Question

A screw gauge
has 50 divisions on
its circular scale. The pitch
of screw gauge is 0.5 mm.
What is the least
count.​

Answer 1

Answer:

According to the question, the screw gauge has 50 divisions on its circular scale.

According to the question, the screw gauge has 50 divisions on its circular scale.The Pitch is calculated as the screw completes one rotation completely, the distance travelled by the screw defines the pitch.

According to the question, the screw gauge has 50 divisions on its circular scale.The Pitch is calculated as the screw completes one rotation completely, the distance travelled by the screw defines the pitch.Thereby, from given data in question we have movement of 1 mm by 2 rotations

⟹ Pitch = distance / rotation

➡ pitch=1/2

➡ pitch=0.5mm

⟹ Least count is given by the ratio of pitch and the divisions present over the circular scale of screw gauge.

Least count is given by the ratio of pitch and the divisions present over the circular scale of screw gauge.Least Count = pitch / number of division over circular scale

➡ least of count=0.5/50

➡ Least Count=0.01mm

⟹ Zero Error of the screw gauge is calculated by the coinciding division of circular scale with zero and least count.

Zero Error of the screw gauge is calculated by the coinciding division of circular scale with zero and least count.Zero Error = Coinciding division x Least Count

➡ Zero error= +4×0.01 mm

➡ Zero Error=+0.04 mm

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