep. A metal block 25 cm long, 16 cm wide and 10 cm high is lowered into the tank and sinks to the bottom. Find the new depth of the water.

ShowHide solution

The block is 10 cm high and the water 20 cm deep, so the block is completely covered.

Volume of the block $= 25 \times 16 \times 10 = 4000 \text{ cm}^{3}$

Inside base area $= 50 \times 40 = 2000 \text{ cm}^{2}$

Let the rise be $x$ cm: $2000x = 4000$ , so $x = 2$ .

New depth $= 20 + 2 = 22$ cm (below the top, 30 cm).

The water is now 22 cm deep.

Question 2

Ravi lowers a stone into a fish tank whose inside is 40 cm long and 25 cm wide. The stone sinks and is completely covered, and the water level rises from 15 cm to 16.8 cm. Find the volume of the stone.

ShowHide solution

Rise $= 16.8 - 15 = 1.8$ cm

Volume of the stone = inside base area × rise $= 40 \times 25 \times 1.8 = 1800 \text{ cm}^{3}$

The volume of the stone is $1800 \text{ cm}^{3}$ .

Question 3

A rectangular container whose inside is 15 cm long, 12 cm wide and 20 cm high holds water 8 cm deep. Hui Min drops in 150 marbles, each of volume $4.2 \text{ cm}^{3}$ . They all sink and are completely covered. Find the new depth of the water.

ShowHide solution

Volume of the marbles $= 150 \times 4.2 = 630 \text{ cm}^{3}$

Inside base area $= 15 \times 12 = 180 \text{ cm}^{2}$

Let the rise be $x$ cm: $180x = 630$ , so $x = 630 \div 180 = 3.5$ .

New depth $= 8 + 3.5 = 11.5$ cm (below the top, 20 cm).

The water is now 11.5 cm deep.

Question 4

Forty identical marbles are dropped into a cylindrical container of internal radius 5 cm holding some water. They all sink and are completely covered, and the water level rises by 1.5 cm. Find the volume of one marble. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

ShowHide solution

Volume of the 40 marbles = inside base area × rise $= \pi \times 5^{2} \times 1.5 = 37.5\pi \text{ cm}^{3}$

Volume of one marble $= 37.5\pi \div 40 = 2.945\ldots \approx 2.95 \text{ cm}^{3}$

The volume of one marble is about $2.95 \text{ cm}^{3}$ .

Question 5

A stone of volume $1350 \text{ cm}^{3}$ is lowered into a fish tank whose inside is 30 cm long and 25 cm wide. It sinks and is completely covered, and the water is then 16 cm deep. How deep was the water before the stone was put in?

ShowHide solution

Inside base area $= 30 \times 25 = 750 \text{ cm}^{2}$

Let the rise be $x$ cm. Volume of the stone = base area × rise: $750x = 1350$ , so $x = 1350 \div 750 = 1.8$ .

Depth before = new depth − rise $= 16 - 1.8 = 14.2$ cm

The water was 14.2 cm deep before the stone was put in.

Question 6

A cylindrical tank of internal radius 15 cm and height 40 cm holds water 30 cm deep. A metal cuboid 20 cm long, 10 cm wide and 12 cm high is lowered in and lies on the bottom. Find the rise in the water level. Use the $\pi$ button on your calculator and give your answer correct to 3 significant figures.

ShowHide solution

The cuboid is 12 cm high and the water 30 cm deep, so it is completely covered.

Volume of the cuboid $= 20 \times 10 \times 12 = 2400 \text{ cm}^{3}$

Inside base area $= \pi \times 15^{2} = 225\pi \text{ cm}^{2}$

Let the rise be $x$ cm: $225\pi x = 2400$

$x = 2400 \div (225\pi) = 3.395\ldots \approx 3.40$

Check: the new level is $30 + 3.395\ldots = 33.4$ cm (to 3 significant figures), below the top of the tank (40 cm), so none overflows.

The water rises by about 3.40 cm.

Question 7

Mr Tan's cylindrical pail has an internal radius of 14 cm and holds water 20 cm deep. A metal block 11 cm long, 8 cm wide and 7 cm high lies on the bottom, completely under water. He lifts the block out. Find the new depth of the water. Take $\pi = \frac{22}{7}$ .

ShowHide solution

Volume of the block $= 11 \times 8 \times 7 = 616 \text{ cm}^{3}$

Inside base area of the pail $= \frac{22}{7} \times 14^{2} = 22 \times 2 \times 14 = 616 \text{ cm}^{2}$

Let the fall be $x$ cm: $616x = 616$ , so $x = 1$ .

New depth $= 20 - 1 = 19$ cm

The water is now 19 cm deep.

Question 8

A tank whose inside is 40 cm long, 30 cm wide and 20 cm high holds water 18 cm deep. Solid metal cubes of edge 5 cm are lowered in one at a time, and each sinks to the bottom. What is the greatest number of cubes that can be put in without any water overflowing?

