JEE Advance - Physics (2022 - Paper 2 Online - No. 17)
Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 \mathrm{~mm}$. 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 divisions |
| Attempt-1: With wire | 4 divisions | 20 divisions |
| Attempt-2: With wire | 4 divisions | 16 divisions |
What are the diameter and cross-sectional area of the wire measured using the screw gauge?
$2.22 \pm 0.02 \mathrm{~mm}, \pi(1.23 \pm 0.02) \mathrm{mm}^{2}$
$2.22 \pm 0.01 \mathrm{~mm}, \pi(1.23 \pm 0.01) \mathrm{mm}^{2}$
$2.14 \pm 0.02 \mathrm{~mm}$, $\pi(1.14 \pm 0.02) \mathrm{mm}^{2}$
$2.14 \pm 0.01 \mathrm{~mm}, \pi(1.14 \pm 0.01) \mathrm{mm}^{2}$
Explanation
$$
\begin{aligned}
& \text {LC }=\frac{0.1}{100}=0.001 \mathrm{~mm} \\\\
& \text {Zero error }=4 \times 0.001=0.004 \mathrm{~mm} \\\\
& \text {Reading } 1=0.5 \times 4+20 \times 0.001-0.004=2.16 \mathrm{~mm} \\\\
& \text {Reading } 2=0.5 \times 4+16 \times 0.001-0.004=2.12 \mathrm{~mm} \\\\
& \text {Mean value }=2.14 \mathrm{~mm} \\\\
& \text {Mean absolute error }=\frac{0.02+0.02}{2}=0.02 \\\\
& \text {Diameter }=2.14 \pm 0.02 \\\\
& \text {Area }=\frac{\pi}{4} \mathrm{~d}^2
\end{aligned}
$$
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