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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?
$2.22 \pm 0.02 mm , \pi(1.23 \pm 0.02) mm ^2$
$2.22 \pm 0.01 mm , \pi(1.23 \pm 0.01) mm ^2$
$2.14 \pm 0.02 mm , \pi(1.14 \pm 0.02) mm ^2$
$2.14 \pm 0.01 mm , \pi(1.14 \pm 0.01) mm ^2$
Solution
$\text { LC }=\frac{0.1}{100}=0.001 mm$
$\text { Zero error }=4 \times 0.001=0.004 mm$
$\text { Reading } 1=0.5 \times 4+20 \times 0.001-0.004=2.16 mm$
$\text { Reading } 2=0.5 \times 4+16 \times 0.001-0.004=2.12 mm$
$\text { Mean value }=2.14 mm$
$\text { Mean absolute error }=\frac{0.02+0.02}{2}=0.02$
$\text { Diameter }=2.14 \pm 0.02$
$\text { Area }=\frac{\pi}{4} d^2$
