A specialized Vernier calliper has its main scale graduated such that 1 main scale division is equal to 0.5 mm. Twenty Vernier scale divisions are equal to nineteen main scale divisions. When the jaws are fully closed

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 1.5175 cm

The least count (LC) of the Vernier calliper is calculated as:

1 \text{ MSD} = 0.5 \text{ mm}

20 \text{ VSD} = 19 \text{ MSD} \Rightarrow 1 \text{ VSD} = \dfrac{19}{20} \times 0.5 \text{ mm} = 0.475 \text{ mm}

\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 0.5 \text{ mm} - 0.475 \text{ mm} = 0.025 \text{ mm} = 0.0025 \text{ cm}

When the jaws are closed, the Vernier zero is to the right of the main scale zero, indicating a positive zero error.

\text{Zero error} = +5 \times \text{LC} = +5 \times 0.0025 \text{ cm} = +0.0125 \text{ cm}

During the measurement, the observed reading is:

\text{Observed reading} = \text{Main scale reading} + (\text{Vernier scale reading} \times \text{LC})

\text{Observed reading} = 1.5 \text{ cm} + (12 \times 0.0025 \text{ cm}) = 1.5 \text{ cm} + 0.0300 \text{ cm} = 1.5300 \text{ cm}

The corrected diameter is obtained by subtracting the zero error from the observed reading:

\text{Corrected reading} = \text{Observed reading} - \text{Zero error}

\text{Corrected reading} = 1.5300 \text{ cm} - 0.0125 \text{ cm} = 1.5175 \text{ cm}

Answer:

1.5175

cm

A specialized Vernier calliper has its main scale graduated such...

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