Question

The circular head of a screw gauge is divided into 100 divisions and the screw moves 1mm ahead in two revolutions of circular head find its
i. pitch
ii. least count​

Answer 1

Answer:

refer to the attachment above for ur answer

Explanation:

hope it helps you ^_^

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Answer 2

Answer :-

• Least count of the screw gauge is 0.005 mm .

Explanation :-

We have :-

→ Number of divisions = 100

→ 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 = 0.5/100

⇒ Least count = 5/1000

⇒ Least count = 0.005 mm