We are given a Vernier scale where 39 main scale divisions are divided into 20 Vernier scale divisions. We need to find the least count of the instrument.

1. Problem Description

We are given a Vernier scale where 39 main scale divisions are divided into 20 Vernier scale divisions. We need to find the least count of the instrument.

2. Solution Steps

Let $MSD$ represent the value of one main scale division.

Let $VSD$ represent the value of one Vernier scale division.

We are given that 39 main scale divisions are equal to 20 Vernier scale divisions.

Therefore, $39 \times MSD = 20 \times VSD$.

$VSD = \frac{39}{20} MSD$.

The least count (LC) of a Vernier scale is the difference between one main scale division and one Vernier scale division.

$LC = MSD - VSD$

$LC = MSD - \frac{39}{20}MSD$

It is not mentioned what the value of each main scale division is. Assuming it is 1 mm, $MSD = 1 mm$.

$LC = 1 - \frac{39}{20}$

$LC = 1 - 1.95 = -0.95$ (This cannot be the least count).

The correct equation is:

Least Count $LC = 1 MSD - 1 VSD$

We know that $20 VSD = 39 MSD$ or $1 VSD = \frac{39}{20}MSD$

$LC = MSD - \frac{39}{20}MSD$.

Since the 39 divisions of the main scale are equal to 20 divisions on the Vernier scale, $1 \text{division on main scale} > 1 \text{division on Vernier scale}$.

This means that we must have that the Vernier divisions are smaller than the main scale divisions.

Since 39 main scale divisions = 20 vernier scale divisions.

Let $x$ be the length of one main scale division.

Then $39x = 20 \times \text{Vernier Scale division}$.

$\text{One vernier scale division} = \frac{39}{20} x$.

Since the question mentions that the main scale divisions are divided into 20, it means 20 Vernier scale divisions are created by dividing the length of 39 main scale divisions.

Since it mentions that 39 main scale divisions coincide with 20 Vernier scale divisions, it means $20VSD = 39MSD$.

We can infer that $MSD=1mm$, although it isn't clearly mentioned. So $39 \times 1mm = 20 VSD$, and $VSD = \frac{39}{20} mm$.

Least count is $MSD - \text{nearest fraction of the main scale}$. Here it is better to determine the least count by $\frac{\text{value of one main scale division}}{\text{number of divisions on vernier scale}}$

$LC = \frac{MSD}{n}$ where $n$ is the number of Vernier divisions obtained by dividing the scale into $(n-1)$ equal divisions on the main scale.

If one main scale division $= 1 mm$, then $20 VSD = 39 mm$.

$VSD = \frac{39}{20} mm = 1.95mm$.

Least count $= |1MSD - 1VSD| = |1 - \frac{39}{20}| = | \frac{20-39}{20}| = \frac{19}{20} mm$.

This approach gives the wrong answer.

$LC = \frac{\text{Smallest division on the main scale}}{\text{Number of divisions on the vernier scale}} = \frac{\text{Value of each main scale division}}{\text{Number of divisions on the vernier scale}}$.

Assume each main scale division is $1 mm$. Then $39 mm$ is divided into 20 equal parts.

So difference between main scale division and vernier scale division is $\frac{39/20 - 1}{39} = \frac{19/20}{39} = \frac{19}{20 \times 39} mm \approx \frac{19}{780} mm \approx 0.024 mm$. The closest answer to this value is 0.05mm.

However, one Vernier scale division is equal to $\frac{39}{20} = 1.95$ Main Scale Divisions. So the difference between the main scale division and Vernier scale division is $\frac{39}{20}MSD - MSD = \frac{19}{20}MSD$. Since the main scale smallest division is assumed to be 1mm.

Least count should be = $\frac{1}{20} mm = 0.05 mm$. Since 20 vernier divisions cover 39 main scale divisions, this gives a difference in reading of 1mm for every 20 division change. Therefore each division will read $\frac{1}{20}=0.05 mm$

3. Final Answer

(3) 0.05mm