Question
How do you account for (a) positive zero error (b) negative zero error, for calculating correct diameter of wires?
✓
Answer
(a) Positive zero error: If the zero line, marked on circular scale, is below the reference line of the main scale, then there is a positive zero error and the correction is negative.
In the figure 5th circular scale division is coinciding with reference line.
$\therefore$ Correction
$=-$ Coinciding division of C.S. $\times$ L.C.
$=-5 \times 0.001 cm=-0.005 cm$
If the observed diameter is 0.557 cm , then:
Corrected diameter
= Observed diameter + Correction
$=0.557 cm-0.005 cm=0.552 cm$
(b) Negative zero error : If the zero line marked on circular scale, is above the reference line of the main scale, then there is a negative error and the correction is positive.
In the figure, there is 96 th division on the circualr scale which coincides with reference line.
$\therefore$ Correction $-+[ n$ - coinciding division of C.S. $\times$ L.C.]
where n is the total number of circular scale divisions.
$\therefore$ Correction $=+[100-96] \times 0.001 cm$
$=0.004 cm$
If observed diameter is 0.557 cm , then :
Corrected diameter
= Observed diameter + Correction
$=0.557 cm+0.004 cm$
$=0.561 cm$
In the figure 5th circular scale division is coinciding with reference line.
$\therefore$ Correction
$=-$ Coinciding division of C.S. $\times$ L.C.
$=-5 \times 0.001 cm=-0.005 cm$
If the observed diameter is 0.557 cm , then:
Corrected diameter
= Observed diameter + Correction
$=0.557 cm-0.005 cm=0.552 cm$
(b) Negative zero error : If the zero line marked on circular scale, is above the reference line of the main scale, then there is a negative error and the correction is positive.
In the figure, there is 96 th division on the circualr scale which coincides with reference line.
$\therefore$ Correction $-+[ n$ - coinciding division of C.S. $\times$ L.C.]
where n is the total number of circular scale divisions.
$\therefore$ Correction $=+[100-96] \times 0.001 cm$
$=0.004 cm$
If observed diameter is 0.557 cm , then :
Corrected diameter
= Observed diameter + Correction
$=0.557 cm+0.004 cm$
$=0.561 cm$
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