The distance moved by the screw of a screw gauge is 2 m m in four rotations and there are 50 divisions on its cap. When nothing is put between its jaws, 30 th division of circular scale coincides with

Question

The distance moved by the screw of a screw gauge is $2 m m$ in four rotations and there are $50$ divisions on its cap. When nothing is put between its jaws, $30^{th}$ division of circular scale coincides with the reference line, with zero of circular scale lying above the reference line. When a plate is placed between the jaws, the main scale reads $2$ divisions and the circular scale reads $20$ divisions. The thickness of the plate is

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Answer:

A

Solution:

Pitch $= \frac{1}{2} mm$
least count $= \frac{0.5 mm}{50} = 0.01 mm$
Reading $= [2 \times \frac{1}{2} + 20 \times 0.01 - (- 30 \times 0.01)] = 1.5 mm$

TestHub · Accepted Answer

Pitch = 1 2 mm least count = 0.5 mm 50 = 0.01 mm Reading = [ 2 × 1 2 + 20 × 0.01 − ( − 30 × 0.01 ) ] = 1.5 mm

Physics - Practical physics Question with Solution | TestHub

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