Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as
Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as
Option 1 - <p>2.121 cm</p>
Option 2 - <p>2.123 cm<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>2.124 cm</p>
Option 4 - <p>2.125 cm</p>
7 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
A
Answered by
7 months ago
Correct Option - 1
Detailed Solution:
The least count (LC) of the screw gauge is the pitch divided by the number of divisions on the circular scale.
LC = Pitch / N = 0.1 cm / 50 = 0.002 cm.
A measurement taken with this instrument must be a multiple of its least count. We check the options:
(A) 2.121 / 0.002 = 1060.5 (Not a multiple)
(B)
Similar Questions for you
From A to B the process is isobaric

= W = 2 × R (600 - 350)
= 500 R
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Physics Units and Measurement 2025
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