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
Draw a neat and labelled diagram of a screw gauge.
Name its main parts and state their functions.
The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of a screw gauge?
Why is the metre length in terms of the wavelength of light considered more accurate?
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.
State the formula for determining a pitch
A micrometre screw gauge has a negative zero error of 7 divisions. While measuring the diameter of a wire the reading on the main scale is 2 divisions and 79th circular scale division coincides with baseline.
If the number of divisions on the main scale is 10 to a centimetre and circular scale has 100 divisions, calculate
- pitch
- observed diameter
- least count
- corrected diameter.
What do you understand by the following term as applied to micrometre screw gauge?
Thimble
