JEE MainPhysicsExperimental Physics
The pitch of a screw gauge is 1.0 mm and there are 100 divisions on its circular scale. When the jaws of the screw gauge are completely closed, the zero mark of the circular scale is 7 divisions above the reference line. When a wire is placed between the jaws to measure its diameter, the main scale reading is 3.0 mm and the 65^ th division of the circular scale coincides with the reference line. The true diameter of
Options
- A3.58 mm
- B3.72 mm
- C3.65 mm
- D4.58 mm
Correct answer
B. 3.72 mm
Step-by-step solution
The least count (LC) of the screw gauge is: LC = Pitch Total circular divisions = 1.0 100 = 0.01 mm When the jaws are closed, the zero mark of the circular scale is above the reference line. This indicates a negative zero error. Zero Error = -7 LC = -7 0.01 = -0.07 mm The measured reading of the wire's diameter is: Measured Value = MSR + ( CSR LC ) Measured Value = 3.0 + (65 0.01) = 3.65 mm The true diameter is obtained by subtracting the zero error from the measured value: True Value = Measured Value - Zero Error