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In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are $50$ divisions on the circular scale, and the main scale moves by $0.5 \,{mm}$ on a complete rotation. For a particular observation the reading on the main scale is $5\, {mm}$ and the $20^{\text {th }}$ division of the circular scale coincides with reference line. Calculate the true reading. (in ${mm}$)
$5.00$
$5.25$
$5.15$
$5.20$
Solution
Least count $({L} \cdot {C})=\frac{0.5}{50}$
True reading $=5+\frac{0.5}{50} \times 20-\frac{0.5}{50} \times 5$
$=5+\frac{0.5}{50}(15)=5.15\, {mm}$
Similar Questions
Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 mm$. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.
| Measurement condition | Main scale reading | Circular scale reading |
| Two arms of gauge touching each other without wire | $0$ division | $4$ division |
| Attempt-$1$: With wire | $4$ division | $20$ division |
| Attempt-$2$: With wire | $4$ division | $16$ division |
What are the diameter and cross-sectional area of the wire measured using the screw gauge?
