- Home
- Standard 11
- Physics
There are $100$ divisions on the circular scale of a screw gauge of pitch $1 \mathrm{~mm}$. With no measuring quantity in between the jaws, the zero of the circular scale lies $5$ divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found the $4$ linear scale divisions are clearly visible while $60$ divisions on circular scale coincide with the reference line. The diameter of the wire is :
$4.65 \mathrm{~mm}$
$4.55 \mathrm{~mm}$
$4.60 \mathrm{~mm}$
$3.35 \mathrm{~mm}$
Solution
$\text { Least count }=\frac{1}{100} \mathrm{~mm}=0.01 \mathrm{~mm}$
$\text { zero error }=+0.05 \mathrm{~mm}$
$\text { Reading }=4 \times 1 \mathrm{~mm}+60 \times 0.01 \mathrm{~mm}-0.05 \mathrm{~mm}$
$=4.55 \mathrm{~mm}$
Similar Questions
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
| 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 ……….. $cm$
