Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
Pitch of the screw= 0.5 mm
No. of divisions on circular scale = 50
`"Least count"="pitch of the screw"/"no. of divisions on the circular scale"`
so, L.C. = `0.5/50=0.01` mm
APPEARS IN
संबंधित प्रश्न
Name the part of the vernier callipers which is used to measure the following
External diameter of a tube
Name the part of the vernier callipers which is used to measure the following
Thickness of a pencil.
Draw a neat and labelled diagram of a screw gauge.
Name its main parts and state their functions.
State one use of a screw gauge.
In the vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided into 10 parts. While measuring the length, the zero of the vernier lies just ahead of the 1.8 cm mark and the 4th division of vernier coincides with a main scale division.
(a) Find the length.
(b) If zero error of vernier callipers is -0.02 cm,
What is the correct length ?
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?
Figure shows a screw gauge in which circular scale has 200 divisions. Calculate the least count and radius of the wire.
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.
