Q4. A vernier has 50 divisions equal to 49 main-scale divisions of 0.5 mm. Find its least count.
LC = 1 MSD/n = 0.5 mm/50 = 0.01 mm.
Q5. A screw gauge of least count 0.01 mm shows 5 mm on the pitch scale and 27 on the circular scale. What is the observed reading?
5 + 27 × 0.01 = 5.27 mm.
Q6. The screw gauge of Q5 has a zero error of −0.04 mm. What is the corrected reading?
Corrected = observed − zero error = 5.27 − (−0.04) = 5.31 mm.