Question

Consider the diameter of a spherical object being measured with the help of a Vernier Callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least count in the M.S. is 0.1 cm and the zero of V.S. is at -0.1 cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is \(M = 5\) cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is:

• A. \( 5.08 \text{ cm} \)
• B. \( 4.98 \text{ cm} \)
• C. \( 5.09 \text{ cm} \)
• D. \( 5.18 \text{ cm} \)

Hint:

Remember the steps for Vernier Callipers: find LC, then ZE and ZC, then the measured reading using MSR and VSC, and finally apply the zero correction. A negative zero error means the zero mark of the Vernier scale is to the left of the main scale zero, and the correction will be positive.

Answer 1

The Correct Option is D

Step 1: Calculate the Least Count (LC)
10 VSD = 9 MSD.
1 MSD = 0.1 cm.
Value of 10 VSD = 9 × 0.1 cm = 0.9 cm.
Value of 1 VSD = 0.9 / 10 cm = 0.09 cm.
Least Count (LC) = 1 MSD - 1 VSD = 0.1 cm - 0.09 cm = 0.01 cm.

Step 2: Determine the Zero Error (ZE) and Zero Correction (ZC)
The zero of the Vernier scale is at -0.1 cm, meaning the zero mark of the Vernier scale is 0.1 cm to the left of the zero mark of the main scale when the jaws are closed.
Zero Error (ZE) = -0.1 cm.
Zero Correction (ZC) = - (ZE) = - (-0.1 cm) = +0.1 cm.

Step 3: Calculate the Measured Reading
Main Scale Reading (MSR) = 5 cm.
Vernier Scale Coincidence (VSC) = 8.
Measured Reading = MSR + (VSC × LC) = 5 cm + (8 × 0.01 cm) = 5 cm + 0.08 cm = 5.08 cm.

Step 4: Apply the Zero Correction
Corrected Reading = Measured Reading + Zero Correction = 5.08 cm + 0.1 cm = 5.18 cm.