Question

In a hypothetical vernier calipers 5 vernier scale divisions matches with 7 main scale division (MSD). When MSD is 1 mm, least-count of this calipers is $\left(\frac{x}{100}\right)$mm. Find the value of x.

Answer 1

20

Answer 2

The problem states that 5 Vernier Scale Divisions (VSD) match with 7 Main Scale Divisions (MSD). This implies that the length occupied by 5 VSD is equal to the length occupied by 7 MSD. Let $V$ be the value of one VSD and $M$ be the value of one MSD. The given relation is: $5V = 7M$

We are given that 1 MSD = 1 mm. Substituting $M = 1$ mm into the equation: $5V = 7 \times 1$ mm $V = \frac{7}{5}$ mm $= 1.4$ mm

For a Vernier caliper to function correctly, the divisions on the Vernier scale ($V$) must be slightly smaller than the divisions on the Main Scale ($M$), i.e., $V < M$. However, the calculation above yields $V = 1.4$ mm, which is greater than $M = 1$ mm. This indicates a potential error or non-standard phrasing in the question.

The Least Count (LC) of a Vernier caliper is defined as: $LC = 1 \text{ MSD} - 1 \text{ VSD} = M - V$

If we proceed with $V = 1.4M$, then $LC = M - 1.4M = -0.4M$, which is physically impossible as the least count cannot be negative.

Given the format of the answer ($\left(\frac{x}{100}\right)$mm), it is highly probable that the question intended a relationship that results in a standard Vernier caliper configuration and a Least Count expressible in the given format. A common scenario is when $N$ VSD match with $(N-1)$ MSD.

Let's assume the question has a typo and was intended to mean: "5 vernier scale divisions match with 4 main scale divisions (MSD)". In this case, the relation would be: $5V = 4M$ $V = \frac{4}{5}M$

Given $M = 1$ mm: $V = \frac{4}{5} \times 1$ mm $= 0.8$ mm. This satisfies the condition $V < M$.

Now, we calculate the Least Count: $LC = M - V$ $LC = 1 \text{ mm} - 0.8 \text{ mm}$ $LC = 0.2 \text{ mm}$

The question states that the least count is $\left(\frac{x}{100}\right)$mm. So, we set our calculated LC equal to the given form: $0.2 \text{ mm} = \left(\frac{x}{100}\right)\text{mm}$ $0.2 = \frac{x}{100}$ $x = 0.2 \times 100$ $x = 20$

This interpretation yields a sensible result and fits the expected answer format.