2021 A/L Physics Past Paper · Question 02 (English Medium)

Answer: (1) $0.5\ \text{mm}$

Given information

We know:

$19$ main scale divisions are divided into $20$ equal vernier scale divisions.

The least count is $0.025\ \text{mm}$.

What the question asks

We want: the length $x$ of one main scale division.

Relevant theory

To coincide mark 1 of the vernier scale with mark 1 of the main scale the distance which the vernier scale should be pushed forward is the difference in length between one part of the vernier scale and one part of the main scale. This length is called the least count of the vernier scale.

least count = length of one part of the main scale - length of one part of the vernier scale

Required equations

Here $x$ is one main scale division and $y$ is one vernier scale division.

$\text{LC} = x - y$

$20y = 19x \quad\Rightarrow\quad y = \dfrac{19}{20}x$

$\text{LC} = x - \dfrac{19}{20}x = \dfrac{x}{20}$

Applying it to this question

The least count is in mm.

So $x$ comes out in mm too.

$0.025 = \dfrac{x}{20}$

$x = 20 \times 0.025$

$0.5$

$x = 0.5$

Reasoning and calculation

The vernier division is a little shorter than the main scale division.

The difference is one twentieth of a main scale division.

That difference is the least count.

So the main scale division is $20$ times the least count.

Option (1).

Checks

$x - 19 * x / 20$

Does it make sense?

A main scale in $0.5\ \text{mm}$ steps is a fine scale.

Dividing it into $20$ parts gives $0.025\ \text{mm}$.

That is a very small least count, as the question wants.

Conclusion

$x = 20 \times 0.025\ \text{mm} = 0.5\ \text{mm}$

The least count is $\dfrac{x}{20} = 0.025\ \text{mm}$, so $x = 0.5\ \text{mm}$.

The answer is (1).

Count the divisions carefully.

Here $19$ main divisions are shared by $20$ vernier divisions.

So the least count is the main scale division divided by $20$.

Exam tip