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JEE Main 2016 · 3 Apr · Shift 1, Physics question 25

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 when the two jaws of the screw gauge are brought in contact, the 45th^{\text{th}} division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th^{\text{th}} division coincides with the main scale line ?

Choose one option
Answer and worked solution

Official answer: B

The least count is 0.550=0.01\frac{0.5}{50}=0.01 mm. With the jaws closed, the 45th division is on the line and the main-scale zero barely shows, so the zero error is −(50−45)×0.01=−0.05-(50-45)\times0.01=-0.05 mm. The reading is 0.5+25×0.01=0.750.5+25\times0.01=0.75 mm, corrected to 0.75+0.05=0.800.75+0.05=0.80 mm.

Topics:Vernier callipers and screw gauge
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