Advertisements
Advertisements
Question
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
Advertisements
Solution
Number of circular scale divisions (C.S.D.) = 50
Distance moved by screw (spindle) on sleeve = 2 mm
Number of complete rotations given = 4
(1) Pitch = `"Distance moved by screw on sleeve"/"No. of complete rotations"`
= `(2 "mm")/4`
= 0.5 mm
(2) Pitch = 0.05 cm
Least count = `"Pitch"/"No. of circular scale divisions"`
= `0.05/50`
= 0.001 cm
APPEARS IN
RELATED QUESTIONS
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 correction if the positive error is 7 divisions when the least count is 0.01 cm.
Which part of vernier callipers is used to measure the internal diameter of a hollow cylinder?
Which part of vernier callipers is used to measure the internal length of a hollow cylinder?
Figure shows a screw gauge in which circular scale has 100 divisions. Calculate the least count and the diameter of a wire.

What do you understand by the following term as applied to micrometre screw gauge?
Thimble
State whether the following statement is true or false by writing T/F against it.
The diameter of a wire can be measured more accurately by using vernier calipers than by a screw gauge.
How will you measure the least count of vernier caliper?
