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
While measuring the length of a rod with a vernier callipers, below shows the position of its scales. What is the length of the rod?

What do you understand by the following term as applied to micrometre screw gauge?
Thimble
How do you account for negative zero error, for calculating the correct diameter of wires?
What is the backlash error?
State whether the following statement is true or false by writing T/F against it.
The least count of a screw gauge can be lowered by increasing the number of divisions on its thimble.
What is the least count in the case of the following instrument?
Spring balance
