The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72^nd division on circular scale coincides with the reference line. The radius of the wire is :

CONCEPT:

A screw gauge is defined as an instrument that measures the diameter or thickness of the wire accurately. The least count of the screw gauge is 0.01 mm and the total reading in the screw gauge is calculated as

Total reading = Linear scale reading + circular scale reading \(×\) Least count

CALCULATIONS:

Given : Pitch = 1 mm

Number of divisions, N = 100

Least count, LC = \(\frac{pitch}{N}\)

Now, on putting the given values we have;

Least count, LC = \(\frac{1 mm}{100}\)

= 0.01 mm

Here the instrument is in positive zero error, therefore we have;

e = + n × LC

⇒ e = + 8 × 0.01

⇒ e =+ 0.08 mm

The linear scale reading = 1 × 1 mm

= 1 mm

The circular scale reading = 72 × 0.01 mm

= 0.72 mm

Therefore, the measured reading is = 1 + 0.72

= 1.72 mm

The true reading is written as = 1.72 - 0.08

= 1.64 mm

Hence option 1) is the correct answer.