Vernier calipers have 1mm marks on the main scale. It has 20 equal divisions on the Vernier scale which matches with 16 main scale divisions. For this Vernier calipers, the least count is:
A. 0.02 mm
B. 0.05 mm
C. 0.1 mm
D. 0.2 mm
Answer
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Hint: The minimum length a device can measure is called its least count. The least count of different devices is different. The minimum the value of the least count of a device, the accurate is it’s measurement. The least count of screw-gauge is 0.01 means it is 100 times more accurate than a meter scale. Similarly we have least count for Vernier calipers.
Formula used:
$L.C = 1MSD - 1VSD$
Complete answer:
Practically there can’t be a measurement where the error becomes totally zero. Every measurement has a certain amount of error which could be human error, instrument error or any other error. But it is a good practice to minimize this error as much as possible. The use of scale in the measurement will result in huge value of error. Whereas if we use Vernier caliper, we reduce the error of measurement.
Now, given main scale reading (M.S.D) = 1 mm
Also $1 VSD = \dfrac{16}{20}MSD$, where VSD means vernier scale reading and MSD means main scale reading.
Thus, using $L.C = 1MSD - 1VSD$
$L.C = 1MSD - \dfrac{16}{20}MSD = \dfrac15 MSD$
$\implies LC = \dfrac15 (1mm) = 0.2 mm$
So, the correct answer is “Option D.
Note:
The least count of screw-gauge is 0.01 mm means it is 100 times more accurate than a meter scale. This means it is 100 times accurate than metre scale. Similarly the least count of Vernier calipers is 0.2 mm means it is 5 times more accurate than a meter scale. Here, we can see that even though the screw gauge is more accurate, it could measure only very thin material. It is used mainly to measure the diameter of wires. Vernier calipers can measure much more than screw gauge. Hence, it has its own importance which should not be neglected.
Formula used:
$L.C = 1MSD - 1VSD$
Complete answer:
Practically there can’t be a measurement where the error becomes totally zero. Every measurement has a certain amount of error which could be human error, instrument error or any other error. But it is a good practice to minimize this error as much as possible. The use of scale in the measurement will result in huge value of error. Whereas if we use Vernier caliper, we reduce the error of measurement.
Now, given main scale reading (M.S.D) = 1 mm
Also $1 VSD = \dfrac{16}{20}MSD$, where VSD means vernier scale reading and MSD means main scale reading.
Thus, using $L.C = 1MSD - 1VSD$
$L.C = 1MSD - \dfrac{16}{20}MSD = \dfrac15 MSD$
$\implies LC = \dfrac15 (1mm) = 0.2 mm$
So, the correct answer is “Option D.
Note:
The least count of screw-gauge is 0.01 mm means it is 100 times more accurate than a meter scale. This means it is 100 times accurate than metre scale. Similarly the least count of Vernier calipers is 0.2 mm means it is 5 times more accurate than a meter scale. Here, we can see that even though the screw gauge is more accurate, it could measure only very thin material. It is used mainly to measure the diameter of wires. Vernier calipers can measure much more than screw gauge. Hence, it has its own importance which should not be neglected.
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