Find the least count of a screw gauge whose least count of linear scale is 1mm and the circular scale is divided in 50 divisions and 2 entire rotations makes to 1 division on linear scale.

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Solution

1mm is least count of the linear scale
the rotary scale has to go through 100 units which is 50 per revolution times 2 revolutions to cover 1 unit of linear scale
therefore the actual least count is 1mm / (50*2) = 1/100 of a mm = 0.01 mm = 0.000001 m

i hope the doubt os clear

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Find the least count of a screw gauge whose least count of linear scale is 1mm and the circular scale is divided in 50 divisions and 2 entire rotations makes to 1 division on linear scale.

1mm is least count of the linear scale the rotary scale has to go through 100 units which is 50 per revolution times 2 revolutions to cover 1 unit of linear scale therefore the actual least count is 1mm / (50*2) = 1/100 of a mm = 0.01 mm = 0.000001 m i hope the doubt os clear

1mm is least count of the linear scale
the rotary scale has to go through 100 units which is 50 per revolution times 2 revolutions to cover 1 unit of linear scale
therefore the actual least count is 1mm / (50*2) = 1/100 of a mm = 0.01 mm = 0.000001 m

i hope the doubt os clear