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Units and Measurements question

2023 · 30 Jan · Shift 1 · Q66
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Units and Measurements question

2023 · 30 Jan · Shift 1 · Q66

JEE MainPhysicsUnits and MeasurementsNumerical+4 / −1
In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by 0.5 mm0.5 \mathrm{~mm}0.5 mm on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while 46th 46^{\text {th }}46th  division the circular scale coincide with the reference line. The diameter of the wire is ‾\underline{\hspace{2cm}}​×10−2 mm\times 10^{-2} \mathrm{~mm}×10−2 mm.
Numerical answer
View written solutionFree

Correct answer: 220

  1. Find the least count of the screw gauge

Given:

  • Number of divisions on circular scale =100=100=100
  • Pitch === distance moved in one complete rotation =0.5 mm=0.5\,\text{mm}=0.5mm

So, least count is

L.C.=0.5100=0.005 mm\text{L.C.} = \frac{0.5}{100} = 0.005\,\text{mm}L.C.=1000.5​=0.005mm
  1. Determine the zero error

When the studs are in contact, the zero of circular scale lies 6 divisions below the reference line.

This means the screw gauge shows a positive zero error of 6 divisions.

Hence,

Zero error=6×0.005=0.03 mm\text{Zero error} = 6 \times 0.005 = 0.03\,\text{mm}Zero error=6×0.005=0.03mm

Therefore, zero correction is

Zero correction=−0.03 mm\text{Zero correction} = -0.03\,\text{mm}Zero correction=−0.03mm
  1. Observed reading with the wire
  • 4 linear scale divisions are visible.
  • In such a screw gauge, one main scale division equals the pitch =0.5 mm=0.5\,\text{mm}=0.5mm.

So main scale reading is

MSR=4×0.5=2.0 mm\text{MSR} = 4 \times 0.5 = 2.0\,\text{mm}MSR=4×0.5=2.0mm

Circular scale reading:

CSR=46×0.005=0.23 mm\text{CSR} = 46 \times 0.005 = 0.23\,\text{mm}CSR=46×0.005=0.23mm

Observed reading:

Observed reading=2.0+0.23=2.23 mm\text{Observed reading} = 2.0 + 0.23 = 2.23\,\text{mm}Observed reading=2.0+0.23=2.23mm
  1. Apply zero correction

True diameter:

True reading=2.23−0.03=2.20 mm\text{True reading} = 2.23 - 0.03 = 2.20\,\text{mm}True reading=2.23−0.03=2.20mm
  1. Express in the required form

We need diameter in the form

‾×10−2 mm\underline{\hspace{2cm}} \times 10^{-2}\,\text{mm}​×10−2mm

Now,

2.20 mm=220×10−2 mm2.20\,\text{mm} = 220 \times 10^{-2}\,\text{mm}2.20mm=220×10−2mm

So the required integer is

220\boxed{220}220​
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