Question

The pitch of a screw gauge is $1mm$ and there are $100$ divisions on its circular scale. When nothing is put in between its jaws, the zero of the circular scale lies $6$ divisions below the reference line. When a wire is placed between the jaws, $2$ linear scale divisions are clearly visible while $62$ divisions on circular scale coincide with the reference line. Determine the diameter of the wire.

• A. Measured reading is 2.56 mm.
• B. Measured reading is 2.62 mm.
• C. True reading is 2.56 mm.
• D. True reading is 2.62 mm.

Answer 1

Answer: (B), (C)

Measured reading is 2.62 mm.
True reading is 2.56 mm.

$LC=\frac{pitch}{N}=\frac{1mm}{100}=0.01mm$
The instrument has a positive zero error.
$e=+n\left(LC\right)=+(6\times 0.01)=+0.06mm$
Lincar scale reading $=2\times \left(1mm\right)=2mm$
Circular scale reading $=62\times \left(0.01mm\right)=0.62mm$
$\therefore$ Measured reading $=2+0.62=2.62mm$
or, True reading $=2.62-0.06$$=2.56mm$