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Question
In the following figure, the pitch of the screw is 1 mm. Calculate:
(i) the least count of screw gauge and
(ii) the reading represented in the figure.
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Solution
Screw gauge in the figure
(i) Pitch of the screw = 1 mm
Divisions on circular scale= 100
least count = pitch of the screw/ divisions on the main scale
L.C. = 1/100 = 0.01 mm= 0.001 cm
(ii) LC. = 0.001 cm
the main scale reading = 2 mm= 0.2 cm
from the figure it is clear that base line coincides with the 97th division of the circular scale.
So, circular scale reading= 97 X 0.001 cm = 0.097 cm Hence, reading shown In the figure= 0.2 + 0.097 = 0.297 cm
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With the help of vernier caliper we can have an accuracy of 0.1 mm and with screw gauge we can have an accuracy of 0.01 mm.
