Vernier Callipers Least Count Calculation
This section explains how to calculate the least count (LC) of a vernier callipers using the provided scale division information.
Provided Data Analysis
We are given the following information about the vernier callipers:
- The number of Vernier Scale Divisions (VSD) that coincide is 20.
- These 20 VSD correspond to the length of 16 Main Scale Divisions (MSD).
- The value of one Main Scale Division (MSD) is $1 \text{ mm}$.
Calculating the Least Count (LC)
The least count of a vernier callipers is the smallest measurement that can be accurately taken using the instrument. It is calculated as the difference between one main scale division and one vernier scale division.
-
Find the length represented by 16 MSD:
Since 1 MSD = $1 \text{ mm}$, the length of 16 MSD is:
$ \text{Length of 16 MSD} = 16 \times 1 \text{ mm} = 16 \text{ mm} $ -
Determine the length of one VSD:
We know that 20 VSD have the same length as 16 MSD.
$ \text{Length of 20 VSD} = \text{Length of 16 MSD} = 16 \text{ mm} $Therefore, the length of a single VSD is:
$ \text{Length of 1 VSD} = \frac{16 \text{ mm}}{20} = 0.8 \text{ mm} $ -
Calculate the Least Count (LC):
The formula for Least Count is: LC = (Value of 1 MSD) - (Value of 1 VSD)
$ \text{LC} = 1 \text{ mm} - 0.8 \text{ mm} = 0.2 \text{ mm} $ -
Convert LC to centimeters:
To express the least count in centimeters, we use the conversion factor $1 \text{ cm} = 10 \text{ mm}$.
$ \text{LC} = \frac{0.2 \text{ mm}}{10 \text{ mm/cm}} = 0.02 \text{ cm} $
Final Result
The least count of the vernier callipers is $0.02 \text{ cm}$.
Similar Questions
In a metre bridge experiment, the positions of the cell, E, and galvanometer, G, are interchanged. We shall observe in the galvanometer :
- Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L, notes down the data of time taken to complete 30 oscillations as $60 \text{ s}$ and hence calculates the length of the simple pendulum as :
(Take $\pi^2 = 9.8$, and $g = 9.8 \text{ m/s}^2$) - One main scale division of a Vernier calliper is equal to $1\text{ mm}$ and the number of divisions on the Vernier scale is $10$. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that $4^{\text{th}}$ Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of a wire to be $1\text{ cm}$, the actual length of the wire is :
- Which of the following measurements require "index correction" ?
Important Questions from Practical Physics
- For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm) :
In a metre bridge experiment, the positions of the cell, E, and galvanometer, G, are interchanged. We shall observe in the galvanometer :
- Savitha, a XI standard student, while conducting an experiment to determine the effective length of a simple pendulum L, notes down the data of time taken to complete 30 oscillations as $60 \text{ s}$ and hence calculates the length of the simple pendulum as :
(Take $\pi^2 = 9.8$, and $g = 9.8 \text{ m/s}^2$) - One main scale division of a Vernier calliper is equal to $1\text{ mm}$ and the number of divisions on the Vernier scale is $10$. When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale in such a way that $4^{\text{th}}$ Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of a wire to be $1\text{ cm}$, the actual length of the wire is :
- Which of the following measurements require "index correction" ?
