JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
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 ........... \(\times 10^{-2} \,cm\)
- A \(160\)
- B \(150\)
- C \(189\)
- D \(180\)
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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