📘 NUMERICALS (Screw Gauge)
Q10. The pitch of a screw gauge is 0.5 mm and the circular scale is divided into 100 parts. What is the least count?
Formula: L.C = Pitch ÷ Number of divisions
Substitution:
= 0.5 ÷ 100
Calculation:
L.C = 0.005 mm
Substitution:
= 0.5 ÷ 100
Calculation:
L.C = 0.005 mm
Q11. The thimble of a screw gauge has 50 divisions. The screw advances 1 mm when rotated through two revolutions. Find:
(i) Pitch
(ii) Least count
(i) Pitch = Distance ÷ Rotations = 1 ÷ 2 = 0.5 mm
(ii) L.C = Pitch ÷ divisions = 0.5 ÷ 50 = 0.01 mm
(ii) L.C = Pitch ÷ divisions = 0.5 ÷ 50 = 0.01 mm
Q12. The pitch of a screw gauge is 1 mm and its circular scale has 100 divisions. While measuring a wire, the main scale reads 2 mm and the 45th division coincides. Find:
(i) Least count
(ii) Diameter of wire
(i) L.C = 1 ÷ 100 = 0.01 mm
(ii) Formula: Reading = MSR + (CSR × L.C)
= 2 + (45 × 0.01)
= 2 + 0.45 = 2.45 mm
(ii) Formula: Reading = MSR + (CSR × L.C)
= 2 + (45 × 0.01)
= 2 + 0.45 = 2.45 mm
Q13. A screw gauge of least count 0.01 mm measures the diameter of a wire. The main scale reads 1 mm and the thimble shows 27 divisions.
(i) Find the diameter in cm
(ii) If zero error is +0.005 cm, find the corrected diameter
(i) Reading = 1 + (27 × 0.01) = 1.27 mm
Convert to cm = 1.27 mm = 0.127 cm
(ii) Correct reading = 0.127 − 0.005 = 0.122 cm
Convert to cm = 1.27 mm = 0.127 cm
(ii) Correct reading = 0.127 − 0.005 = 0.122 cm
Q14. A screw gauge has 50 divisions on its circular scale and moves 1 mm in two rotations. When closed, the 4th division coincides with reference line. Find:
(i) Pitch
(ii) Least count
(iii) Zero error
(i) Pitch = 1 ÷ 2 = 0.5 mm
(ii) L.C = 0.5 ÷ 50 = 0.01 mm
(iii) Zero error = 4 × 0.01 = 0.04 mm
(ii) L.C = 0.5 ÷ 50 = 0.01 mm
(iii) Zero error = 4 × 0.01 = 0.04 mm
Q15. Two micrometer screw gauges are shown in the figures. Find the actual reading.
(a) L.C = 0.001 cm, zero error = +0.05 mm
📌 Insert Figure Here (Micrometer Diagram)
Formula: Reading = MSR + (CSR × L.C)
Correct reading = Observed − Zero error
(Substitute values from diagram to get final answer)
Correct reading = Observed − Zero error
(Substitute values from diagram to get final answer)
Q16. Fig. 1.15 shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances 1 division on circular scale in one rotation. Find:
(i) Pitch
(ii) Least count
(iii) Diameter
📌 Insert Figure Here (Fig. 1.15)
(i) Pitch = distance moved in one rotation
(ii) L.C = Pitch ÷ number of divisions
(iii) Reading = MSR + (CSR × L.C)
(Substitute values from diagram to get final answer)
(ii) L.C = Pitch ÷ number of divisions
(iii) Reading = MSR + (CSR × L.C)
(Substitute values from diagram to get final answer)
Q17. A screw gauge has pitch 0.5 mm. What should be the number of divisions on its circular scale to measure up to 0.001 mm accuracy?
Formula: L.C = Pitch ÷ divisions
Rearranging:
Divisions = Pitch ÷ L.C
= 0.5 ÷ 0.001 = 500 divisions
Rearranging:
Divisions = Pitch ÷ L.C
= 0.5 ÷ 0.001 = 500 divisions
