Question

A travelling microscope has 20 divisions per cm on the main scale while its vernier scale has total 50 divisions and 25 vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?

• A. 0.001 cm
• B. 0.002 mm
• C. 0.002 cm
• D. 0.005 cm

Answer 1

The Correct Option is C

To determine the least count of the travelling microscope, we need to understand how the main scale and vernier scale relate to each other.

1. The main scale of the microscope has 20 divisions per cm. Thus, each division on the main scale is: \(\frac{1}{20} \, \text{cm} = 0.05 \, \text{cm}\).
2. The vernier scale has 50 divisions, and 25 of these vernier scale divisions are equal to 24 main scale divisions. Thus, the length of one division on the vernier scale is: \(\frac{24 \times 0.05 \, \text{cm}}{25} = \frac{1.2}{25} \, \text{cm} = 0.048 \, \text{cm}\). 
3. The least count (LC) of the vernier caliper is calculated as the difference between one main scale division and one vernier scale division: \(\text{Least Count} = \text{Main Scale Division} - \text{Vernier Scale Division}\) \(\text{Least Count} = 0.05 \, \text{cm} - 0.048 \, \text{cm} = 0.002 \, \text{cm}\).

Therefore, the least count of the travelling microscope is 0.002 cm which corresponds to the correct option.

This means that the smallest measurement that can be accurately read using this microscope is 0.002 cm.

Answer 2

\(1\, \text{MSD} = \frac{1}{20}\, \text{cm}\)

\(1\, \text{VSD} = \frac{24}{25} \times \frac{1}{20}\, \text{cm}\)
\(∴\) Least count = 1 MSD – 1 VSD
\(=\frac{1}{20}(1 - \frac{24}{25})\)
\(=\frac{1}{20} \times \frac{1}{25}\)
\(= 0.002 cm\)
So, the correct option is (C): 0.002 cm