Question

the radius of sphere is measured as 5.2 plus minus 0.2 CM the percentage error in measurement of volume of the square is ​

Answer 1

Given:

Radius of the sphere = $5.2$ ± $0.2\,cm$

To find:

Percentage error in the measurement of volume of the sphere.

Solution:

We know that,

$Volume\, of\, the\,sphere \,(V)=\frac{4}{3} \pi\,r^{3}$  

We also know that,

$Percentage\, error\, in\,volume=\frac{\Delta\,V}{V} \times100$

Now, to find the expression for ΔV, Differentiate the expression for volume of the sphere with respect to $r$, i.e. differentiate $V$ with respect to $r$

$\therefore \Delta\,V=\frac{d}{dr} (\frac{4}{3} \pi\,r^{3} )$

⇒$\Delta\,V=3\times\frac{4}{3} \pi\,r^{2}\Delta\,r =4\pi\,r^{2}\Delta\,r$

We multiplied $\Delta r$ after differentiation because the radius is varying between the given tolerance limits of ±$0.2cm$

Now,

$Percentage\, error\, in\,volume=\frac{4\pi\,r^{2} \Delta\,r}{\frac{4}{3}\pi\,r^{3} } \times100$

                                            $=3\times\frac{\Delta\,r}{r}\times100$

                                           $=3\times\frac{0.2}{5.2}\times100$

                                           $= 11.538\%$

Therefore, the percentage error in measurement of the volume of the sphere is $11.538\%$