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JEE Main 202125 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Morning Shift
Subject
Physics
Chapter
Experimental Physics

JEE Main 2021 · 25 February · Morning Shift · Physics · Experimental Physics

The pitch of the screw gauge is 1mm and there are 100 divisions on...

The pitch of the screw gauge is 1mm1mm and there are 100100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 88 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd{72}^{\text{nd}} division on circular scale coincides with the reference line. The radius of the wire is

Options

  1. A1.64mm1 . 64mm
  2. B0.82mm0 . 82mm
  3. C0.90mm0 . 90mm
  4. D1.80mm1 . 80mm

Correct answer

B. 0.82mm0 . 82mm

Step-by-step solution

Least count =1mm100=0.01mm= \dfrac{1mm}{100} = 0 . 01mm

zero error =+8×LC=+0.08mm= + 8 \times LC = + 0 . 08mm

True reading (Diameter)

=(1mm+72×LC)−= \left(1mm + 72 \times LC\right) - (Zero error)

=(1mm+72×0.01mm)−0.08mm= \left(1mm + 72 \times 0 . 01mm\right) - 0 . 08mm

=1.72mm−0.08mm= 1 . 72mm - 0 . 08mm

=1.64mm= 1 . 64mm

therefore, radius =1.642=0.82mm.= \dfrac{1 . 64}{2} = 0 . 82mm .

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