JEE Main 2026 — Experimental Physics Question with Solution

Least count of the Vernier calipers is given by $\text{LC}=\frac{1\text{ MSD}}{\text{Number of VSD}}$.

Substituting the given values, $\text{LC}=\frac{1\text{ mm}}{10}=0.1\text{ mm}=0.01\text{ cm}$.

Since the zero of the Vernier scale is shifted to the right of the zero of the main scale, the zero error is positive.

The zero error is calculated as $+(n\times \text{LC})$, where $n$ is the coinciding Vernier division.

Given that the $7^{\text{th}}$ Vernier division coincides with a main scale reading, $n=7$.

Zero error $=+(7\times 0.01\text{ cm})=+0.07\text{ cm}$.

Therefore, the Vernier caliper has a $0.07\text{ cm}$ positive zero error.

Answer: $0.07$ cm positive zero error