In a vernier calliper N divisions of vernier scale coincides with (. N - 1 .) divisions of the main scale (in which length of one division is 1 mm ). The least count of the instrument should be (in cm ):

Question

In a vernier calliper N divisions of vernier scale coincides with (. N - 1 .) divisions of the main scale (in which length of one division is 1 mm ). The least count of the instrument should be (in cm ): (A) (1/N - 1) (B) (1/10 N) (C) N - 1 (D) N

Tardigrade Admin · Accepted Answer

Least count 1 M S D - 1 V S D = 1 M S D - (. (N - 1/N) .) M S D ∵ N V S D = (. N - 1 .) M S D ∴ 1 V S D = (. (N - 1/N) .) M S D ∴ Least count = (1/N) M S D = (. (1/N) .) × (. (1/10) .) c m = (. (1/10 N) .)

Q. In a vernier calliper $N$ divisions of vernier scale coincides with $(N - 1)$ divisions of the main scale (in which length of one division is $1 mm$ ). The least count of the instrument should be (in $c m$ ):

$\frac{1}{N - 1}$

$\frac{1}{10 N}$

$N - 1$

$N$

Least count $1 MS D - 1 V S D$
$= 1 MS D - (\frac{N - 1}{N}) MS D$
$∵ N V S D = (N - 1) MS D$
$∴ 1 V S D = (\frac{N - 1}{N}) MS D$
$∴$ Least count $= \frac{1}{N} MS D = (\frac{1}{N}) \times (\frac{1}{10}) c m = (\frac{1}{10 N})$

Q. In a vernier calliper N divisions of vernier scale coincides with (N−1) divisions of the main scale (in which length of one division is 1mm ). The least count of the instrument should be (in cm ):

N−11​

10N1​

N−1

N

Least count 1MSD−1VSD =1MSD−(NN−1​)MSD ∵NVSD=(N−1)MSD ∴1VSD=(NN−1​)MSD ∴ Least count =N1​MSD=(N1​)×(101​)cm=(10N1​)

Q. In a vernier calliper $N$ divisions of vernier scale coincides with $(N - 1)$ divisions of the main scale (in which length of one division is $1 mm$ ). The least count of the instrument should be (in $c m$ ):

$\frac{1}{N - 1}$

$\frac{1}{10 N}$

$N - 1$

$N$

Least count $1 MS D - 1 V S D$
$= 1 MS D - (\frac{N - 1}{N}) MS D$
$∵ N V S D = (N - 1) MS D$
$∴ 1 V S D = (\frac{N - 1}{N}) MS D$
$∴$ Least count $= \frac{1}{N} MS D = (\frac{1}{N}) \times (\frac{1}{10}) c m = (\frac{1}{10 N})$