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 :

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

Tardigrade Admin · Accepted Answer

Reading = Main scale reading +(Vernier Scale div

\times L . C .

)
Least count

= \frac{1 , M.S. division}{Total Vernier divisions}

= \frac{0.1}{10} = 0.01 c m

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 c m;

d_{2} = 0.5 + (4 \times 0.01) = 0.54 c m;

d_{3} = 0.5 + (6 \times 0.01) = 0.56 c m

\overset{ˉ}{d} = \frac{d _{1} + d _{2} + d _{3}}{3} = 0.56 c m

\Rightarrow

Mean corrected diameter

= \overset{ˉ}{d} -

Zero error

= 0.56 - (- 0.03) = 0.59 c m

$0.56 c m$

$0.59 c m$

$0.53 c m$

$0.52 c m$

0.56cm

0.59cm

0.53cm

0.52cm

$0.56 c m$

$0.59 c m$

$0.53 c m$

$0.52 c m$