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प्रश्न
A micrometre screw gauge having a positive zero error of 5 divisions is used to measure diameter of a wire, when reading on the main scale is 3rd division and the 48th circular scale division coincides with baseline. If the micrometer has 10 divisions to a centimetre on the main scale and 100 divisions on a circular scale, calculate
- Pitch of screw
- Least count of screw
- Observed diameter
- Corrected diameter.
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उत्तर
(1) Micrometre has 10 divisions in one centimetre on the main scale.
∴ Pitch = `"Unit"/"No. of divisions in unit"`
Pitch = `(1 "cm")/10`
= 0.1 cm
(2) No. of circular scale divisions = 100
∴ Least count (L.C.) = `"Pitch"/"No. of circular divisions"`
= `(0.1 "cm")/100`
= 0.001 cm
(3) Reading on main scale is 3rd division.
⇒ Main scale reading = 3 mm = 0.3 cm.
Circular scale reading = 48 div.
Observed diameter = M.S. reading + L.C. × C.S. reading
= 0.3 + 0.001 × 48
= 0.3 + 0.048
= 0.348 cm
(4) Positive zero error = +5 divisions
∴ Correction = −(Error × L.C.)
= −(5 × 0.001)
= −0.005 cm
∴ Corrected diameter = Observed diameter + Correction
= 0.348 − 0.005
= 0.343 cm
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