JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2023
- Hard
- Numerical
In a screw gauge, there are \(100\) divisions on the circular scale and the main scale moves by \(0.5\,mm\) on a complete rotation of the circular scale. The zero of circular scale lies \(6\) divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, \(4\) linear scale divisions are clearly visible while \(46^{\text {th }}\) division the circular scale coincide with the reference line. The diameter of the wire is \(...........\times 10^{-2}\,mm\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(22\)
Step-by-step Solution
Detailed explanation
\(\text { Least count }=\frac{\text { Pitch }}{\text { No. of circular divisions }}\)
\(=\frac{0.5\,mm }{100}\)
\(\text { Least count }=5 \times 10^{-3}\,mm\)
\(\text { Positive Error }= MSR + CSR ( LC )\)
\(=0\,mm +6\left(5 \times 10^{-3}\,mm \right)\)
\(\text { Reading of Diameter }= MSR + CSR ( LC )-\)
\(\text { Positive zero error }\)
\(=4 \times 0.5\,mm +\left(46\left(5 \times 10^{-3}\right)\right)-6\left(5 \times 10^{-3}\right)\,mm\)
\(=2\,mm +40 \times 5 \times 10^{-3}\,mm =2.2\,mm\)
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