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If an error of k%k\% is made in measuring the radius of a sphere, then percentage error in its volume is

হানি নাটস

Volume of sphere V=43πr3V=\dfrac { 4 }{ 3 } \pi { r }^{ 3 }
dVdr=4πr2\displaystyle \dfrac{dV}{dr}=4\pi r^{2}

Given, percentage error in measuring radius =k%
Δrr=k100\Rightarrow \displaystyle \dfrac { \Delta r }{ r } =\dfrac { k }{ 100 }

Δr=kr100\Rightarrow \displaystyle { \Delta r }=\dfrac { kr }{ 100 }

Now, approximate error in measuring V=dV=(dVdr)Δr\displaystyle =dV= (\dfrac{dV}{dr}){ \Delta r}

=k1004πr3=3k100V=3k%\displaystyle = \dfrac{k}{100} {4\pi r^{3}} = \dfrac{3k}{100}V =3k\% of V

Percentage error in measuring V=3k%V =3k\%

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