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.


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Answer 1

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SOLUTION

GIVEN

• DiViSiON = 20
• Screw moves 1 mm ahead in 2 rotation.

CALCULATION

• Pitch = Distance / No. of Rotation
• Pitch = 1mm / 2
• Pitch = 0.5 mm

• Least Count = Pitch / No. of Division
• Least Count = 0.5 / 20
• Least Count = 0.025 mm

RESULTS

• Least Count of screw gauge is 0.025 mm

Answer 2

Answer:

SOLUTION

GIVEN

DiViSiON = 20

Screw moves 1 mm ahead in 2 rotation.

CALCULATION

Pitch = Distance / No. of Rotation

Pitch = 1mm / 2

Pitch = 0.5 mm

Least Count = Pitch / No. of Division

Least Count = 0.5 / 20

Least Count = 0.025 mm

RESULTS

Least Count of screw gauge is 0.025 mm