While measuring diameter of wire using screw gauge the following… | JEE Main 2024 (6 Apr, Shift 1)

Q24JEE Main 2024 · 6 APR · SHIFT 1Medium

Least Count, Vernier Callipers & Screw Gauge

While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is $1\text{ mm}$ and circular scale reading is equal to 42 divisions. Pitch of screw gauge is $1\text{ mm}$ and it has 100 divisions on circular scale. The diameter of the wire is $\frac{x}{50}\text{ mm}$ . The value of $x$ is:

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Answer: (b)

LC $= \dfrac{1\text{ mm}}{100} = 0.01\text{ mm}$
Diameter $= 1 + 42\times 0.01 = 1.42\text{ mm}$
$\dfrac{x}{50} = 1.42 \Rightarrow x = 71$ . So the answer is (b).

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Least Count, Vernier Callipers & Screw Gauge question from JEE Main 2024 (6 Apr, Shift 1)

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