Advertisements
Advertisements
प्रश्न
Figure shows a screw gauge in which circular scale has 200 divisions. Calculate the least count and radius of the wire.
Advertisements
उत्तर
No. of circular scale divisions = 200
Pitch = 1 mm
Least count (L.C.) = `"Pitch"/"No. of circular divisions"`
= `(1 "mm")/200`
= 0.005 mm
L.C. = 0.0005 cm
Main scale reading = 5 mm = 0.5 cm
(C.S.D.) circular scale reading = 34 divisions
Observed diameter of wire = Main scale reading + L.C. × C.S.D.
= 0.5 + 0.0005 × 34
= 0.5 + 0.0170
= 0.5170 cm
Redius of wire = `"diameter"/2`
= `0.5170/2`
= 0.2585 cm
APPEARS IN
संबंधित प्रश्न
Describe in steps, how would you use a vernier callipers to measure the length of a small rod?
A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two revolutions. When the flat end of the screw is in contact with the stud, the zero of the circular scale lies below the base line and 4th division of the circular scale is in line with the base line. Find
(i) The pitch,
(ii) The least count and
(iii) The zero error of the screw gauge
State the correction if the positive error is 7 divisions when the least count is 0.01 cm.
Name the measuring employed to measure the diameter of a pencil.
Consider the following case where the zero of vernier scale and the zero of the main scale are clearly seen. If L.C. of the vernier calipers is 0.01 cm, write the zero error and zero correction of the following.
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.
