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JEE Main201815 Apr 2018Morning ShiftPhysicsExperimental PhysicsActual

In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 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:

Options

  1. A0.0430 ~cm
  2. B0.3150 ~cm
  3. C0.4300 ~cm
  4. D0.2150 ~cm

Correct answer

D. 0.2150 ~cm

Step-by-step solution

Least count = Value of 1 part on main scale Number of parts on vernier scale = 0.25 5 100 ~cm =5 10⁻⁴ ~cm Reading =4 0.05 ~cm +30 5 10⁻⁴ ~cm =(0.2+0.0150) cm =0.2150 ~cm (Thickness of wire)

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