If the radius of a sphere is measured as 9\,{cm} with an error of 0.03\,{cm} , then find the approximating error in calculating its volume.

Correct Answer is : 9.72π cm39.72π\,{cm}³9.72πcm3

Correct Answer is : 9.72π cm39.72π\,{cm}³9.72πcm3

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BITSAT 2021 Paper

Section: Mathematics

Question : 124 of 150

Marks: +1, -0

9\,\text{cm}

0.03\,\text{cm}

Solution:

Let

r

be the radius of the sphere and

\Delta r

be the error in measuring the radius. Then,

r=9\,\text{cm}

and

\Delta r=0.03\,\text{cm}

Let V be the volume of the sphere. Then,

V=\frac{4}{3}\pi r^{3}

\Rightarrow \frac{dV}{dr}=4\pi r^{2}

⇒

\left(\frac{dV}{dr}\right)_{r=9}=4\pi \times 9^{2}=324\pi

Let

\Delta V

be the error in

V

due to error

\Delta r

r

Then,

\Delta V=\frac{dV}{dr}\Delta r

\Rightarrow \Delta V=324\pi \times 0.03

=9.72\pi\,\text{cm}^{3}

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