In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division $= 0.05 \text{ mm}$, then the least count of the vernier callipers is ______

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Question

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division $= 0.05 \text{ mm}$, then the least count of the vernier callipers is _______ mm.

0.02

0.005

0.002

0.05

The correct answer is

0.005

Vernier Callipers Least Count Calculation

This section details the calculation for the least count (LC) of a vernier callipers based on the provided instrument specifications.

Given Information

Number of Vernier Scale Divisions (VSD) = 50
Number of Main Scale Divisions (MSD) equivalent to 50 VSD = 48
Value of one Main Scale Division (1 MSD) = 0.05 mm

Least Count Formula

The least count represents the smallest measurement the instrument can accurately measure. It is calculated as the difference between the value of one main scale division and one vernier scale division.

$\text{Least Count (LC)} = 1 \text{ MSD} - 1 \text{ VSD}$

Calculation Steps

Relate Vernier Scale Divisions to Main Scale Divisions:
The problem states that 50 VSD are equal in length to 48 MSD.

From this, we find the value of 1 VSD in terms of MSD:

$1 \text{ VSD} = \frac{48}{50} \times 1 \text{ MSD}$

$1 \text{ VSD} = 0.96 \times 1 \text{ MSD}$
Calculate the value of 1 VSD in millimeters:
Substitute the value of 1 MSD (0.05 mm) into the equation from Step 1:

$1 \text{ VSD} = 0.96 \times 0.05 \text{ mm}$

$1 \text{ VSD} = 0.048 \text{ mm}$
Compute the Least Count (LC):
Apply the least count formula using the values of 1 MSD and 1 VSD:

$\text{LC} = 0.05 \text{ mm} - 0.048 \text{ mm}$

$\text{LC} = 0.002 \text{ mm}$

Result

Following the standard calculation method, the least count for the vernier callipers is determined to be 0.002 mm.

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