Question

In an experiment with Vernier calipers of least count 0.1 mm, when two jaws are joined together the zero of Vernier scale lies right to the zero of the main scale and 6th division of Vernier scale coincides with the main scale division. While measuring the diameter of a spherical bob, the zero of Vernier scale lies in between 3.2 cm and 3.3 cm marks, and 4th division of Vernier scale coincides with the main scale division. The diameter of bob is measured as

Hint:

Always remember: "Actual = Observed - (Zero Error)". If zero is to the right, error is positive (subtract). If zero is to the left, error is negative (effectively add).

Answer 1

Step 1: Understanding the Question:
We need to calculate the actual diameter of a bob using Vernier caliper readings, accounting for a positive zero error.

Step 2: Key Formula or Approach:
1. Least Count (LC) = 0.1 mm = 0.01 cm.
2. Zero Error (ZE) = \(+(\text{Vernier division} \times \text{LC})\) [since zero is to the right].
3. Observed Reading = Main Scale Reading (MSR) + (Vernier Scale Reading (VSR) \(\times\) LC).
4. Actual Reading = Observed Reading - Zero Error.

Step 3: Detailed Explanation:
Part 1: Calculate Zero Error
The zero of Vernier is to the right, so it's a positive zero error.
ZE = \(+6 \times 0.01 = +0.06 \text{ cm}\).
Part 2: Calculate Observed Reading
MSR = 3.2 cm (since zero lies between 3.2 and 3.3).
VSR = 4.
Observed Reading = \(3.2 + (4 \times 0.01) = 3.24 \text{ cm}\).
Part 3: Calculate Actual Diameter
Actual Diameter = \(3.24 - 0.06 = 3.18 \text{ cm}\).

Step 4: Final Answer:
The diameter of the bob is 3.18 cm.