Question

Diameter of a steel ball is measured using a Vernier cllipers 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 measurement for a ball are given as.

S. 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
• B. 0.53
• C. 0.52
• D. 0.59

Answer 1

The correct option is D 0.59

The LC of vernier calipers is given as $L.C.=1M.S.D.-1V.S.D.$
Where MSD and VSD are main and vernier scale division:
$MSD=0.1$ cm

Given $10VSD$ coincides with $9MSD$
$10VSD=9MSD\Rightarrow VSD=0.9MSD=0.09cm$

Least count is given as $0.1cm$$-0.09cm$ $=0.01cm$

The reading of scale is given as Main Scale reading + L.C.$\times VSD-error$
Reading 1 $=0.5cm+0.01\times 8-(-0.03)cm$ $=0.61cm$
Reading 2 $=0.5cm+0.01\times 4-(-0.03)cm$ $=0.57cm$
Reading 3 $=0.5cm+0.01\times 6-(-0.03)cm$ $=0.59cm$

Taking average of three readings diameter $=\frac{0.61+0.57+0.59}{3}=0.59cm$