One main scale division of a Vernier calliper is equal to 1 mm and the number of divisions on the Vernier scale is 10 . When both the jaws touch each other, the Vernier scale shifts to the left of zero of the main scale

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 1.04 cm

The least count (LC) of the Vernier calliper is given by:

\text{LC} = \dfrac{1 \text{ mm}}{10} = 0.1 \text{ mm} = 0.01 \text{ cm}

Since the zero of the Vernier scale shifts to the left of the zero of the main scale, the instrument has a negative zero error.

The zero error is calculated as:

\text{Zero error} = -4 \times \text{LC} = -4 \times 0.01 \text{ cm} = -0.04 \text{ cm}

The actual length of the wire is obtained by subtracting the zero error from the measured length:

\text{Actual length} = \text{Measured length} - \text{Zero error}

\text{Actual length} = 1.00 \text{ cm} - (-0.04 \text{ cm}) = 1.04 \text{ cm}

Answer:

1.04

cm

One main scale division of a Vernier calliper is equal to 1 mm and...

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