In a vernier calliper, N divisions of vernier scale coincide with (.N-1.) divisions of main scale (in which 1 division represents 1mm ). The least count of the instrument in cm . should be

Question

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

Tardigrade Admin · Accepted Answer

L.C=M.S.D-V.S.N Given (.N-1.)M.S.D=N(.V.S.D.) V.S.D = ((. N - 1 .)/N)M.S.D L.C=M.S.D - ((. N - 1 .)/N)M.S.D =M.S.D((1/ N))

Q. In a vernier calliper, $N$ divisions of vernier scale coincide with $(N - 1)$ divisions of main scale (in which $1$ division represents $1 mm$ ). The least count of the instrument in $c m$ . should be

$N$

$N - 1$

$\frac{1}{10 N}$

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

$L . C = M . S . D - V . S . N$
Given $(N - 1) M . S . D = N (V . S . D)$
V.S.D = $\frac{( N - 1 )}{N} M . S . D$
$L . C = M . S . D$ - $\frac{( N - 1 )}{N} M . S . D$
$= M . S . D (\frac{1}{N})$

Q. In a vernier calliper, N divisions of vernier scale coincide with (N−1) divisions of main scale (in which 1 division represents 1mm ). The least count of the instrument in cm . should be

N

N−1

10N1​

N−11​

L.C=M.S.D−V.S.N Given (N−1)M.S.D=N(V.S.D) V.S.D = N(N−1)​M.S.D L.C=M.S.D - N(N−1)​M.S.D =M.S.D(N1​)

Q. In a vernier calliper, $N$ divisions of vernier scale coincide with $(N - 1)$ divisions of main scale (in which $1$ division represents $1 mm$ ). The least count of the instrument in $c m$ . should be

$N$

$N - 1$

$\frac{1}{10 N}$

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

$L . C = M . S . D - V . S . N$
Given $(N - 1) M . S . D = N (V . S . D)$
V.S.D = $\frac{( N - 1 )}{N} M . S . D$
$L . C = M . S . D$ - $\frac{( N - 1 )}{N} M . S . D$
$= M . S . D (\frac{1}{N})$