In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and $7^{th}$ Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has:
Step 1: Understanding the Question:
The problem describes a Vernier caliper with a zero error. We need to determine the type (positive or negative) and magnitude of this error based on the given information. Step 2: Key Formula or Approach:
1. Determine Least Count (LC): LC = (Value of 1 Main Scale Division) - (Value of 1 Vernier Scale Division). For a standard caliper, 10 VSD = 9 MSD.
2. Identify Error Type: If the Vernier scale zero is to the right of the main scale zero, the error is positive. If it's to the left, the error is negative.
3. Calculate Error Magnitude: Zero Error = (Coinciding Vernier Division) $\times$ (Least Count). Step 3: Detailed Explanation: Part 1: Calculate the Least Count (LC)
Given: 1 Main Scale Division (MSD) = 1 mm.
There are 10 Vernier Scale Divisions (VSD). It is standard that these 10 VSDs are equal in length to 9 MSDs.
So, 10 VSD = 9 MSD = 9 mm, which means 1 VSD = 0.9 mm.
Least Count (LC) = 1 MSD - 1 VSD = 1 mm - 0.9 mm = 0.1 mm.
In centimeters, LC = 0.01 cm. Part 2: Determine the Zero Error
The problem states that the zero of the Vernier scale is shifted to the right of the zero of the main scale. This condition corresponds to a positive zero error.
The coinciding Vernier division is the 7th one.
The magnitude of the zero error is calculated as:
Zero Error = $7 \times \text{LC} = 7 \times 0.01 \text{ cm} = 0.07$ cm.
Therefore, the instrument has a positive zero error of 0.07 cm. Step 4: Final Answer:
The caliper has 0.07 cm positive zero error.
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