QJEEM18Ph09ce: JEE Main 2018 Physics - Class 12 JEEM18Ph09ce Solution: In a screw gauge, 5 complete rotations of the screw cause it to …

Question

Verified Solution • 2025–26 Session moderate

In a screw gauge,

5

complete rotations of the screw cause it to move a linear distance of

0.25

c m .

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 :

Accepted Answer

5 complete rotations = 0.25 cms

So, 1 complete rotation of screw = 0.05 cm

∴

1 main scale division = 0.05 cm

1 circular scale =

\frac{0.05}{100}

= 5

\times

10^{$-$ 4} cm

Thickness of a wire

= 4 main scale and 30 circular scale divisions

= 4

\times

0.05 + 30

\times

5

\times

10 ^{$-$ 4}

= 0.2150 cm.

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