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Subjects
Science State Board
Class 9
Measurement
Textbook questions

14. Answer in detail - II

Exercise condition:

8 m.

1. How will you find the thickness of a one rupee coin?

Solution:

To find the thickness of a one rupee coin,

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:

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

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

$=$

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

So,

$Zero\ correction\ (ZC)\ =$

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

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

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

So,

$Zero\ correction\ (ZC)\ =$

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

Place the coin between the two studs.
Rotate the head until the coin 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 coin is given by . Repeat the experiment for different positions of the coin.
Tabulate the readings.
The average of the last column readings gives the thickness of the coin.

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