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Units and Measurements question

2025 · 24 Jan · Shift 1 · Q70
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Units and Measurements question

2025 · 24 Jan · Shift 1 · Q70

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
For an experimental expression y=32.3×112527.4y=\frac{32.3 \times 1125}{27.4}y=27.432.3×1125​, where all the digits are significant. Then to report the value of yyy we should write
  1. A
    y=1326.186y=1326.186y=1326.186
  2. B
    y=1326.2y=1326.2y=1326.2
  3. C
    y=1326.19y=1326.19y=1326.19
  4. D
    y=1330y=1330y=1330
View written solutionFree

Correct answer: D

  1. Given expression

    y=32.3×112527.4y=\frac{32.3\times 1125}{27.4}y=27.432.3×1125​

  2. Count significant figures in each measured number

    • 32.332.332.3 has 3 significant figures.
    • 112511251125 has 4 significant figures.
    • 27.427.427.4 has 3 significant figures.
  3. Rule for multiplication/division

    In multiplication or division, the result should have the same number of significant figures as the quantity with the least number of significant figures.

    Here, the least number of significant figures is 3.

    So, final value of yyy must be reported to 3 significant figures.

  4. Calculate the numerical value

    First multiply:

    32.3×1125=36337.532.3\times 1125 = 36337.532.3×1125=36337.5

    Now divide:

    y=36337.527.4y=\frac{36337.5}{27.4}y=27.436337.5​

    y≈1326.1861…y\approx 1326.1861\ldotsy≈1326.1861…

  5. Round to 3 significant figures

    1326.186≈1.33×1031326.186\approx 1.33\times 10^31326.186≈1.33×103

    Writing this in ordinary decimal form to 3 significant figures:

    y=1330y=1330y=1330

  6. Check options

    • A: 1326.1861326.1861326.186 → too many significant figures
    • B: 1326.21326.21326.2 → too many significant figures
    • C: 1326.191326.191326.19 → too many significant figures
    • D: 133013301330 → corresponds to 333 significant figures in this context

Therefore, the correct option is D.

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