Question:easy

If the percentage error in the measurement of radius is \(2\%\), then the error in measurement of volume of a sphere is

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For quantities of the form \[ y=kx^n, \] the percentage error in \(y\) is approximately \[ n\times (\text{percentage error in }x). \] Here \(V\propto r^3\), so the error gets multiplied by \(3\).
Updated On: Jul 29, 2026
  • \(6\%\)
  • \(8\%\)
  • \(4\%\)
  • \(10\%\)
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The Correct Option is A

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