A screw gauge (pitch 0.5 text mm , 50 circular divisions) measures a sheet. Initially, with jaws closed, the 45^ th division aligns with the main scale line, and the zero of the main scale is barely visible. If the measurement shows a main scale reading of 0.5 text mm and the 25^ th circular division aligns, what is the sheet's thickness?

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EEJ MAIN Physics QUESTION #7035

Question 1

A screw gauge (pitch $0.5 \text{ mm}$, $50$ circular divisions) measures a sheet. Initially, with jaws closed, the $45^{th}$ division aligns with the main scale line, and the zero of the main scale is barely visible. If the measurement shows a main scale reading of $0.5 \text{ mm}$ and the $25^{th}$ circular division aligns, what is the sheet's thickness?

$0.80 \text{ mm}$✔️
$0.70 \text{ mm}$
$0.50 \text{ mm}$
$0.75 \text{ mm}$

Correct Answer Explanation

First, calculate Least Count ($LC$): $LC = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm}$.

Zero Error: The $45^{th}$ division coincides and the zero is barely visible, indicating a negative zero error. Error $= (45 - 50) \times 0.01 = -0.05 \text{ mm}$.

Observed Reading: $MSR + (CSR \times LC) = 0.5 + (25 \times 0.01) = 0.75 \text{ mm}$.

True Thickness: Observed Reading $-$ Zero Error $= 0.75 - (-0.05) = 0.80 \text{ mm}$.

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