A) 0.50 mm
B) 0.75 mm
C) 0.80 mm
D) 0.70 mm
Correct Answer: C
Solution :
[c] 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 mm |
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