Question

Diameter of a steel ball is measured using a Vernier callipers which has divisions of $0.1\, cm$ on its main scale (MS) and $10$ divisions of its vernier scale (VS) match $9$ divisions on the main scale. Three such measurements for a ball are given as :

SI.NO.	MS (cm)	VS Divisions
1	0.5	8
2	0.5	4
3	0.5	6

If the zero error is — $0.03\, cm$, then mean corrected diameter is :

• A. $0.56\, cm$
• B. $0.59\, cm$
• C. $0.53\, cm$
• D. $0.52\, cm$

Answer 1

Answer: (B)

Reading = Main scale reading +(Vernier Scale div$ \times L.C.$)
Least count $=\frac{1, \text { M.S. division}}{\text {Total Vernier divisions}}$
$=\frac{0.1}{10}=0.01\, cm$
Let $d_{1}, d_{2}$ and $d_{3}$ be the three readings of the diameter
$d_{1}=0.5+(8 \times 0.01)=0.58\, cm ;$
$d_{2}=0.5+(4 \times 0.01)=0.54\, cm ;$
$d_{3}=0.5+(6 \times 0.01)=0.56\, cm$
$\bar{d}=\frac{d_{1}+d_{2}+d_{3}}{3}=0.56\, cm$
$\Rightarrow $ Mean corrected diameter
$=\bar{d}-$ Zero error $=0.56-(-0.03)=0.59\, cm$