Question

# A screw gauge with  a pitch of 0.5mm 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 division concides 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 0.5mm and the 25th division coincides with the main scale line?

A
0.75 mm
B
0.80 mm
C
0.70 mm
D
0.50 mm

Solution

## The correct option is B 0.80 mm$$LC=\dfrac { \text{pitch} }{ \text{circular scale divisions} } =\dfrac { 0.5 }{ 50 } mm=0.01mm$$Negative zero error =  $$-5\times (0.01)mm=-0.05mm$$Measured value = $$0.5mm+{ 25×0.01–(–0.05) }mm=0.8mm$$Physics

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