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प्रश्न
Define the least count of vernier callipers. How do you determine it?
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उत्तर
The least count of vernier callipers is equal to the difference between the values of one main scale division and one vernier scale division.
Let n divisions on vernier callipers be of length equal to that of (n - 1) divisions on the main scale and the value of 1 main scale division be x. Then,
Value of n divisions on vernier = (n - 1) x
Alternatively, value of 1 division on vernier = `((n - 1)"x")/n`
Least count = `x - ((n - 1)"x")/n = x/n`
L.C. = `"(Value of one main scale division)"/"(Total no. of divisions on vernier callipers)"`
Value of one main scale division = 1 mm
Total no. of divisions on vernier = 10
Therefore, L.C. = `(1 "mm")/10` = 0.1 mm = 0.01 cm
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संबंधित प्रश्न
Name the part of the vernier callipers which is used to measure the following
Depth of a small bottle
State one use of a screw gauge.
Fig., below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on the main scale when circular head is rotated once.
Find:
(i) Pitch of the screw gauge,
(ii) Least count of the screw gauge and
(iii) The diameter of the wire.
(a) A vernier scale has 20 divisions. It slides over the main scale, whose pitch is 0.5 mm. If the number of divisions on the left hand of the zero of vernier on the main scale is 38 and the 18th vernier scale division coincides with the main scale, calculate the diameter of the sphere, held in the jaws of vernier callipers.
(b) If the vernier has a negative error of 0.04 cm, calculate the corrected radius of the sphere.
State whether the following statement is true or false by writing T/F against it.
The ratchet of a screw gauge is used to measure the depth of a beaker.
What is the least count in the case of the following instrument?
vernier calipers
In a vernier calliper, (N + 1) divisions of the vernier scale coincide with N divisions of the main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is ______.
