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Division on the main scale when the circular head is rotated once. Calculate:(i) pitch of the screw gauge,(ii) Least count of the screw gauge, and (iii) The diameter of the wire.

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Question

The following Figure shows the reading obtained while measuring the diameter of a wire using a screw gauge. The screw advances by 1 division on the main scale when the circular head is rotated once. Calculate:
(i) the pitch of the screw gauge,
(ii) Least count of the screw gauge, and

(iii) The diameter of the wire.

Answer in Brief
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Solution

Since the screw advances by 1 division on the main scale when the circular head is rotated once
So, the pitch of the screw = 1mm
The number of divisions= 50
Least count = pitch/number of divisions

= 1/50 = 0.02 mm
Now for the reading shown in the figure,
The main scale reading = 4 mm

From the figure, it is clear that base Line coincides with the 47th division of the circular scale.
So, circular scale reading = 47 x 0.02 mm = 0.94 cm

Hence, reading shown in the figure = 4+0.94 = 4.94 cm

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Chapter 1: Measurement - Exercise 2 [Page 30]

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Frank Physics [English] Class 9 ICSE
Chapter 1 Measurement
Exercise 2 | Q 28 | Page 30
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