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प्रश्न
(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.
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उत्तर
No. of divisions on vernier scale = 20
Pitch 0.5 mm
Least count = `"Pitch"/"No. of divisions on the vernier scale"`
L.C. = `0.5/20` mm
L.C. = 0.025 mm
L.C. = 0.0025 cm
(a) There is 38 number of main scale divisions on the left of the zero of the vernier scale.
⇒ Main scale reading = `38/2` mm = 1.9 cm
Vernier scale division coinciding with the main scale = 18th
Diameter of sphere = Main scale reading + L.C. × V.S.D.
= 1.9 + 0.0025 × 18
= 1.9 + 0.0450
= 1.945 cm
(b) Negative error = −0.04 cm
⇒ Correction = −(−0.04) = +0.04 cm
Corrected diameter of sphere = Observed reading + Correction
= 1.945 + (+0.04)
= 1.945 + 0.04
= 1.985 cm
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संबंधित प्रश्न
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.
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(b) If zero error of vernier callipers is -0.02 cm,
What is the correct length ?
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(a) A vernier scale has 10 divisions. It slides over the main scale, whose pitch is 1.0 mm. If the number of divisions on the left hand of zero of the vernier scale on the main scale is 56 and the 8th vernier scale division coincides with the main scale, calculate the length in centimetres.
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Negative zero error
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