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JEE Main 20264 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Morning Shift
Subject
Physics
Chapter
Experimental Physics

JEE Main 2026 · 4 April · Morning Shift · Physics · Experimental Physics

In a screw gauge when the circular scale is given five complete...

In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.52.5 mm. If the circular scale has 100100 divisions, the least count of screw gauge is _____ mm.

Options

  1. A1×10−21 \times 10^{-2}
  2. B1×10−31 \times 10^{-3}
  3. C5×10−25 \times 10^{-2}
  4. D5×10−35 \times 10^{-3}

Correct answer

D. 5×10−35 \times 10^{-3}

Step-by-step solution

Pitch of the screw gauge is the linear distance moved in one complete rotation.

Pitch=2.55=0.5 mm\text{Pitch} = \dfrac{2.5}{5} = 0.5 \text{ mm}

Least count of the screw gauge is given by the ratio of the pitch to the total number of divisions on the circular scale.

Least count=PitchNumber of divisions on circular scale\text{Least count} = \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}

Least count=0.5100=0.005 mm\text{Least count} = \dfrac{0.5}{100} = 0.005 \text{ mm}

Least count=5×10−3 mm\text{Least count} = 5 \times 10^{-3} \text{ mm}

Answer: 5×10−35 \times 10^{-3}
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