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प्रश्न
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.
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उत्तर
(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
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संबंधित प्रश्न
Name the instrument which has the least count
0.1 mm
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 100
What is the need for measuring length with vernier callipers?
State the formula for calculating length if:
The reading of the main scale is known and the number of vernier scale divisions coinciding with the main scale is known.
For what range of measurement is micrometre screw gauge used?
What do you understand by the following term as applied to screw gauge?
Positive zero error
State whether the following statement is true or false by writing T/F against it.
A metre scale can measure a length of 6.346 cm.
