Diameter of a steel ball is measured using a Vernier calipers 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:<p></p> <table> <thead> <tr> <th>S. No.</th> <th>MS(cm)</th> <th>VS divisions</th> </tr> </thead> </table> <ol> <li><pre><code> | 0.5 | 8 </code></pre> </li> <li><pre><code> | 0.5 | 4 </code></pre> </li> <li><pre><code> | 0.5 | 6 </code></pre> </li> </ol> <p>If the zero error is -0.03 cm, then the mean corrected diameter is?</p>

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Question:

Diameter of a steel ball is measured using a Vernier calipers 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:

S. No.	MS(cm)	VS divisions

```
 | 0.5     | 8
```
```
 | 0.5     | 4
```
```
 | 0.5     | 6
```

If the zero error is -0.03 cm, then the mean corrected diameter is?

0.53

0.56

0.59

0.52

Solution:

The LC of vernier calipers is given as
L.C.=1M.S.D.×V.S.D.
Where MSD and VSD are main and vernier scale division:
MSD=0.1cm
VSD=9/10=0.9cm
L.C.=0.1×10/9=0.01cm
Measurement 1:
Diameter = 0.5cm + 8×0.01cm = 0.58cm
Corrected diameter = 0.58cm + (-0.03)cm = 0.55cm
Measurement 2:
Diameter = 0.5cm + 4×0.01cm = 0.54cm
Corrected diameter = 0.54cm + (-0.03)cm = 0.51cm
Measurement 3:
Diameter = 0.5cm + 6×0.01cm = 0.56cm
Corrected diameter = 0.56cm + (-0.03)cm = 0.53cm
Mean corrected diameter = (0.55+0.51+0.53)/3 = 0.53cm