Q51 / 75 · JEE Main 2026 (05 April Shift 1) · Physics
In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and $7^{\text{th}}$ Vernier division coincides with a main scale reading. If the value of $1$ main scale division is $1$ mm and there are $10$ Vernier scale divisions, then the Vernier caliper has
- A$0.07$ cm negative zero error
- B$0.7$ cm negative zero error
- Correct answerC$0.07$ cm positive zero error
- D$0.7$ cm positive zero error
Solution
Least count of the Vernier calipers is given by $\text{LC} = \dfrac{1 \text{ MSD}}{\text{Number of VSD}}$.
Substituting the given values, $\text{LC} = \dfrac{1 \text{ mm}}{10} = 0.1 \text{ mm} = 0.01 \text{ cm}$.
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
Substituting the given values, $\text{LC} = \dfrac{1 \text{ mm}}{10} = 0.1 \text{ mm} = 0.01 \text{ cm}$.
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
Hence, the correct answer is option: 'C'.