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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
Explain the terms : Least count of a screw gauge.
How are they determined?
Name the instrument which has the least count
0.1 mm
Define metre in terms of the wavelength of light.
Define the term pitch.
State the formula for determining least count for a vernier callipers.
State the correction if the negative error is 7 divisions when the least count is 0.01 cm.
A micrometre screw gauge has a positive zero error of 7 divisions, such that its main scale is marked in 1/2 mm and the circular scale has 100 divisions. The spindle of the screw advances by 1 division complete rotation.
If this screw gauge reading is 9 divisions on the main scale and 67 divisions on the circular scale for the diameter of a thin wire, calculate
- Pitch
- L.C.
- Observed diameter
- Corrected diameter
Find the volume of a book of length 25 cm, breadth 18 cm and height 2 cm in m3.
In the following figure, the pitch of the screw is 1 mm. Calculate:
(i) the least count of screw gauge and
(ii) the reading represented in the figure.

