A) 0.75 mm
B) 0.80 mm
C) 0.70mm
D) 0.50mm
Correct Answer: B
Solution :
[b] Least count \[LC=\frac{pitch}{no.of\,div.on\,circular\,scale}\] \[\Rightarrow \] \[LC=\frac{0.5mm}{50}=0.01mm\] When jaws are closed, the zero error will be = main scale reading + (circular scale reading) (Least count) =-0.5mm + (45) (0.01) Hence Zero error e=-0.05mm When the sheet is placed between the jaws: Measured thickness = 0.5 mm+(25) (0.01)=0.75mm Hence actual thickness or true reading = observed reading + zero error =0.75 mm - (-0.05) = 0.80 mmYou need to login to perform this action.
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