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JEE Main 20163 April
Exam
JEE Main
Year
2016
Date
3 April
Subject
Physics
Chapter
Experimental Physics

JEE Main 2016 · 3 April · Physics · Experimental Physics

A screw gauge with a pitch of 0.5mm and a circular scale with 50...

A screw gauge with a pitch of 0.5mm0.5mm and a circular scale with 5050 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th{45}^{th} division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5mm0 .5 mm and the 25th{25}^{th} division coincides with the main scale line?

Options

  1. A0.70mm0.70mm
  2. B0.50mm0.50mm
  3. C0.75mm0.75mm
  4. D0.80mm0.80mm

Correct answer

D. 0.80mm0.80mm

Step-by-step solution

LC=0.550=0.01mmLC = \dfrac{0.5}{50} = 0.01mm

Zero error =0.50−0.45=−0.05= 0.50 - 0.45 = - 0.05

Thickness =(0.5+25×0.01)+0.05= \left(0.5 + 25 \times 0.01\right) + 0.05

=0.50+0.25+0.05= 0.50 + 0.25 + 0.05

=0.80mm= 0.80mm

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