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PhysicsPractical physicsMiscellaneousEasy2 minQB
PhysicsEasynumerical

A screw gauge with a pitch of0.5 mmand a circular scale with50divisions is used to measure the thickness of a thin sheet of aluminum. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the45thdivision coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness (inmm) of the sheet if the main scale reading is0.5 mmand the30thdivision coincides with the main scale line?

Answer:
0.85
Solution:

Least count, LC=0.550=0.01 mm

Zero error, ZE=50-45×0.01=-0.05 mm

Main Scale Reading MSR=0.5 mm
Thickness, t=0.5 mm +30×0.01 mm - -0.05 mm

t=0.85 mm

Stream:JEESubject:PhysicsTopic:Practical physicsSubtopic:Miscellaneous
⏱ 2mℹ️ Source: QB

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