ShowHide solution

Space above the water $= 40 \times 30 \times (20 - 18) = 2400 \text{ cm}^{3}$

Volume of one cube $= 5^{3} = 125 \text{ cm}^{3}$ ; each cube lies on the bottom, 5 cm high, well under the water.

$2400 \div 125 = 19.2$

19 cubes push up $19 \times 125 = 2375 \text{ cm}^{3}$ of water, which fits; a 20th would make $2500 \text{ cm}^{3}$ , more than the $2400 \text{ cm}^{3}$ of space, and water would spill. So round down.

At most 19 cubes can be put in without any water overflowing.

Question 9

A tank whose inside is 30 cm long, 20 cm wide and 25 cm high holds water 22 cm deep. A metal block 15 cm long, 10 cm wide and 16 cm high is lowered in and sinks. (a) How much water overflows? (b) The block is then lifted out. Find the new depth of the water.

ShowHide solution

(a) Space above the water $= 30 \times 20 \times (25 - 22) = 1800 \text{ cm}^{3}$

Volume of the block $= 15 \times 10 \times 16 = 2400 \text{ cm}^{3}$ , more than the space, so the water rises to the brim (25 cm; the 16 cm block is covered) and the rest spills.

Volume that overflows $= 2400 - 1800 = 600 \text{ cm}^{3}$ , which is 600 mL.

(b) The block was completely under water. Let the fall be $x$ cm. Volume of the block = base area × fall:

$600x = 2400$ , so $x = 4$ .

New depth $= 25 - 4 = 21$ cm.

The level rose only 3 cm (from 22 cm to the brim) but falls 4 cm: the 600 mL that spilt is gone, so the water ends 1 cm lower than it started. Check: water left $= 30 \times 20 \times 22 - 600 = 12\,600 \text{ cm}^{3}$ , and $600 \times 21 = 12\,600$ .

(a) 600 mL of water overflows. (b) The water is then 21 cm deep.

Question 10

A cylindrical tank of internal radius 10 cm and height 25 cm holds water 12 cm deep. A solid metal cylinder of radius 5 cm and height 20 cm is stood upright on the bottom of the tank. Find the new depth of the water.

ShowHide solution

Volume of water $= \pi \times 10^{2} \times 12 = 1200\pi \text{ cm}^{3}$

Suppose the metal cylinder stands out of the water (this is checked at the end). Then the rise rule does not hold: the water stands on the tank's base less the cylinder's base:

area $= \pi \times 10^{2} - \pi \times 5^{2} = 100\pi - 25\pi = 75\pi \text{ cm}^{2}$

Let the new depth be $d$ cm: $75\pi d = 1200\pi$ , so $75d = 1200$ and $d = 16$ .

Check: 16 cm is less than the metal cylinder's height of 20 cm, so it does stand out of the water, as supposed (and 16 cm is below the top of the tank).

The water is now 16 cm deep.

Question 11

A tank whose inside is 30 cm long, 20 cm wide and 25 cm high holds water 12 cm deep. A metal block with a square base of side 10 cm and a height of 13 cm is stood upright on the bottom. Find the new depth of the water, correct to 3 significant figures.

ShowHide solution

Volume of water $= 30 \times 20 \times 12 = 7200 \text{ cm}^{3}$

Suppose the block stands out of the water. The water stands on $600 - 100 = 500 \text{ cm}^{2}$ , so $500d = 7200$ and $d = 14.4$ .

Check: 14.4 cm is more than the block's height of 13 cm, so the block would be under water after all. So the block cannot stand out of the water: it is completely covered.

So use the rise rule. Volume of the block $= 10 \times 10 \times 13 = 1300 \text{ cm}^{3}$ .

Let the rise be $x$ cm: $600x = 1300$ , so $x = 2.166\ldots$

New depth $= 12 + 2.166\ldots = 14.166\ldots \approx 14.2$ cm, which is above the block (13 cm), as it must be, and below the top (25 cm).

The water is now about 14.2 cm deep, and the block is covered.

Key points

A solid that sinks and is completely covered: volume of the solid = inside base area × rise.
The rule needs a tank of uniform cross-section standing on its base, its inside base area, and a solid that sinks and ends completely under water.
New depth = depth before + rise; the depth before = new depth − rise.
The rise measures the volume of a solid with no formula: a stone, a handful of marbles.
Only the space above the water is free: if the solid's volume is more, the water rises to the brim and the rest overflows.
Taking a covered solid out makes the level fall: volume of the solid = inside base area × fall. After an overflow the fall is more than the rise was, because the spilt water is gone.
A block stood in the water and not covered (a heavy block that stays on the bottom, with the same cross-section all the way up): the water stands on (tank's base − block's base); check that the new depth is below the top of the block and below the top of the tank.
How many solids before the water overflows: divide the free space by one solid's volume and round down.
Use the value of $\pi$ the question gives; bracket a divisor with $\pi$ in it; round only at the end.