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In a screw gauge, five complete rotations move the screw 0.25 cm. There are 100 circular scale divisions. A wire reading gives 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is

Asked in JEE Main Online 2018 · Total reading

Answer: (2) 0.2150 cm

Step-by-step solution

Pitch =(0.25)/5=0.05 cm; L.C. =(0.05)/(100)=0.0005 cm.

Main scale =4×0.05=0.20 cm; circular =30×0.0005=0.015 cm.

Thickness =0.20+0.015=0.215 cm.

Why the other options are wrong

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