While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is 1 mm and circular scale reading is equal to 42 divisions. Pitch of screw gauge is 1 mm and it has 100 divisions on circular scale. The diameter of

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Solution

Q.
While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is 1 mm and circular scale reading is equal to 42 divisions. Pitch of screw gauge is 1 mm and it has 100 divisions on circular scale. The diameter of the wire is $\frac{x}{50}$ mm. The value of $x$ is [2024]

1 142

2 21

3 42

4 71

Ans.
(4)

MSR = 1 mm, CSR = 42, pitch = 1 mm

$L C = \frac{pitch}{No . of CSD} = (\frac{1}{100}) = 0.01 mm$

Reading by screw gauge = Main scale reading + Least count $\times$ circular scale reading

Diameter = MSR + LC $\times$ CSD = 1 mm + 0.1 mm $\times$ 42 = 1.42 mm

Diameter = 1.42 mm = $\frac{x}{50} ∴ x = 71$

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