Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as

Solution

Physics sir · 3 steps

Find the least count first: a correct reading must be a whole number of least counts.

1. Least count. One full turn moves the screw by one pitch, 0.1 cm, and that turn is split into 50 circular divisions.

   LC = pitch/(circular divisions) = (0.1 cm)/50 = 0.002 cm

2. Reading formula. Reading = main scale reading + (circular division number) × LC. The circular division is a whole number, so the part after 2.1 cm must be an exact multiple of 0.002 cm.

   Option	Part beyond 2.1 cm	Divided by 0.002 cm	Possible?
   (a) 2.123 cm	0.023 cm	11.5	no
   (b) 2.124 cm	0.024 cm	12	yes
   (c) 2.125 cm	0.025 cm	12.5	no
   (d) 2.121 cm	0.021 cm	10.5	no

3. So. Only 2.124 cm matches a whole circular division (the 12th), so the thickness is correctly recorded as 2.124 cm, option (b).

The idea to keep. Every reading from a screw gauge or vernier is a whole multiple of its least count. Work out the LC, then reject any value that would need a fractional division.

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