Physics And Measurement Question 205
Question: A screw gauge with a pitch of 0.5 mm 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 coincides 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 0.5 mm and the 25th division coincides the main scale line?
Options:
A) 0.75 mm
B) 0.80 mm
C) 0.70mm
D) 0.50mm
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Least count $ LC=\frac{pitch}{no.ofdiv.oncircularscale} $
$ \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 mm
