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PUMPA - SMART LEARNING

எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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Subjects
Science State Board
Class 9
Measurement
Textbook questions

13. Answer in detail - I

Exercise condition:

8 m.

Explain a method to find the thickness of a hollow tea cup.

To find the thickness of a hollow teacup,

Determine the pitch, the least count and the zero error of the screw gauge.

Pitch:

\(Pitch\ of\ the\ screw\ =\ \frac{Distance\ moved\ by\ the\ Main\ scale}{No.\ of\ rotations\ Head\ scale}\)
\(Pitch\ of\ the\ screw\ =\ \frac{Distance\ moved\ by\ the\ pitch}{No.\ of\ rotations\ by\ Head\ scale}\)
\(Pitch\ of\ the\ screw\ =\ \frac{Distance\ moved\ by\ the\ Head\ scale}{No.\ of\ rotations\ by\ pitch}\)

Least count:

\(Least count\ (LC)\ =\) \(mm\)

Zero error:

\(Positive\ zero\ error\ (ZE)\ =\)

\(+\ (n \times LC)\ mm\)
\(+\ (\frac {n}{LC})\ mm\)

\(=\)

\(+\ (\frac {n}{0.01})\ mm\)
\(+\ (\frac {n}{0.1})\ mm\)
\(+\ (n \times 0.01)\ mm\)
\(+\ (n \times 0.1)\ mm\)

So, \(Zero\ correction\ (ZC)\ =\)

\(-\ (\frac {n}{0.01})\ mm\)
\(+\ (\frac {n}{0.1})\ mm\)
\(-\ (n \times 0.01)\ mm\)
\(+\ (n \times 0.1)\ mm\)

\(Negative\ zero\ error\ (ZE)\ =\)

\(-\ ({(100-n)} \times 0.01)\ mm\)
\(-\ (\frac {(100-n)}{0.01})\ mm\)
\(-\ (\frac {(100-n)}{0.1})\ mm\)
\(-\ ({(100-n)} \times 0.1)\ mm\)

So, \(Zero\ correction\ (ZC)\ =\)

\(-\ (\frac {(100-n)}{0.01})\ mm\)
\(-\ (\frac {(100-n)}{0.1})\ mm\)
\(-\ ({(100-n)} \times 0.01)\ mm\)
\(-\ ({(100-n)} \times 0.1)\ mm\)

Place the teacup between the two studs.
Rotate the head until the teacup is held firmly but not tightly, with the help of the ratchat.
Observe the reading of the pitch scale crossed by the head scale (PSR) and the head scale division that coincides with the pitch scale axis (HSC).
The thickness of the teacup is given by . Repeat the experiment for different positions of the teacup.
Tabulate the readings.
The average of the last column readings gives the thickness of the teacup.

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