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 zer

Question

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 in such a way that 4 t h Vernier division coincides with a division of the main scale. If this Vernier calliper measures the length of a wire to be 1 cm, the actual length of the wire is :NEET 2026 (Re-NEET)

TestHub · Accepted Answer

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

$LC = \frac{1 mm}{10} = 0.1 mm = 0.01 cm$

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:

$Zero error = - 4 \times LC = - 4 \times 0.01 cm = - 0.04 cm$

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

$Actual length = Measured length - Zero error$

$Actual length = 1.00 cm - (- 0.04 cm) = 1.04 cm$

Answer: $1.04$ cm

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