Question

The circular head of the screw gauge is divided into 20 divisions and the screw moves 1 mm ahead in two rotations of the head of the screw. Find its least count
Plz answer fast plz​

Answer 1

Answer :-

Least count of the screw gauge is 0.025 mm .

Explanation :-

We have :-

→ Number of divisions = 20

→ Distance moved = 1 mm

→ Number of rotations = 2

To find :-

→ Least count of the screww gauge .

______________________________

Firstly, let's calculate the pitch of the screw gauge from the given values.

Pitch = Distance moved/No of rotations

⇒ Pitch = 1/2

⇒ Pitch = 0.5 mm

Now, we know that least count of a screw gauge is given by :-

Least count = Pitch/No of divisions

⇒ Least count = 0.5/20

⇒ Least count = 5/200

⇒ Least count = 0.025 mm

Answer 2

Given :-

The circular head of the screw gauge is divided into 20 divisions and the screw moves 1 mm ahead in two rotations of the head of the screw.

To Find :-

Least count

Solution :-

Screw is moving 1mm in two rotation

So, in a rotation it will move = 1/2 = 0.5 mm

Now

Finding the least count

Least count = Distance covered in 1 rotation/Total division in it

Least count = 0.5/20

Least count = 5/20 × 1/10

Least count = 5/200

Least count = 0.025 mm

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