Question

The vernier scale of travelling microscope has 50 divisions which coincides with 49 main scale division. If each main scale division is 0.5 mm, then the least count of the microscope is

• A. 0.01 mm
• B. 0.01 cm
• C. 0.5 cm
• D. 0.5 mm

Answer 1

The Correct Option is A

To find the difference between one main scale division and one vernier scale division, we use the formula:

$$\text{Difference} = \frac{\text{Measurement of one main scale division}}{\text{Number of divisions on the main scale}}$$ 

Substituting the given values:

$$\text{Difference} = \frac{0.5 \, \text{mm}}{49} = 0.0102 \, \text{mm}$$

Therefore, the least count of the microscope is: 0.0102 mm.

Since the options are provided in different units, we can convert this to centimeters:

$$\text{Least count} = 0.0102 \, \text{mm} = 0.0102 \, \text{cm}$$

The correct answer is: (A) 0.01 mm.

Answer 2

Given: 1 MSD = 0.5 mm

Also, 50 VSD = 49 MSD, so we can find 1 VSD: 

$1 \, \text{VSD} = \frac{50}{49} \, \text{MSD}$

Now, the least count is given by:

$$\text{Least Count} = 1 \, \text{MSD} - 1 \, \text{VSD} = 1 \, \text{MSD} - \frac{50}{49} \, \text{MSD}$$

Substitute the value of 1 MSD = 0.5 mm:

$$\text{Least Count} = 1 \, \text{MSD} - \left( \frac{50}{49} \times 1 \, \text{MSD} \right) = \frac{50}{49} \times 0.5 \, \text{mm} = 0.01 \, \text{mm}$$

Therefore, the least count is 0.01 mm.