Question

There are four Vernier scales, whose specifications are given in column $1$ and the least count is given in Column$2$. Match the columns $1$and $2$with the correct specification and corresponding least count (s=value of main scale division, n=number of marks on Vernier). Assume (n-1) main scale division is equal to n Vernier divisions.

Column $1$	Column $2$
$i.s = 1mm,n = 10$	$a.0.05mm$
$ii.s = 0.5mm,n = 10$	$b.0.01mm$
$iii.s = 0.5mm,n = 20$	$c.0.1mm$
$iv.s = 1mm,n = 100$	$d.0.025mm$

(A) $i.c,ii.a,iii.d,iv.b$
(B) $i.c,ii.a,iii.b,iv.a$
(C) $i.b,ii.a,iii.d,iv.c$
(D) $i.d,ii.a,iii.d,iv.c$

Answer 1

Hint Least count of an instrument is the smallest value the instrument can measure. Here, for a Vernier scale, the least count can be precisely calculated by dividing the smallest reading on the main scale by the total number of divisions in the Vernier scale.

Complete Step by step solution
Vernier callipers is an instrument to measure length. The least count of the Vernier callipers is the smallest reading it can take accurately.
Mathematically, the formula of least count is given as,
$Least\ Count = \dfrac{Smallest\ reading\ on\ main\ scale}{No\ of\ division\ on\ vernier\ scale}$
In this problem, we have been provided with these two parameters, Smallest reading on the main scale (i.e. $s$) and the number of divisions on the Vernier scale (i.e. $n$).
Calculating the least count, we get
For the first Vernier scale $\left( {i.s = 1mm,n = 10} \right)$, we have
$L.C = \dfrac{s}{n}$
$ \Rightarrow L.C = \dfrac{1}{{10}}$
Solving this, we get
$L.C = 0.1mm$
For the second Vernier scale $\left( {ii.s = 0.5mm,n = 10} \right)$, we have
$L.C = \dfrac{s}{n}$
$ \Rightarrow L.C = \dfrac{{0.5}}{{10}}$
Solving this, we get
$L.C = 0.05mm$
For the third Vernier scale $\left( {iii.s = 0.5mm,n = 20} \right)$ , we have $L.C = \dfrac{s}{n}$
$ \Rightarrow L.C = \dfrac{{0.5}}{{20}}$
Solving this, we get
$L.C = 0.025mm$
For the fourth Vernier scale $\left( {iv.s = 1mm,n = 100} \right)$ , we have
$L.C = \dfrac{s}{n}$
$ \Rightarrow L.C = \dfrac{{0.1}}{{100}}$
Solving this, we get
$L.C = 0.001mm$
Now, matching these answers to the Column$2$, we get
$i.c,ii.a,iii.d,iv.b$

Thus, Option (A) is correct.

Note Take care of the fact that the concept of least count is the same for every measuring instrument but the formula to numerically calculate it may vary from instrument to instrument. The smaller the least count of an instrument, the more accurate is the measurement.