Physics – Instructor Exam

Metric constants for the Physics calculations

Metric constants for the Physics calculations

One of the components of the IDC is the exam. Practising past questions is always a good idea that way you are prepared. The calculations are much easier in Metric – so use Metric.

Some points to note:

  1. Make sure you know whether you are working in salt or fresh water
  2. Draw a diagram to help you (it can make it much easier if you can visualise what you are doing)
  3. Know the difference between the different pressure types
  4. Read, re-read, and then re-read again

So here are some example questions with explained working

Example 1

If an object weighs 105 kilograms is neutrally buoyant in salt water, what is the volume of water that the object displaces? (Give your answer to the nearest litre)

The question states that the object is neutrally buoyant, that means that the amount of water it displaces is equal to its weight. eg the downward force = the upward force

So as 1 litre of sea water weighs 1.03 kg we can easily convert the 105 kg into litres. 105 / 1.03 = 101.94 litres

That means that the object displaces 102 litres (to the nearest litre)

Example 2

If a neutrally buoyant object displaces 102 litres of sea water, what will happen to the object when is placed in fresh water?

First we need to know how much the object weighs. 1 litre of sea water weighs 1.03 kg.

102 litres x 1.03 = 105 kg

That means the object exerts a downwards force of 105 kg.

1 litre of fresh water weighs 1 kg. So if the object displaces 102 litres the weight of that water is 102 kg.

That means the displaced water exerts an upward force of 102 kg.

The upward force is less than the downward force, that means the object will sink.

Some definitions:

Gauge Pressure = this ignores the atmospheric pressure. So at sea level the gauge would read zero on an empty cylinder.

Absolute Pressure = this is the gauge pressure plus atmospheric pressure (ata).

Ambient Pressure = means “surrounding pressure” and may be expressed as absolute or gauge pressure.

Example 3

What is the absolute (ata) pressure in 15 metres of salt water?

Pressure increases by 1 atm every 10 metres in sea water.

The gauge pressure would be equal to 15 / 10 = 1.5 atm.

To convert the gauge pressure to absolute we just add on 1 atm. 1.5 + 1 = 2.5 ata

Example 4

What is the gauge pressure in 24 metres of fresh water?

Pressure increases by 1 atm every 10.3 metres in fresh water.

The gauge pressure would be equal to 24 / 10.3 = 2.33 atm.

Example 5

You take 5.5 litres of gas in a flexible container from 9 metres in seawater to a depth of 32 metres. What will be the new volume of the gas?

Start by working out the volume of the gas at the surface.

So what is the absolute pressure at 9 metres?

Pressure increases by 1 atm every 10 metre in salt water. So at 9 metres the gauge pressure is 9 / 10 = 0.9 atm.

That makes the absolute pressure 1.9 ata (remember from above that we add 1 onto the gauge pressure to get the absolute pressure).

Remember that gas expands. So 5.5 x 1.9 = 10.45 litres

Now work out the absolute pressure at 32 metres. 32 / 10 = 3.2 atm. 1 + 3.2 = 4.3 ata.

That that we have worked out the absolute pressure we can work out the volume at 32 metres.

10.45 / 4.3 = 2.43 litres.

Example 6

An object weighs 40 kg and displaces 40 litres of water. Disregarding the minimal displacement of the lead, how much lead weight do you need to make it 15kg negatively buoyant in sea water?

1 litre of seawater weighs 1.03 kg

The object has a downwards force of 40kg

40 litres of seawater gives an upwards force of 40 x 1.03 = 41.2 kg

The upwards force is greater than the downwards force so it is positively buoyant. To make it neutrally buoyant we must add 1.2kg.

Then to make it 15kg negatively buoyant we add another 15kg. This brings the total lead added to 16.2kg

Example 7

An object weighs 125kg and displaces 75 litres of seawater. How much additional seawater do you need to displace to make the object 20kg positively buoyant. ?

1 litre of seawater weighs 1.03 kg

So 75 litres of sea water weighs 75 x 1.03 = 77.25 kg

That means it is negatively buoyant by 47.75 kg (125 – 77.25).

So to make it neutrally buoyant one must add 47.75 kg and then to make it positively buoyant we must add another 20 kg. Making a total of 67.75 kg.

67.75 / 1.03 = 65.8 litres.

Example 8

A diver is using Enriched Air Nitrox that is 32% Oxygen, 68% Nitrogen. What will be the partial pressure of oxygen and nitrogen at a depth of 17 metres in seawater?

We need to work out the absolute pressure at 17 metres in seawater.

1 atm for every 10 metres of seawater.So the gauge pressure at 17m is 17 / 10 = 1.7 atm. The absolute pressure is 2.7 ata.

For Oxygen: 0.32 x 2.7 = 0.864 ataFor Nitrogen: 0.68 x 2.7 = 1.836 ata

Example 9

A scuba cylinder containing 1.5% of carbon dioxide at the surface is taken down to 25 metres. What is the percentage of carbon dioxide in the cylinder?

This is a trick question. The percentage of a gas mix will not change at depth. 32% nitrox will always we 32% Oxygen. What changes is the partial pressure. The question doesn’t ask for the partial pressure.

So the answer is 1.5%.

Example 10

A 150 kg engine displaces 100 litres of water lies on the bottom in 22 metres of seawater. What is the minimum amount of water that must be displaced from a lifting device to being the engine to the surface?

Downward force is 150 kg

1 litre of seawater weighs 1.03 kg.

Upward force is 100 * 1.03 = 103 kg

So the engine is negatively buoyant by 150 – 103 = 47 kg.

Now we work out how many litres of seawater we would require.

1 litre of seawater weighs 1.03 kg

47 / 1.03 = 45.6 litres

Don’t be fooled by the 22 metre depth. That isn’t relevant for this question so don’t let it catch you out.

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