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Different types of lengths are measured by using different types of instruments. Lengths like the length of cloth or length of a line can be measured by using measuring tape, a metre scale or a foot rule. But these instruments cannot be used to measure the diameter of a metal sphere or a cylinder. To measure the diameter of a cylinder we can use paper strip method and wooden block method:
(i) The correct way to read a ruler is shown in the figure below. The eye must be positioned vertically above the mark to avoid error due to Parallax.
(ii) Paper strip method: Wind a strip of paper closely round the object once and prick the overlapping position with a pin (Shown in figure below).
A method for measuring diameter of a cylinder
Unwind the paper strip and measure the distance between the two pinholes.
This measure is the measure of the circumference, since circumference = (
x diameter).
Hence now the diameter can be calculated.
(iii) Wooden block method: Place the sphere or the cylinder between two blocks in contact with a ruler as shown in figure below.
A simple method for measuring diameter of a sphere
Read the distance between the two blocks on the ruler accurately. (The line of sight should be vertical.)
The meter scale enables us to measure the length to the nearest millimeter only. Engineers and scientists need to measure much smaller distances accurately. For this a special type of scale called Vernier scale is used.
Vernier Calliper
The Vernier scale consists of a main scale graduated in centimeters and millimeters. On the Vernier scale 0.9 cm is divided into ten equal parts. The least count or the smallest reading which you can get with the instrument can be calculated as under:
Least count = one main scale (MS) division - one vernier scale (VS) division.
= 1 mm - 0.09 mm
= 0.1 mm
= 0.01 cm
The least count of the vernier
= 0.01 cm
The Vernier calliper consists of a main scale fitted with a jaw at one end. Another jaw, containing the vernier scale, moves over the main scale. When the two jaws are in contact, the zero of the main scale and the zero of the vernier scale should coincide. If both the zeros do not coincide, there will be a positive or negative zero error.
After calculating the least count place the object between the two jaws.
Record the position of zero of the vernier scale on the main scale (3.2 cm in figure below).
Principle of Vernier
You will notice that one of the vernier scale divisions coincides with one of the main scale divisions. (In the illustration, 3rd division on the vernier coincides with a MS division).
Reading of the instrument = MS div + (coinciding VS div x L.C.)
= 3.2 + (3 x 0.01)
= 3.2 + 0.03
= 3.23 cm
To measure the inner and outer diameter of a hollow cylinder or ring, inner and outer callipers are used. Take measurements by the two methods as shown in figure below.
Micrometer screw-gauge is another instrument used for measuring accurately the diameter of a thin wire or the thickness of a sheet of metal.
It consists of a U-shaped frame fitted with a screwed spindle which is attached to a thimble.
Screw-gauge
The screw has a known pitch such as 0.5 mm. Pitch of the screw is the distance moved by the spindle per revolution. Hence in this case, for one revolution of the screw the spindle moves forward or backward 0.5 mm. This movement of the spindle is shown on an engraved linear millimeter scale on the sleeve. On the thimble there is a circular scale which is divided into 50 or 100 equal parts.
When the anvil and spindle end are brought in contact, the edge of the circular scale should be at the zero of the sleeve (linear scale) and the zero of the circular scale should be opposite to the datum line of the sleeve. If the zero is not coinciding with the datum line, there will be a positive or negative zero error as shown in figure below.
Zero error in case of screw gauge
While taking a reading, the thimble is turned until the wire is held firmly between the anvil and the spindle.
The least count of the micrometer screw can be calculated using the formula given below:
Least count
= 0.01 mm
The wire whose thickness is to be determined is placed between the anvil and spindle end, the thimble is rotated till the wire is firmly held between the anvil and the spindle. The rachet is provided to avoid excessive pressure on the wire. It prevents the spindle from further movement. The thickness of the wire could be determined from the reading as shown in figure below.
Reading = Linear scale reading + (Coinciding circular scale x Least count)
= 2.5 mm + (46 x 0.01)
= (2.5 + 0.46) mm
= 2.96 mm
Relationship in the Metric system of length
1 kilometer (km) = 103 m
1 centimeter (cm) = 10-2 m
1 millimeter (mm) = 10-3 m
Example 1:
The diagram shown is a section of Vernier Calliper. Find the least count of the instrument and also the final reading, which is the thickness of a metal sheet.
Suggested answer :
Least count of the Vernier
= 0.01 cm
Reading of the instrument = Main scale reading
+ (coinciding v.s. div x least count)
= 4.3 + (8 x 0.01)
= 4.3 + 0.08
= 4.38 cm
Example 2:
In a Vernier calliper 1 cm of the main scale is divided into 20 equal parts. 19 divisions of the main scale coincide with 20 divisions on the vernier scale. Find the least count of the instrument.
Suggested answer :
The value of one main scale division
Number of divisions on vernier scale = 20
Least count of the vernier scale
= 0.025 cm
Example 3:
The circular scale (head scale) of a screw gauge is divided into 100 equal parts and it moves 0.5 mm ahead in one revolution. Find the pitch and the least count.
Suggested answer :
Pitch = distance moved in one revolution
= 0.5 mm
Least count
= 0.005 mm
Example 4:
The accompanying diagram represents a screw gauge. The circular scale is divided in to 50 divisions and the linear scale is divided into millimeters. If the screw advances by 1 mm when the circular scale makes 2 complete revolutions, find the least count of the instrument and the reading of the instrument in figure below.
Suggested answer :
Pitch of the screw
= 0.5 mm
Least count
= 0.01 mm
Reading = L.S. reading + (coinciding circular scale x least count)
= 3.5 mm + (32 x 0.01)
= 3.5 + 0.32
= 3.82 mm
Example 5:
Two simple pendulums are of lengths 40 cm and 1.6 m respectively. What will be the ratio of their time periods?
Suggested answer :
Since
therefore,
or T1 : T2 = 1 : 2
Example 6 :
The diagram below shows part of the main scale and vernier of a calliper, which is used to measure the diameter of a metal ball. Find the radius of the ball (sphere).
Suggested answer :
Least count
= 0.01 cm
Reading (diameter) = M.S. reading + (coinciding V.S. reading x Least count)
= 4.3 cm + (7 x 0.01)
= 4.3 + 0.07
= 4.37 cm
Diameter = 4.37 cm
Radius
= 2.185 cm
= 2.18 cm
To the required number of significant figures.
Example 7:
The diagram shown below shows a part of the linear scale and head scale (circular scale) of a micrometer screw which is used to measure the thickness of a glass plate. Calculate the thickness (pitch=0.5 mm), Total number of divisions on head scale = 50.
Suggested answer :
Pitch = 0.5 mm
Least count
= 0.01 mm
Thickness = Linear S.D + (Circular S.D. x Least count)
= 3.5 mm + (11 x 0.01) mm
= 3.61 mm.