Question

A screw gauge has a pitch 0.5 mm. Find the number of divisions on circular scale to read correct up to 0.001mm
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Answer 1

here is your ans. so simple

Answer 2

Pitch of Screw Gauge=0.5 mm

Let Circular Scale Division be=x

We have to find the Total CSD so that We can read Measurement upto 0.001 mm

Least Count Should be =0.001 mm

As We Know.

$\mathbf{LC=\frac{Pitch}{Number\: of \:CSD}}$

$0.001 = \frac{0.5}{x} \\ \\ 0.001x = 0.5 \\ \\ x = \frac{0.5}{0.001} \\ \\ x = \frac{5}{10} \div \frac{1}{1000} \\ \\ x = \frac{5}{10} \times 1000 \\ \\ x = 500$

C.S.D=500

Graduation Provided on main Scale is Pitch.

Like when Screw show 6mm on main Scale after rotation of 6 circular Scale.

Pitch=Distance moved by main scale÷Distance moved on Circular Scale

=6÷6

=1 mm

Means Graduation on main Scale is 1mm