🔢 Vernier Caliper & Screw Gauge Numericals

🔢 Vernier Callipers & Screw Gauge Numericals

Q1: A Vernier caliper has 10 divisions on the Vernier scale which match with 9 divisions of the main scale. Find the least count.

LC = (1 MSD - 1 VSD) = (1 mm - 0.9 mm) = 0.1 mm

Q2: Main scale reading = 2.3 cm; Vernier scale division = 6; LC = 0.01 cm. Find total reading.

Total = 2.3 + (6 × 0.01) = 2.36 cm

Q3: Screw gauge pitch = 1 mm, divisions = 100. What is the least count?

LC = Pitch / Divisions = 1 mm / 100 = 0.01 mm

Q4: Pitch = 0.5 mm, Divisions = 50. Calculate the LC of the screw gauge.

LC = 0.5 mm / 50 = 0.01 mm

Q5: Main scale reading = 3.0 mm, circular scale reading = 47, LC = 0.01 mm. Find total reading.

Total = 3.0 + (47 × 0.01) = 3.47 mm

Q6: Zero error of Vernier = -0.02 cm. Total reading = 2.34 cm. Find corrected reading.

Corrected = 2.34 + 0.02 = 2.36 cm

Q7: Screw gauge zero error = +0.01 mm. Measured length = 2.58 mm. Correct reading?

Corrected = 2.58 - 0.01 = 2.57 mm

Q8: A wire is measured using a screw gauge. Main scale = 1.5 mm, Circular scale = 25, LC = 0.01 mm.

Total = 1.5 + (25 × 0.01) = 1.75 mm

Q9: In a Vernier caliper, 1 VSD = 0.09 cm, 1 MSD = 0.1 cm. Find the least count.

LC = 1 MSD - 1 VSD = 0.1 - 0.09 = 0.01 cm

Q10: Circular scale = 35, LC = 0.01 mm, Zero error = -0.02 mm. Main scale = 2.5 mm.

Total = 2.5 + 0.35 - 0.02 = 2.83 mm