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Measurement
Example 2.3 of both the sleeve and the thimble,
What is the correct reading for the taking into account the least count
micrometre screw gauges shown in of the instrument.
Figure 2.17?
(b) Least count (LC) of a micrometre
(a)
(a) (b) screw gauge given = 0.01 mm.
(b)
Main scale reading (MSR) = 1.6 cm
or 16 mm.
Circular scale reading (CSR) = 48
Figure 2.17
divisions.
Solution The reading = MSR + CSR × LC
From Figure 2.17 (a)
Main scale reading = 7.50 mm = 16 mm + 48 × 0.01 mm
= 16.48 mm
Thimble scale reading = 0.38 mm
Reading = 7.88 mm. The correct reading
= 16.48 mm + 0.03 mm
From Figure 2.17 (b) = 16.51 mm.
Main scale reading = 7.50 mm
Thimble scale reading = 0.22 mm Exercise 2.2
Reading = 7.72 mm.
1. What is the reading of the
Example 2.4 micrometre screw gauge in Figure
(a) Explain how the components of 2.18?
the screw gauge, including the
sleeve and thimble readings, as
well as the least count, are utilized
to determine the precise diametre
of the rod. What factors should be
considered to ensure the accuracy of
this measurement? Figure 2.18
(b) In the process of measuring the 2. What is the difference between a
diametre of a rod using a micrometre micrometre having 100 divisions
screw gauge with a least count of and one with 50 divisions?
0.01 mm, the sleeve shows a reading 3. Write limitations of using a
of 1.6 cm, and the thimble indicates micrometre screw gauge to measure
48 divisions. Calculate the accurate length.
diametre of the rod. 4. The actual diametre of a lead ball
is 3.21 mm. Determine the reading
Solution that would have been obtained if:
(a) To calculate the accurate diametre of (a) a micrometre screw
the rod using the screw gauge, we gauge was used.
need to consider the contributions (b) a vernier calliper was used.
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Physics Form 1 Final.indd 39 16/10/2024 20:55
