Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm.
Reason R :
Least Count = Pitch/(Total divisions on circular scale)
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Physics sir · 4 steps

Find the pitch first: it is the distance moved in ONE rotation, not in five.

1. Find the pitch. Pitch is the distance the screw moves along the main scale in one full rotation. Five rotations cover 5 mm, so one rotation covers 5/5 = 1 mm.

   LC = Pitch/(Divisions on circular scale) = (1 mm)/50 = 0.02 mm

2. Convert to cm. 0.02 mm = 0.002 cm. The assertion says 0.001 cm, so A is wrong.
3. Check the reason. The formula in R is the standard definition of least count of a screw gauge, so R is correct.
4. So. A is not correct but R is correct, which is option (a).

The idea to keep. For a screw gauge, first get pitch = distance moved ÷ number of rotations. Then LC = pitch ÷ circular divisions. Convert mm to cm only at the very end.

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