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JEE Main Physics Class 11 Physical world and Measurements Part 2 Questions
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© examsnet.com
Question : 98
Total: 100
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.
[25 Jul 2021 Shift 1]
Your Answer:
Validate
Solution:
For (A)
Reading = MSR + CSR + Error
0.322 = 0.300 + CSR + 5 × LC
0.322 = 0.300 + CSR + 0.005
CSR = 0.017
For B
Reading = MSR + CSR + Error
0.322 = 0.200 + CSR + 0.092
CSR = 0.030
Difference = 0.030 – 0.017 = 0.013 cm
Division on circular scale
=
0.013
0.001
=
13
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