Page 41 - Physics

Measurement
Use the screw clamp to secure the object
in position, shown in Figure 2.12.
Figure 2.12: Measuring the diametre of
a steel ball bearing
4. Read and record the reading on the main
scale which is to the left of the zero cm
mark of the vernier scale. This value is
to the nearest tenth of a centimetre.
5. Observe along the vernier scale and
record the mark which coincides with
a mark on the main scale.
6. Read and note down the value on the
vernier scale. This gives the digit in the
hundredth place of the measurement.
7. Add the values in steps 4 and 6 to get
your correct reading. Therefore, the
length of the object = main scale reading
+ vernier scale reading.
Example 2.1
Using Figure 2.13, determine the
diametre of the object that is placed
between the jaws of the vernier calliper.
Main scalein scale
Ma
vernier scale
vernier scale
Figure 2.13
Physics Form 1 Final.indd 35
Solution
From Figure 2.13,
Main scale reading = 3.1 cm
Vernier scale reading = 2 divisions ×
0.01 cm = 0.02 cm
Total reading
= Main scale reading + Vernier scale reading
= 3.1 cm + 0.02 cm
= 3.12 cm
The total reading = 3.12 cm
Therefore,
the diametre of the object = 3.12 cm.
Example 2.1
The jaws of the vernier calliper make
contact with the inner wall of the
calorimetre without exerting any
pressure and the zero scale of the vernier
reading on the main scale as 3.4 cm,
the 6 vernier scale division coinciding
th
with a main scale division, and a vernier
constant of 0.01 cm.
(a) How might the presence of a negative
zero error in the vernier scale of the
calliper impact the determination of
the actual internal diametre of the
calorimeter?
(b) Calculate the precise internal
diametre of the calorimetre when
we notice that the vernier scale has
a negative zero error of -0.03 cm.
Solution
(a) To calculate the actual internal
diametre of the calorimetre while
accounting for the negative zero
error in the vernier scale.
(b) Main scale reading = 3.4 cm
Least count (L.C) = 0.01 cm
Vernier coincidence = 6
35
16/10/2024 20:55