JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2016
- Medium
- MCQ
A screw gauge with a pitch of \(0.5 \ mm\) and a circular scale with \(50\) divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that wen the two jaws of the screw gauge are brought in contact, the \(45^{th} \)division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet (in \(mm\)) if the main scale reading is \(0.5\ mm\) and the \(25^{th}\) division coincides with the main scale line
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(0.80\)
Step-by-step Solution
Detailed explanation
\(L.C = \frac{{0.5}}{{50}} = 0.001\,mm\)
zero error \(= 5 \times 0.001 = 0.05\) mm negative
zero error \(= 5 \times 0.001 = 0.05\) mm negative
Reading \(= \left( {0.5 + 25 \times 0.01} \right) + 0.05 = 0.80\) mm
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