Advertisements
Advertisements
Question
A micrometre screw gauge has a negative zero error of 7 divisions. While measuring the diameter of a wire the reading on the main scale is 2 divisions and 79th circular scale division coincides with baseline.
If the number of divisions on the main scale is 10 to a centimetre and circular scale has 100 divisions, calculate
- pitch
- observed diameter
- least count
- corrected diameter.
Advertisements
Solution
(1) The number of divisions on the main scale are 10 to a centimetre
⇒ Pitch = `"Unit"/"No. of divisions in unit"=(1 "cm")/10` = 0.1 cm
(2) No. of circular scale divisions = 100
∴ Least count (L.C.) = `"Pitch"/"No. of circular scale divisions"`
= `0.1/100` cm
= 0.001 cm
(3) Main scale reading = 2 division
⇒ Main scale reading = 2 × Pitch = 0.2 cm
Circular scale reading = 79 division
∴ Observed diametre = M.S. reading + L.C. × C.S. reading
= 0.2 + 0.001 × 79
= 0.2 + 0.079 cm
= 0.279 cm
(4) Negative zero error = 7 division
∴ Correct = −(−7 × L.C.)
= −(−7 × 0.001) cm
= +0.007 cm
Correct diametre = Observed diametre + Correction
= 0.279 + 0.007
= 0.286 cm
APPEARS IN
RELATED QUESTIONS
Name the part of the vernier callipers which is used to measure the following
Thickness of a pencil.
Fig., below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on the main scale when circular head is rotated once.
Find:
(i) Pitch of the screw gauge,
(ii) Least count of the screw gauge and
(iii) The diameter of the wire.
Why is the metre length in terms of the wavelength of light considered more accurate?
Is it possible to increase the degree of accuracy by mathematical manipulations? Support your answer by an example.
The final result of the calculation in an experiment is 125,347,200. Express the number in term of significant place when accuracy is between 1 and 1000
Name the measuring employed to measure the thickness of a paper.
What is the least count in the case of the following instrument?
Stopwatch
