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JEE Main 2021 (18 Mar, Shift 2) · Errors in Measurement

Errors in Measurement question from JEE Main 2021 (18 Mar, Shift 2)

Q81JEE Main 2021 · 18 MAR · SHIFT 2Medium

Errors in Measurement

The radius of a sphere is measured to be (7.50 ±\pm 0.85) cm. Suppose the percentage error in its volume is x.

The value of x, to the nearest integer, is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 34

  1. V=43πr3⇒ΔVV=3ΔrrV = \frac{4}{3}\pi r^3 \Rightarrow \frac{\Delta V}{V} = 3\frac{\Delta r}{r}
  2. =3×0.857.50×100%=34%= 3 \times \frac{0.85}{7.50} \times 100\% = 34\%
  3. Answer: x=34x = 34.
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