Specific objectives.

• Use vernier calipers and micrometer screw gauge to measure length.
• Estimate the diameter of an oil molecule.
• Solve numerical problems in measurement.

MEASUREMENT.

• Involves the quantity of length.
• The instruments used to measure length are;
1. Ruler
2. Meter rule
• Tape measure
• Other instruments are vernier calipers and micrometer screw gauge which measure very small length e.g. diameter of a wire or steel ball, diameter of thickness of a paper or internal diameter of a pipe.

Zone error.

• An error is the difference between the correct value and the value measured or obtained.
• Most of the instruments have a point or a zero mark.
• When the pointer or the zero mark coincide, the instrument is said to have no error. If the pointer or zero mark do not coincide, that is, is either to the left or the right of the main mark, the instrument has an error.
• The instrument used(micrometer screw gauge and vernier calipers) have vertical scale (m.s.a) and horizontal scale(v.s.r).

Positive zero error.

• It occurs when the pointer or zero mark is on the right side of the main zero mark.
• The measured mass is more than the main length.

• Check the line in the vernier scale that coincide with the one on the main scale. Multiply this number by 1 cm to grt the accuracy of the vernier callipers. The error is corrected by subtracting it from the measured length.

Correct length = Final reading – Positive zero error.

Negative zero error.

• It occurs when the pointer or the zero mark is to the left of the main zero mark.
• The measured length is less than the actual length/diameter.
• It is corrected by adding the zero error to the reading.

Correct value = Final reading + Zero error.

Accuracy.

• Is the closeness os a measurement to the correct value of quantity to be measured.
• The accuracy of a measurement depends on both the sensitivity of the instrument and the size of the quantity to be measured.

Vernier callipers.

• It is an instrument that measures accurately to 01 cm.
• It is used to measure external and internal diameter of objects.
• It has an accuracy of 01 cm/0.1mm.
• It consists a pair of steel jaws which closes until they touch the object in the desired position.
• It consists of;

2 scales i.e the main scale on the steel frame graduated in cm.

Vernier scale on the sliding jaw which has 10 dividions ond of total length 0.9 cm.

• Each division is 0.1 cm. The difference between the main scale and vernier scale division is 0.1 – 0.09 = 0.01. This is the accuracy of the instrument which is referred to as the least count.

How to use vernier callipers.

EXP:TO MEASURE THE DIAMETER OF A GLASS MARBLE.

Requirements.

1. Vernier callipers.
2. Glass marble.

Procedure.

1. Note the zero error of the instrument.
2. Open the jaws of the instrument and placethe marble bbetween the jaws. They should touch gently.
3. Note the main scale reading of the vernier scale.
4. Repeat the procedure for a small marble and find their volumes.

Results.

1. The zero error of the instrument ________ cm.
2. The main scale reading __________cm.

Micrometer screw gauge.

• It is an instrument used to measure small length or diameter e.g. thickness of a paper, wire, ball bearing e.t.c.
• It is a very accurate instrument with an accuracy of 01mm.

• It consists of;
1. U- frame carrying an anvil at one end.
2. A thimble which carries a circular rotating scale known as thimble scale and a spindle which can move forward and backwards when the thimble rotates. The scale is divided into 50 or 100 equal divisions.
• The sleeve which carries linear scale in mm- main scale and has 0.5 mm marks.
1. The ratchet at the end of the thimble which prevent the user from exacting high pressure on an object whose diameter is being measured. The pitch of the micrometer is the distance moved by the spindle in one complete rotation of the thimble.
• There are two types of micrometer screw gauge which are determined by the thimble scale. One type has a pitch of of 0.5 mm and the other of 1.0mm.
 Pitch of 0.5mm Pitch of 1.0 mm One complete turn is 0.5 mm One complete turn is 1 mm Each division is 0.5/50 Each division is 1/100 Has 50 divisions Has 100divisions

• The sleeve reading gives the units and the first two decimal places while the thimble reading gives the third decimal places.

How it works.

• Open the jaws of the instrument and place the object between the anvil and the spindle.
• Turn the thimble until the object is gripped between the anvil and the spindle.
• Sleeve has a scale of 0.5 mm which of it rep one complete turn of the screw.
• The sleeve reading gives the unit and the first two decimal places while the thimble reading gives the third to decimal place.

Precautions.

• Wipe clean the faces of anvil and spindle to remove dust particles which would cause false reading.
• Check for the zero error and record it using the zero adjusting screw which should be added or subtracted from the final answer.
• The object should/must not be ovvergripped i.e it should just be gripped slightly.

Advantages of micrometer screw gauge over vernier callipers.

1. Can mesure very small diameter of 0.01 mm.
2. Has a higher accuracy of 0.01 mm.

Advanntages of vernier callipers over micrometer screw gauge.

1. Can measure both internal and external diameter.
2. Cn measure long length.

Comparison beteen vernier callipers over micrometer screw gauge.

 Micrometer screw gauge Vernier callipers Can only measure external diameter Can measure both inside and outside diameter Can measure diameter of 0.01 mm Measure diameter of 0.1 mm Maximum length measured is 25 cm Maximum length measured is 5 cm Has an acuracy of 0.01 mm Has an accuracy of 0.1 mm

The size of an oil molecule.

• A molecule is a very small particle of matter which cannot be measured directly unless with very sophisticated apparatus such as an electron microscope.

EXP: TO MEASURE A VOLUME OF AN OUI DROP.

Requirements.

1. Burette.
2. Oil.
3. Clamp and stand.
4. Beaker.

Procedure.

1. Half fill the burette with oil and mark the initial volume V1.
2. Release 20 drops of oil and note the final reading V2

Results.

1. Initial volume V1 = 36.2 ml
2. Final volume V2 = 36.8 ml
3. Volume of 20 drops V3 = V2 – V1 = 36.8 – 36.2 = 0.6 ml
4. Volume of 1 drop = V3/20 = 0.6/20 = 0.03 ml
5. Repeat the experiment and find the average. Assume that the oil drop is spherical.

EXP 2: TOESTIMATE THE SIZE OF AN OIL MOLECULE.

Requirements.

1. Oil.
2. Clean water.
3. Petridish.
4. Lycopodium powder or chalk dust.
5. 2 wooden splints.

Procedure.

1. Put clean water in a petridish and sprindle lycopodim powder (chalk dust) on the surface of water.
2. Release one drop of oil at the surfaceof water at the center.The oil spread into a circular clear patch pushing aside the powder.
3. Place the two splints at the boundaries ao the patch and measure the distance between them, that is the diameter of the oil patch.

Results.

Average diameter =   D1 + D2 + D3           radius = __D___

3                                                        2

Area of oil patch

Volume of oil atch

Volume of oil drop

Volume of oil drop = volume of oil patch where h is the thickness of 1 oil molecule.

h = _Volume of oil drop__

Area of oil patch

Assumption.

1. The oil spread on the watersurface in form ofa patch whose height is that of 1 molecule.
2. The oil drop is a perfect sphere.
3. The purpose of lycopodium powder or chalk dust is to make the oil patch visible.

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