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The relative error in the determination of the surface area of a sphere is α. Then the relative error in the determination of its volume is :

  • Option 1)

    \frac{3}{2}\alpha

     

     

     

  • Option 2)

    \frac{2}{3}\alpha

  • Option 3)

    \frac{5}{2}\alpha

  • Option 4)

    \alpha

 

Answers (1)

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As we learned 

 

Relative error /Fractional error -

Ratio of mean absolute error to mean value of the quantity measured

- wherein

\dpi{100} \frac{\Delta\bar{ a}}{a_{m}}

\Delta\bar{a} - mean\: absolute\: error

a_{m}- \: absolute\: error

 

V = \frac{4\pi}{3}r^{2} \Rightarrow \frac{dv}{dr} = 4\pi r^{2}

 

 

 

A = 4\pi r^{2} \Rightarrow \frac{dA}{dr}= 8\pi r

\frac{dv}{v} = 3\frac{dr}{r} and \frac{dA}{A}= 2.\frac{dr}{r}

\therefore \frac{dv}{v}=\frac{3}{2}\cdot\frac{dA}{A}= \frac{3}{2} \alpha


Option 1)

\frac{3}{2}\alpha

 

 

 

Option 2)

\frac{2}{3}\alpha

Option 3)

\frac{5}{2}\alpha

Option 4)

\alpha

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Plabita

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