A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5mm is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively :

Solution

Physics sir · 3 steps

Least count is pitch over divisions, and a circular zero sitting ahead of the line means positive error.

1. Pitch. One full turn of the circular scale moves the screw 0.5 mm along the pitch scale, so the pitch is 0.5 mm.

   LC = pitch/(circular divisions) = (0.5 mm)/50 = 0.01 mm = 10 μm

2. Nature of zero error. With the jaws closed, the circular zero should sit exactly on the reference line. Here it has already gone 4 divisions past it, so every reading comes out too large. That is a positive zero error of 4 × 0.01 = 0.04 mm.
3. So. The zero error is positive and the least count is 10 μm, which is option (c).

The idea to keep. Least count of a screw gauge is pitch divided by circular divisions. If the circular zero is ahead of the reference line at closed jaws, readings are too big, so the zero error is positive: subtract it.

Answer from the NTA final answer key. Solution by PrepWiser.