Distance travelled by linear scale of screw gauge during two full rotation of circular scale is 1mm and circular scale has 50 divisions. In an experiment to measure thickness of a plate, six divisions of main scale are clearly visible and 28th division of circular scale coincides with reference line. Moreover when studs touch each other, zero of circular scale lies 4 division below the reference line, thickness of the plate will be:

Option 1 - <p>3.28 mm<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>2.74 mm<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>3.32 mm<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>3.24 mm</p>
13 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago
Correct Option - 4
Detailed Solution:

Pitch p = 0.5 mm
Device has +ve zero error
R = 6p + 28p/50 - 4p/50 = 6p + 24p/50; 3mm + (24/50) * 0.5 = 3.24 mm

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Physics Units and Measurement 2025

Physics Units and Measurement 2025

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