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In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of $0.25$ $\mathrm{cm}$. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:
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The correct answer is:
$0.2150 \mathrm{~cm}$
$0.2150 \mathrm{~cm}$
Least count
$$
=\frac{\text { Value of } 1 \text { part on main scale }}{\text { Number of parts on vernier scale }}
$$
$$
=\frac{0.25}{5 \times 100} \mathrm{~cm}=5 \times 10^{-4} \mathrm{~cm}
$$
Reading $=4 \times 0.05 \mathrm{~cm}+30 \times 5 \times 10^{-4} \mathrm{~cm}$ $=(0.2+0.0150) \mathrm{cm}=0.2150 \mathrm{~cm}$ (Thickness of wire)
$$
=\frac{\text { Value of } 1 \text { part on main scale }}{\text { Number of parts on vernier scale }}
$$
$$
=\frac{0.25}{5 \times 100} \mathrm{~cm}=5 \times 10^{-4} \mathrm{~cm}
$$
Reading $=4 \times 0.05 \mathrm{~cm}+30 \times 5 \times 10^{-4} \mathrm{~cm}$ $=(0.2+0.0150) \mathrm{cm}=0.2150 \mathrm{~cm}$ (Thickness of wire)
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