2025 Units and MeasurementsVernier Callipers and Zero Error Easy Question 5 of 53 in Units and Measurements

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 division in the M.S. is $0.1\text{ cm}$ and the zero of V.S. is at $x = 0.1\text{ cm}$ when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is $M = 5\text{ cm}$ and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is:

A 5.18 cm
B 5.08 cm
C 4.98 cm
D 5.00 cm

The answer is C.

Solution

Observed reading $= 5 + 8 \times 0.01 = 5.08\text{ cm}$. With positive zero error $+0.1\text{ cm}$, true diameter $= 5.08 - 0.10 = 4.98\text{ cm}$.

Least Count of the Vernier calliper: $LC = 1\text{ MSD} - 1\text{ VSD} = 0.1\text{ cm} - \frac{9}{10}(0.1\text{ cm}) = 0.01\text{ cm}$. Observed reading: $MSR + (VSD \times LC) = 5\text{ cm} + (8 \times 0.01\text{ cm}) = 5.08\text{ cm}$. When the jaws are closed, the zero of the Vernier scale lies at $+0.1\text{ cm}$, which constitutes a positive zero error of $+0.10\text{ cm}$. Corrected reading: $\text{True reading} = \text{Observed reading} - \text{Zero error} = 5.08\text{ cm} - (+0.10\text{ cm}) = 4.98\text{ cm}$.

Concept Zero Error Correction

Units and Measurements in past papers

53 questions from this chapter have appeared across 25 exam years.