Student A and student B used two screw gauges of equal pitch and 100… | JEE Main 2021 (25 Jul, Shift 1)

Q46JEE Main 2021 · 25 JUL · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

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.

Numerical answer type. Work it out, then open the answer.

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Answer: 13

LC $= \dfrac{0.1\text{ cm}}{100} = 0.001\text{ cm}$
Gauge A: reference O at division 5, zero error $= +0.005\text{ cm}$ ; it reads $0.322 + 0.005 = 0.327\text{ cm}$ , so CSR $= 27$
Gauge B: reference O at division 92, zero error $= -(100-92)\times 0.001 = -0.008\text{ cm}$ ; it reads $0.322 - 0.008 = 0.314\text{ cm}$ , so CSR $= 14$
Difference $= 27 - 14 = 13$ .

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Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2021 (25 Jul, Shift 1)

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