JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2021
- Hard
- Numerical
Student \(A\) and Student \(B\) used two screw gauges of equal pitch and \(100\) equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is \(0.322\, {cm}\). The absolute value of the difference between the final circular scale readings observed by the students \(A\) and \(B\) is .... . [Figure shows position of reference \('O'\) when jaws of screw gauge are closed] Given pitch \(=0.1 \,{cm}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(13\)
Step-by-step Solution
Detailed explanation
For \(A\)
Reading \(=\) MSR \(+\) CSR \(+\) Error
\(0.322=0.300+C S R+5 \times L C\)
\(0.322=0.300+C S R+0.005\)
\(C S R=0.017\)
For \(B\)
Reading \(=\) MSR \(+\) CSR \(+\) Error
\(0.322=0.200+C S R+0.092\)
\(C S R=0.030\)
Difference \(=0.030-0.017=0.013 cm\)
Division on circular scale \(=\frac{0.013}{0.001}=13\)
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