2024 A/L Physics Past Paper · Question 02 (English Medium)

Answer: (1) $0.025\,\text{mm}$

Given information

What we know:

The main scale $1.0\,\text{cm} = 10\,\text{mm}$ has $20$ subdivisions.

The length of $19$ main scale subdivisions is divided equally into $20$ vernier scale divisions.

What the question asks

We want: the least count of the calliper.

What each letter means:

$x$ ($\text{mm}$): the length of one main scale division.

$y$ ($\text{mm}$): the length of one vernier scale division.

$\text{LC}$ ($\text{mm}$): the least count.

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.

The quote says that the least count is the difference in length between one part of the main scale and one part of the vernier scale. So we need the length of one part of each scale.

For example, two friends walk side by side along a path. One takes a slightly longer step than the other. After each step the friend with the shorter step falls a little behind, and the distance between them grows by the same small amount at each step.

That small distance is the difference of one step and one step, like the least count.

Required equations

$\text{LC} = x - y$

$x = \dfrac{10}{20}$

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

Applying it to this question

$0.5$

$x = \dfrac{10}{20}$

$0.475$

$y = \dfrac{19 \times 0.5}{20}$

$0.025$

$\text{LC} = x - y = 0.5 - 0.475$

Reasoning and calculation

$L / n - (19 * L / n) / 20$

Now match it with the five options.

Option (1) is $0.025\,\text{mm}$. It matches.

Option (2), $0.050\,\text{mm}$, is what we get by taking one main division as $1\,\text{mm}$. Option (5), $0.50\,\text{mm}$, is the main division itself. The others do not follow from $\text{LC} = \dfrac{x}{20}$ with $x = 0.5\,\text{mm}$.

Checks

Check the units. $x$ and $y$ are lengths in $\text{mm}$, so $x - y$ is a length in $\text{mm}$, as a least count must be.

Does it make sense?

The vernier scale is meant to read finer than the main scale.

The least count $0.025\,\text{mm}$ is $20$ times smaller than the main scale division $0.5\,\text{mm}$, because one main division is shared by $20$ vernier divisions.

Conclusion

$\text{LC} = 0.025\,\text{mm}$

The least count of the calliper is $\text{LC} = 0.025\,\text{mm}$.

The answer is (1).

For a vernier with $n$ divisions equal to $n - 1$ main divisions, the least count is one main division divided by $n$.

First find the length of one main division from the scale, as the question may not use $1\,\text{mm}$.

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