If the ratio of base radius and height of a cone is 1 : 2 and percentage error in radius $\lambda \%$, then the error in its volume, is

Solution:

let $r$ be the base radius and $h$ be the height of the

cone. Then, $2r=h.$ Let $V$ be the volume of the cone. Then,

$\begin{array}{l}V=\frac{1}{3}\pi r^{2}h=\frac{4\pi r^3}{3}\\ \Rightarrow \frac{dV}{dr}=4\pi r^2\\ \therefore \mathrm{\Delta }V=\frac{dV}{dr}\mathrm{\Delta }r\\ \Rightarrow \mathrm{\Delta }V=4\pi r^2\mathrm{\Delta }r\\ \Rightarrow \frac{\mathrm{\Delta }V}{V}\times 100=\frac{4\pi r^2}{4}\pi r^3\times 100=3\left(\frac{\mathrm{\Delta }r}{r}\times 100\right)=3\lambda \end{array}$