JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2022
- Medium
- Numerical
The vernier constant of Vernier callipers is \(0.1 \,mm\) and it has zero error of \((-0.05) \,cm\). While measuring diameter of a sphere, the main scale reading is \(1.7 \,cm\) and coinciding vernier division is \(5\). The corrected diameter will be ........... \(\times 10^{-2} \,cm\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(180\)
Step-by-step Solution
Detailed explanation
Measured diameter \(= MSR + VSR \times VC\)
\(=1.7+0.01 \times 5\)
\(=1.75\)
Corrected \(=\) Measured - Error
\(=1.75-(-0.05)\)
\(=1.80 \,cm\)
\(=180 \times 10^{-2} \,cm\)
\(180\)
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