Question

In a Vernier caliper, \(N+1\)  divisions of vernier scale coincide with  \(N\) divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:

• A. \(\frac{1}{10N}\)
• B. \(\frac{1}{100(N+1)}\)
• C. \(100N\)
• D. \(10N+1\)

Answer 1

The Correct Option is B

The vernier constant of a Vernier caliper represents the smallest measurable quantity and is equivalent to its least count. The problem states that \(N+1\) divisions on the Vernier scale align with \(N\) divisions on the main scale. Given that 1 Main Scale Division (MSD) equals 0.1 mm, then \(N\) main scale divisions are equal to \(N \times 0.1 \, \text{mm} = 0.1N \, \text{mm}\). Consequently, each Vernier scale division (VSD) measures \( \frac{0.1N}{N+1} \, \text{mm}\). The vernier constant (VC) is the difference between one MSD and one VSD: \( VC = \text{MSD} - \text{VSD} = 0.1 \, \text{mm} - \frac{0.1N}{N+1} \, \text{mm} \). Simplifying this yields \( VC = \frac{0.1(N+1) - 0.1N}{N+1} = \frac{0.1}{N+1} \, \text{mm} \). To express the vernier constant in centimeters, we convert mm to cm (1 mm = 0.1 cm): \( VC = \frac{0.1 \times 0.1}{N+1} \, \text{cm} = \frac{0.01}{N+1} \, \text{cm} \). As a fraction, this is \( \frac{1}{100(N+1)} \, \text{cm} \). Therefore, the vernier constant in centimeters is \(\frac{1}{100(N+1)}\).