Advertisements
Advertisements
Question
A micrometre screw gauge has a positive zero error of 7 divisions, such that its main scale is marked in 1/2 mm and the circular scale has 100 divisions. The spindle of the screw advances by 1 division complete rotation.
If this screw gauge reading is 9 divisions on the main scale and 67 divisions on the circular scale for the diameter of a thin wire, calculate
- Pitch
- L.C.
- Observed diameter
- Corrected diameter
Advertisements
Solution
(1) Pitch = `1/2` mm
= 0.5 mm
= 0.05 cm
(2) No. of circular scale divisions = 100
∴ Least count (L.C.) = `"Pitch"/"No. of circular scale divisions"`
L.C. = `(0.05 "cm")/100`
= 0.0005 cm
(3) Main scale reading = 9 divisions = `9xx1/2` mm
= 4.5 mm
= 0.45 cm
Circular scale reading = 67 div.
∴ Observed diameter = M.S. reading + L.C. × C.S. reading
= 0.45 + 0.0005 × 67
= 0.45 + 0.0335
= 0.4835 cm
(4) Positive zero error = 7 divisions
Correction = −(Error × L.C.)
= −(7 × 0.0005) cm
= −0.0035 cm
∴ Corrected diameter = Observed diameter + Correction
= 0.4835 + (−0.0035)
= 0.4835 − 0.0035
= 0.4800 cm
APPEARS IN
RELATED QUESTIONS
A vernier callipers has a zero error of + 0.06 cm. Draw a neat labelled diagram to represent it.
Name the two scales of a vernier callipers and explain how it is used to measure length correct up to 0.01 cm.
What do you mean by zero error of a screw gauge? How is it accounted for?
(a) A vernier scale has 20 divisions. It slides over the main scale, whose pitch is 0.5 mm. If the number of divisions on the left hand of the zero of vernier on the main scale is 38 and the 18th vernier scale division coincides with the main scale, calculate the diameter of the sphere, held in the jaws of vernier callipers.
(b) If the vernier has a negative error of 0.04 cm, calculate the corrected radius of the sphere.
Define the term least count as applied to a vernier callipers.
The circular scale of a screw gauge has 100 divisions. Its spindle moves forward by 2.5 mm when given five complete turns. Calculate
- pitch
- least count of the screw gauge
A micrometre screw gauge has a negative zero error of 8 divisions. While measuring the diameter of a wire the reading on the main scale is 3 divisions and the 24th circular scale division coincides with baseline.
If the number of divisions on the main scale are 20 to a centimetre and circular scale has 50 divisions, calculate
- pitch
- observed diameter.
- least count
- corrected diameter.
Consider the following case where the zero of vernier scale and the zero of the main scale are clearly seen. If L.C. of the vernier calipers is 0.01 cm, write the zero error and zero correction of the following.
