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प्रश्न
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.
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उत्तर
Here, zero of the vernier scale lies on the right of the main scale.
Also, the 5th vernier division is coinciding with some main scale division
Thus, zero error = 0 + 5 x L.C. = -0.05 cm
Hence zero correction = 0.05 cm
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संबंधित प्रश्न
A screw gauge has a least count 0.001 cm and zero error +0.007 cm. Draw a neat diagram to represent it.
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.
The pitch of a screw gauge is 1 mm and the circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on the circular scale coincides with the base line. Find
(i) The least count and
(ii) The diameter of the wire.
Define metre according to the old definition.
(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 the formula for determining least count for a vernier callipers.
What is the backlash error?
