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 :
- \(\frac{1}{2}\)
- \(\frac{10}{11}\)
- \(\frac{50}{11}\)
- \(\frac{5}{11}\)
The Correct Option is D
Approach Solution - 1
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:
\[ LC = \frac{5 \, \text{units}}{11} \]
Approach Solution -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.
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