Advertisements
Advertisements
प्रश्न
The pitch of a screw gauge is 1 mm and the circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on the circular scale coincides with the base line. Find
(i) The least count and
(ii) The diameter of the wire.
Advertisements
उत्तर
Pitch of the screw gauge = 1mm
No. of divisions on the circular scale = 100
(i) L.C. = Pitch/No. of divisions on the circular head
= (1/100) mm
= 0.01 mm or 0.001 cm
(ii) Main scale reading = 2mm = 0.2 cm
No. of division of circular head in line with the base line (p) = 45
Circular scale reading = (p) × L.C.
= 45 x 0.001 cm
= 0.045 cm
Total reading = M.s.r. + circular scale reading
= (0.2 + 0.045) cm
= 0.245 cm
APPEARS IN
संबंधित प्रश्न
Name the part of the vernier callipers which is used to measure the following
Thickness of a pencil.
Name the instrument which has the least count
0.1 mm
The main scale of vernier callipers has 10 divisions in a centimetre and 10 vernier scale divisions coincide with 9 main scale divisions. Calculate
- pitch
- L.C. of vernier callipers.
In a vernier callipers, 19 main scale divisions coincide with 20 vernier scale divisions. If the main scale has 20 divisions in a centimetre, calculate
- pitch
- L.C. of vernier callipers.
State the formula for determining a pitch
