Question

The vernier of a circular scale is divided into 30 divisions, which coincides with 29 main scale divisions. If each main scale division is $\left( {\dfrac{1}{2}} \right)^\circ $ , find the least count of the instrument, $\left( {1^\circ = 60'} \right)$ .
A) $0 \cdot 1'$
B) $1 \cdot 0'$
C) $10 \cdot 0'$
D) $30 \cdot 0'$

Answer 1

Hint: The given instrument consists of a vernier scale which is a circular scale and a main scale. The least count of this instrument will be the difference between the value of one main scale division and the value of one vernier scale division. As the number of divisions that coincide with the vernier scale is given we can easily obtain the least count of the vernier.

Formula used:
-The least count of a vernier is given by, $L.C = 1{\text{m}}{\text{.s}}{\text{.d}} - 1{\text{v}}{\text{.s}}{\text{.d}}$ where ${\text{1m}}{\text{.s}}{\text{.d}}$ is one main scale division and $1{\text{v}}{\text{.s}}{\text{.d}}$ is one vernier scale division.

Complete step by step solution:
Step 1: List the parameters given about the vernier.
The number of divisions on the vernier scale is given to be $n = 30{\text{div}}$ .
The number of divisions on the main scale that coincides with the $n$ number of divisions on the vernier scale is given to be $29{\text{div}} = n - 1$ .
i.e., $30{\text{v}}{\text{.s}}{\text{.d}} = 29{\text{m}}{\text{.s}}{\text{.d}}$
Then the value of one vernier scale division will be $1{\text{v}}{\text{.s}}{\text{.d}} = \dfrac{{n - 1}}{n} = \dfrac{{29}}{{30}}{\text{m}}{\text{.s}}{\text{.d}}$ .
Also, it is given that the value of one main scale division is $1{\text{m}}{\text{.s}}{\text{.d}} = \left( {\dfrac{1}{2}} \right)^\circ $ .

Step 2: Express the relation for the least count of the instrument.
The least count of the given instrument can be expressed as $L.C = 1{\text{m}}{\text{.s}}{\text{.d}} - 1{\text{v}}{\text{.s}}{\text{.d}}$ -------- (1)
Substituting for $1{\text{v}}{\text{.s}}{\text{.d}} = \dfrac{{29}}{{30}}{\text{m}}{\text{.s}}{\text{.d}}$ in equation (1) we get,
$L.C = 1{\text{m}}{\text{.s}}{\text{.d}} - \dfrac{{29}}{{30}}{\text{m}}{\text{.s}}{\text{.d}} = \dfrac{1}{{30}}{\text{m}}{\text{.s}}{\text{.d}}$ ---------- (2)
Substituting the value of one main scale division as $1{\text{m}}{\text{.s}}{\text{.d}} = \left( {\dfrac{1}{2}} \right)^\circ $ in equation (2) we get, $L.C = \left( {\dfrac{1}{{30}} \times \dfrac{1}{2}} \right)^\circ = \left( {\dfrac{1}{{60}}} \right)^\circ $ . The obtained least count is in degrees.
Now if we were to express the least count in minutes we obtain $L.C = \left( {\dfrac{1}{{60}}} \right) \times 60' = 1'$ as $1^\circ = 60'$ .
$\therefore $ The least count of the given instrument is obtained to be $L.C = 1'$ .

So the correct option is B.

Note: The least count of the instrument refers to the smallest distance that can be accurately measured with the vernier scale. The least count of a vernier is also referred to as the vernier constant. It is worth to remember that the main scale reading with which the vernier scale coincides has no connection with the reading.