Advertisements
Advertisements
प्रश्न
Figure shows a screw gauge in which circular scale has 100 divisions. Calculate the least count and the diameter of a wire.

Advertisements
उत्तर
No. of circular scale division = 100
Pitch =0.5 mm
= `(0.5 "mm")/100`
= 0.005 mm
L.C. = 0.005 mm = 0.0005 cm
Main scale reading = 4.5 mm = 0.45 cm
(C.S.D.) circular scale reading = 73 division
Observed diameter of wire = Main scale reading + L.C. × C.S.D.
= 0.45 + 0.0005 × 73
= 0.45 + 0.0365
= 0.4865 cm
APPEARS IN
संबंधित प्रश्न
Define the term 'Vernier constant'.
What do you mean by zero error of a screw gauge? How is it accounted for?
The least count of a vernier calipers is :
The diameter of a thin wire can be measured by
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.
What do you understand by the following term as applied to micrometre screw gauge?
Thimble scale
What is the least count in the case of the following instrument?
Spring balance
On a vernier calliper, 25 vernier scale divisions are equal in length to 24 main scale divisions. One main scale division is 1 mm. Find the least count of the instrument.
