A) 0.50 mm
B) 0.75 mm
C) 0.80 mm
D) 0.70 mm
Correct Answer: C
Solution :
Least count\[\text{=}\frac{\text{pitch}}{\text{no}\text{.}\,\text{of}\,\text{division}\,\text{on}\,\text{circular}\,\text{scale}}\]\[\text{=}\frac{0.5mm}{50}\]LC = 0.001 mm ?ve zero errow \[=-5\times LC=-0.005mm\] Measured value = main scale reading + screw gauge reading zero error \[=0.\text{5mm}+\{\text{25}\times 0.00\text{1}(0.0\text{5})\}\text{mm}\] = 0.8 mmYou need to login to perform this action.
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