Advertisements
Advertisements
प्रश्न
The circular scale of a screw gauge has 100 divisions. Its spindle moves forward by 2.5 mm when given five complete turns. Calculate
- pitch
- least count of the screw gauge
Advertisements
उत्तर
Number of circular scale divisions = 100
Distance moved by spindle (screw) = 2.5 mm
No. of complete rotations given = 5
(1) Pitch = `"Distance moved by screw on sleeve"/"No. of complete rotations"`
= `2.5/5`
= 0.5 mm
= 0.05 cm
(2) Least count = `"Pitch"/"No. of circular scale divisions"`
= `0.05/100` cm
= 0.0005 cm
APPEARS IN
संबंधित प्रश्न
Draw a neat and labelled diagram of a vernier callipers. Name its main parts and state their functions.
Name the two scales of a vernier callipers and explain how it is used to measure length correct up to 0.01 cm.
Draw a neat and labelled diagram of a screw gauge.
Name its main parts and state their functions.
A vernier callipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to 9 mm. When the two jaws are in contact, the zero of the vernier scale is ahead of the zero of the main scale and the 3rd division of the vernier
scale coincides with a main scale division.
Find : (i) The least count and
(ii) The zero error of the vernier callipers.
Why is the metre length in terms of the wavelength of light considered more accurate?
What do you understand by the following term as applied to screw gauge?
Positive zero error
The main scale reading while measuring the thickness of a rubber ball using Vernier caliper is 7 cm and the Vernier scale coincidence is 6. Find the radius of the ball.
