RE-NEET 2026 Physics Vernier Callipers MCQ Question

This question was dropped by NTA — every candidate is awarded the marks.

The instrument reads 1 cm, and 1 MSD = 1 mm with 10 vernier divisions, so the least count is $\frac{1\ \text{mm}}{10} = 0.01$ cm. With the jaws closed the vernier zero lies to the left of the main-scale zero and the 4th vernier division coincides, so the instrument carries a zero error before any measurement is taken.

Which correction that implies depends on how the left-shifted zero is counted: reading the coinciding division directly gives a $-0.04$ cm error and an actual length of $1.04$ cm, while counting from the far end $(10-4)$ gives $-0.06$ cm and $0.94$ cm. The question does not fix that convention, and no single option follows from it.