Page 42 - Physics

Physics for Secondary Schools
Reading
= Main scale Reading + Vernier coincidence
× LC
= 3.4 cm + 6 × 0.01cm
= 3.4 cm + 0.06 cm
= 3.46 cm.
Corrected reading
= reading – zero error
= 3.46 cm – (-0.03 cm)
= 3.54 cm + 0.03 cm
= 3.57 cm.
Exercise 2.1
1. Draw both the main and vernier
scales of a vernier calliper to
show a reading of 0.36 cm.
2. What is the reading shown in
Figure 2.14?
Figure 2.14
3. (a) What is the signifi cance of the
design of the vernier calliper and
its readings when measuring
the depth of the beaker, and
how does the alignment of
vernier scale divisions with the
main scale affect the accuracy
of the measurement?
(b) In a scenario where a thin
metallic strip on a vernier
36
Physics Form 1 Final.indd 36
calliper descends from the upper edge
to the lower edge and makes contact
with the surface of a beaker, the main
scale displays a reading of 6.4 cm, with
a vernier constant of 0.1 mm, and the 4 th
vernier scale division aligns precisely
with a main scale division. Determine
the precise depth of the beaker, assuming
there is no zero end error.
Acti vity 2.2
Aim: To measure the external
diametre of a test tube using
a vernier calliper.
Materials: Test tube and vernier calliper
Procedure
1. Close the jaws of the vernier calliper
and observe the zero cm mark of the
main and vernier scales. Record your
observations.
2. Open the jaws and place the test tube
between them.
3. Close the jaws gently until the test tube
is held fi rmly. Record your observations.
4. Repeat steps 1 to 3 using other parts
along the test tube.
5. Obtain three readings. Record your
observations.
Questions
(a) Calculate the average of the readings
obtained.
(b) Were the four obtained readings equal?
(c) Why was it important to take four
readings and not one?
Student’s Book Form One
16/10/2024 20:55