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
below:If the zero error is – 0.03 cm, then mean corrected diameter is
(a) 0.52 cm (b) 0.59 cm
(c) 0.56 cm (d) 0.53 cm

Answer 1

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

below :

S NO. main scale VS divisions

1. 0.5 8

2. 0.5 4

3. 0.5 6

To find : if the zero error is -0.03 cm then mean of corrected diameters is ...

solution : given 10 divisions of its Vernier scale match 9 divisions on the main scale.

i.e., 10 VSD = 9 MSD

so, 1 VSD = 0.9 MSD

so, least count = MSD - VSD

= MSD - 0.9 MSD = 0.1 MSD

= 0.1 × 0.1 cm

= 0.01 cm

now the reading of scale is given by,

main scale reading + LC × VSD - zero error

1st reading, 0.5 + 0.01 × 8 - (-0.03) = 0.61 cm

2nd reading, 0.5 + 0.01 × 4 - (-0.03) = 0.57 cm

3rd reading, 0.5 + 0.01 × 6 - (-0.03) = 0.59 cm

Now the mean = (0.61 + 0.57 + 0.59)/3

= 0.59 cm

Therefore the mean of corrected diameters is 0.59 cm