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Measurement and Experimentation — PHYSICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9PHYSICSMeasurement and Experimentation4 Marks
Question
A micrometre screw gauge has a negative zero error of 8 divisions. While measuring the diameter of a wire the reading on main scale is 3 divisions and 24th circular scale division coincides with base line.
If the number of divisions on the main scale are 20 to a centimetre and circular scale has 50 divisions, calculate
1. pitch
2. observed diameter.
3. least count
4. corrected diameter.
✓

Answer

(i) The number of divisions on the main scale are 20 to a centimetre
$
\Rightarrow \text { Pitch }=\frac{\text { Unit }}{\text { No. of divisions in unit }}=\frac{1 cm}{20}=0.05 cm
$
(ii) No. of circular scale divisions $=50$
$
\begin{array}{l}
\therefore \text { Least count (L.C.) }=\frac{\text { Pitch }}{\text { No. of circular scale divisions }} \\
\quad=\frac{0.05}{50} cm \\
\text { L.C. }=0.001 cm
\end{array}
$
(iii) Main scale reading $=3$ division
$\Rightarrow$ Main scale reading $=3 \times$ Pitch $=3 \times 0.05=0.15 cm$
Circular scale reading $=24$ division
$\therefore$ Observed diameter $=$ M.S. reading + L.C. $\times$ C.S. reading
$
\begin{array}{l}
=0.15+0.001 \times 24 \\
=0.15+0.024=0.174 cm
\end{array}
$
(iv) Negative zero error $=8$ division
Correction $=-(-8 \times$ L.C. $)$
$
=-(-8 \times 0.001) cm=+0.008 cm
$
Correct diameter $=$ Observed diameter + Correction
$
=0.174+0.008=0.182 cm
$

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