JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
In a screw gauge, \(5\) complete rotations of the screw cause it to move a linear distance of \(0.25\, cm\). There are \(100\) circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of \(4\) main scale divisions and \(30\) circular scale divisions . Assuming negligible zero error, the thickness of the wire is
- A \(0.0430\,cm\)
- B \(0.3150\,cm\)
- C \(0.43 00\,cm\)
- D \(0.2150\, cm\)
Answer & Solution
Correct Answer
(D) \(0.2150\, cm\)
Step-by-step Solution
Detailed explanation
In one rotation scale moves \(\frac{0.25}{5}=0.05 cm\) Least count \(=0.05 \times 10^{-2} cm\) For 4 main scale division \(=4 \times 0.05=0.2 cm\) For circular scale divosion \(=30 \times 0.05 \times 10^{-2}=1.5 \times 10^{-2} cm\) Thickness of wire \(=0.2+0.015=0.2150 cm\)
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