A screw gauge has a pitch of 0.5 text mm and 100 circular divisions. When tightened without an object, the zero of the circular scale is 3 divisions below the mean line. For a thin sheet, the readings are: Main Scale = 5.5 text mm , Circular Scale = 48. Find the sheet thickness.

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EEJ MAIN Physics QUESTION #7038

Question 1

A screw gauge has a pitch of $0.5 \text{ mm}$ and $100$ circular divisions. When tightened without an object, the zero of the circular scale is $3$ divisions below the mean line. For a thin sheet, the readings are: Main Scale $= 5.5 \text{ mm}$, Circular Scale $= 48$. Find the sheet thickness.

$5.755 \text{ mm}$
$5.950 \text{ mm}$
$5.725 \text{ mm}$✔️
$5.740 \text{ mm}$

Correct Answer Explanation

$LC = \frac{0.5}{100} = 0.005 \text{ mm}$.

Zero Error: Since the zero is 3 divisions below the mean line, it is a positive zero error. Zero Error $= +3 \times 0.005 = +0.015 \text{ mm}$.

Observed Reading: $5.5 + (48 \times 0.005) = 5.5 + 0.240 = 5.740 \text{ mm}$.

Actual Thickness: Observed Reading $-$ Zero Error $= 5.740 - 0.015 = 5.725 \text{ mm}$.

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