Advertisements
Advertisements
प्रश्न
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
उत्तर
(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
संबंधित प्रश्न
Name the part of the vernier callipers which is used to measure the following
External diameter of a tube
Explain the terms : Least count of a screw gauge.
How are they determined?
In a vernier callipers, 19 main scale divisions coincide with 20 vernier scale divisions. If the main scale has 20 divisions in a centimetre, calculate
- pitch
- L.C. of vernier callipers.
State the correction if the positive error is 7 divisions when the least count is 0.01 cm.
What do you understand by the following term as applied to screw gauge?
Zero error
Find the volume of a book of length 25 cm, breadth 18 cm and height 2 cm in m3.
In a vernier calliper, (N + 1) divisions of the vernier scale coincide with N divisions of the main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is ______.
