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
If the zero error is $- 0.03\ cm,$ then mean corrected diameter is $........... cm$
| S.No. | $MS\ (cm)$ | $VS$ divisions |
| $(1)$ | $0.5$ | $8$ |
| $(2)$ | $0.5$ | $4$ |
| $(3)$ | $0.5$ | $6$ |
JEE MAIN 2015, Medium
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Lets count $=\frac{0.1}{10}=0.01 \ cm$
$d_1=0.5+8 \times 0.01+0.03=0.61 \ cm$
$d_2=0.5+4 \times 0.01+0.03=0.57 \ cm$
$d_3=0.5+6 \times 0.01+0.03=0.59 \ cm$
Mean diameter $=\frac{0.61+0.57+0.59}{3}=0.59 \ cm$
$d_1=0.5+8 \times 0.01+0.03=0.61 \ cm$
$d_2=0.5+4 \times 0.01+0.03=0.57 \ cm$
$d_3=0.5+6 \times 0.01+0.03=0.59 \ cm$
Mean diameter $=\frac{0.61+0.57+0.59}{3}=0.59 \ cm$
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