Question

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is :

• A. \(\frac{1}{2}\)
• B. \(\frac{10}{11}\)
• C. \(\frac{50}{11}\)
• D. \(\frac{5}{11}\)

Answer 1

The Correct Option is D

Given:

\[ 10 \, \text{MSD} = 11 \, \text{VSD} \]

1 VSD (Vernier Scale Division) is equivalent to:

\[ 1 \, \text{VSD} = \frac{10}{11} \, \text{MSD} \]

The least count (LC) of the Vernier caliper is given by:

\[ LC = 1 \, \text{MSD} - 1 \, \text{VSD} \]

Substituting the values:

\[ LC = 1 \, \text{MSD} - \frac{10}{11} \, \text{MSD} = \frac{1}{11} \, \text{MSD} \]

Given that 1 MSD corresponds to 5 units:

\[ LC = \frac{5 \, \text{units}}{11} \]

Answer 2

10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. Each division on the main scale is of 5 units. We need to find the least count of the instrument.

Concept Used:

The least count (L.C.) of a Vernier calliper is given by:

\[ \text{L.C.} = \frac{\text{Value of one main scale division (MSD)}}{\text{Number of divisions on the Vernier scale}} \]

However, this standard formula applies when n Vernier scale divisions equal (n-1) main scale divisions. In the general case, the least count is:

\[ \text{L.C.} = 1 \ \text{MSD} - 1 \ \text{VSD} \]

where VSD is the value of one Vernier scale division.

Step-by-Step Solution:

Step 1: Determine the value of one Main Scale Division (MSD).

\[ \text{1 MSD} = 5 \ \text{units} \]

Step 2: Find the value of one Vernier Scale Division (VSD).

Given: 10 divisions on the main scale = 11 divisions on the Vernier scale.

\[ 10 \ \text{MSD} = 11 \ \text{VSD} \] \[ 1 \ \text{VSD} = \frac{10}{11} \ \text{MSD} \] \[ 1 \ \text{VSD} = \frac{10}{11} \times 5 \ \text{units} = \frac{50}{11} \ \text{units} \]

Step 3: Calculate the Least Count (L.C.).

The least count is the difference between one MSD and one VSD.

\[ \text{L.C.} = 1 \ \text{MSD} - 1 \ \text{VSD} \] \[ \text{L.C.} = 5 - \frac{50}{11} \] \[ \text{L.C.} = \frac{55}{11} - \frac{50}{11} = \frac{5}{11} \ \text{units} \]

Thus, the least count of the Vernier calliper is \( \frac{5}{11} \) units.