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Units and Measurements question

2025 · 28 Jan · Shift 1 · Q74
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Units and Measurements question

2025 · 28 Jan · Shift 1 · Q74

JEE MainPhysicsUnits and MeasurementsNumerical+4 / −1
A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm , respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be x100\frac{x}{100}100x​ where xxx is ‾\underline{\hspace{2cm}}​ .
Numerical answer
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Correct answer: 3

  1. Given data
  • Length of sheet: l=5 mml = 5\,\text{mm}l=5mm
  • Breadth of sheet: b=2.5 mmb = 2.5\,\text{mm}b=2.5mm
  • Pitch of screw gauge: 0.75 mm0.75\,\text{mm}0.75mm
  • Circular scale divisions: 151515
  1. Least count of screw gauge

The least count is

LC=pitchnumber of circular divisions=0.7515=0.05 mm\text{LC} = \frac{\text{pitch}}{\text{number of circular divisions}} = \frac{0.75}{15} = 0.05\,\text{mm}LC=number of circular divisionspitch​=150.75​=0.05mm

So the maximum absolute error in each measurement is

Δl=Δb=0.05 mm\Delta l = \Delta b = 0.05\,\text{mm}Δl=Δb=0.05mm
  1. Area of the rectangular sheet
A=l bA = l\,bA=lb

For multiplication, the maximum fractional error is

ΔAA=Δll+Δbb\frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b}AΔA​=lΔl​+bΔb​

Substitute the values:

ΔAA=0.055+0.052.5\frac{\Delta A}{A} = \frac{0.05}{5} + \frac{0.05}{2.5}AΔA​=50.05​+2.50.05​ ΔAA=0.01+0.02=0.03\frac{\Delta A}{A} = 0.01 + 0.02 = 0.03AΔA​=0.01+0.02=0.03

Thus,

ΔAA=3100\frac{\Delta A}{A} = \frac{3}{100}AΔA​=1003​

So,

x=3x = 3x=3
  1. Comparison with stored answer

Stored correct answer = 333

Our derived answer also gives x=3x=3x=3.

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