Experimental Physics

JEE Main level, numerical answer, medium

39 of 70

Question 39 of 70

Student AA and student BB used two screw gauges of equal pitch and 100100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322cm0.322cm. The absolute value of the difference between the final circular scale readings observed by the students AA and BB is _______.
[Figure shows position of reference
OO when jaws of screw gauge are closed]
Given pitch
=0.1cm=0.1cm.

Show answer and solution

Answer: 13

For AA,
Reading
=MSR+CSR+Error=MSR+CSR+Error
0.322=0.300+CSR+5×LC0.322=0.300+CSR+5\times LC
0.322=0.300+CSR+0.0050.322=0.300+CSR+0.005
CSR=0.017CSR=0.017
For
BB,
Reading
=MSR+CSR+Error=MSR+CSR+Error
0.322=0.200+CSR+0.0920.322=0.200+CSR+0.092
CSR=0.030CSR=0.030
Difference
=0.030−0.017=0.013cm=0.030-0.017=0.013cm
Division on circular scale
=0.0130.001=13=\dfrac{0.013}{0.001}=13

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