A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm; 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as:

Option 1 - <p>(5.5375 ± 0.0739)mm<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>(5.5375 ± 0.0740)mm<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>(5.538 ± 0.074)mm<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>(5.54 ± 0.07)mm</p>
3 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
5 months ago
Correct Option - 4
Detailed Solution:

Since significant figures show the degree of correctness of any measurement, so in any mathematical calculation we cannot increase the number of significant digits. Because the four reading has 3 significant digits so the answer should also have 3 significant digits only.

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Physics Units and Measurement 2025

Physics Units and Measurement 2025

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