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In Fig. the pitch of the screw is 1 mm....

In Fig. the pitch of the screw is 1 mm. Find : (i) the least count of screw gauge and (ii) the reading represented in the diagram.

Text Solution

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(i) Given, pitch of the screw = 1 mm
Number of divisions on the circular scale = 100
`therefore` L.C. of the screw gauge
`=("Pitch")/("Number of divisions on circular scale") =(1 mm)/(100)`
`=0.01 ` mm or 0.001 cm.
(ii) Main scale reading = 2 mm = 0.2 cm
Since 97th division of head scale coincides with base line, i.e., p = 97
Circular scale reading = `p xx L.C.`
`= 97 xx 0.001 cm= 0.097 cm`
Total reading = main scale reading + circular scale reading
`=0.2 cm + 0.097 cm = 0.297cm
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