JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
The pitch of the screw gauge is \(1\, mm\) and there are \(100\) divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies \(8\) divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while \(72^{\text {nd }}\) division on circular scale coincides with the reference line. The radius of the wire is.........\(mm\)
- A \(1.64\)
- B \(0.82\)
- C \(1.80\)
- D \(0.90\)
Answer & Solution
Correct Answer
(B) \(0.82\)
Step-by-step Solution
Detailed explanation
Least count \(=\frac{1 mm }{100}=0.01 mm\) zero error \(=+8 \times LC =+0.08 mm\) True reading (Diameter) \(=(1 mm +72 \times LC )-(\) Zero error \()\) \(=(1 mm +72 \times 0.01 mm )-0.08 mm\) \(=1.72 mm -0.08 mm\) \(=1.64 mm\) therefore, radius \(=\frac{1.64}{2}=0.82 mm\)
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