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
Describe in steps, how would you use a vernier callipers to measure the length of a small rod?
Name the instrument which can accurately measure the following
The internal diameter of the neck of a water bottle
The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions.
(i) What is the pitch of the screw gauge?
(ii) What is the least count of the screw gauge?
The final result of the calculation in an experiment is 125,347,200. Express the number in term of significant place when accuracy is between 1 and 1000
In figure for vernier callipers, calculate the length recorded.
When does a vernier callipers has the negative error?
What do you understand by the following term as applied to screw gauge?
Positive zero error
The level of water in a measuring cylinder is 12.5 ml. When a stone is lowered in it, the volume is 21.0 ml. Find the volume of the stone.
State whether true or false. If false, correct the statement.
With the help of vernier caliper we can have an accuracy of 0.1 mm and with screw gauge we can have an accuracy of 0.01 mm.
