JEE Physics 2025 Question with Solution

Given: The main scale has $300$ divisions equal to $15 \, \text{cm}$. The vernier scale has $25$ divisions equal to $24$ main scale divisions.

Find: The least count of the travelling microscope.

The least count of a vernier scale is the difference between the value of one main scale division and the value of one vernier scale division.

First, value of one main scale division is

$$\text{1 MSD} = \frac{15 \, \text{cm}}{300} = 0.05 \, \text{cm}$$

Since $25$ vernier divisions are equal to $24$ main scale divisions,

$$25 \, \text{VSD} = 24 \, \text{MSD}$$

So, one vernier scale division is

$$\text{1 VSD} = \frac{24}{25} \times 0.05 \, \text{cm} = 0.048 \, \text{cm}$$

Hence, the least count is

$$LC = \text{1 MSD} - \text{1 VSD} = 0.05 \, \text{cm} - 0.048 \, \text{cm} = 0.002 \, \text{cm}$$

Therefore, the least count of the travelling microscope is $0.002 \, \text{cm}$. The correct option is B.

Given: $300$ main scale divisions equal $15 \, \text{cm}$, and $25$ vernier divisions equal $24$ main scale divisions.

Find: The least count.

First calculate one main scale division:

$$\text{1 MSD} = \frac{15}{300} = 0.05 \, \text{cm}$$

For a direct vernier where $25$ vernier divisions equal $24$ main scale divisions, the least count is

$$LC = \frac{\text{1 MSD}}{25} = \frac{0.05}{25} = 0.002 \, \text{cm}$$

This works because the difference between $1$ main scale division and $1$ vernier scale division becomes one-$25^{\text{th}}$ of a main scale division.

Therefore, the least count is $0.002 \, \text{cm}$, so the correct option is B.