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प्रश्न
The least count of a vernier callipers is 0.0025 cm and it has an error of + 0.0125 cm. While measuring the length of a cylinder, the reading on the main scale is 7.55 cm, and the 12th vernier scale division coincides with the main scale. Calculate the corrected length.
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उत्तर
Least count (L.C.) = 0.0025 cm
Error = +0.0125 cm
Correction = – (Error)
= – (+0.0125)
= – 0.0125 cm
Main scale reading = 7.55 cm
Vernier scale division (V.S.D.) coinciding with main scale = 12th
Length recorded = Main scale reading + L.C. × V.S.D.
= 7.55 + 0.0025 × 12
= 7.55 + 0.0300
= 7.58 cm
Correct length = Length recorded + Correction
= 7.58 + (−0.0125)
= 7.58 – 0.0125
= 7.5675 cm
= 7.567 cm
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संबंधित प्रश्न
State three uses of the vernier callipers.
The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions.
(i) What is the pitch of the screw gauge?
(ii) What is the least count of the screw gauge?
How can the least count of a screw gauge be decreased?
Define metre according to the old definition.
What is the least count in the case of the following instrument?
vernier calipers
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.
What is the zero error of a screw gauge?
