Question

The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to

Answer 1

Answer:

200 divisions

Step-by-step explanation:

Question being incomplete, we assume that the specimen being measured by gauge is a 5 μm wire, thus calculating,

The minimum number of divisions on its scale = ?

Least count = Pitch / Number of divisions on the circular scale

Least count = 5 x 10⁻⁶ m

Pitch = 10⁻³ m

Number of divisions = Pitch / least count = 10⁻³ / 5 x 10⁻⁶ = 200 divisions