JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2019
- Hard
- MCQ
The pitch and the number of divisions, on the circular scale, for a given screw gauge are \(0.5\,mm\) and \(100\) respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies \(3\) divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet, are \(5.5\,mm\) and \(48\) respectively, the thickness of this sheet is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(5.725\,mm\)
Step-by-step Solution
Detailed explanation
\(L C=\frac{P i t c h}{\text {No.ofdivision}}\)
\(L C=0.5 \times 10^{-2} mm\)
+ve error \(=3 \times 0.5 \times 10^{-2} mm\)
\(=1.5 \times 10^{-2} mm =0.015 mm\)
Reading \(= MSR + CSR -(+\text { ve error })\)
\(=5.5 m m+\left(48 \times 0.5 \times 10^{-2}\right)-0.015\)
\(=5.5+0.24-0.015=5.725 mm\)
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