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
Draw a neat and labelled diagram of a screw gauge.
Name its main parts and state their functions.
A screw has a pitch equal to 0.5 mm. What should be the number of divisions on its head so as to read correctly up to 0.001 mm with its help?
The circular scale of a screw gauge has 50 divisions. Its spindle moves by 2 mm on the sleeve, when given four complete rotations calculate
- pitch
- least count
Figure shows a screw gauge in which circular scale has 200 divisions. Calculate the least count and radius of the wire.

Find the volume of a book of length 25 cm, breadth 18 cm and height 2 cm in m3.
Name the measuring employed to measure the diameter of a needle.
What is the least count in the case of the following instrument?
vernier calipers
What is the least count in the case of the following instrument?
Thermometer
An instrument that is used to measure the diameter of a cricket ball is ______.
