JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2026
- Medium
- MCQ
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
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Answer & Solution
Correct Answer
(C) \(0.07\) cm positive zero error
Step-by-step Solution
Detailed explanation
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
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