Question

Q). The error in measurements of diameter and height of a cylinder are 2 % and 3 % respectively. Find percentage error in its volume.

Answer 1

Dear Student,


$$\begin{gathered} V=\frac{4}{3}{\mathrm{\pi R}}^3 \\ \mathrm{error}\mathrm{in}\mathrm{diameter}\mathrm{and}\mathrm{radius}\mathrm{remain}\mathrm{same}\mathrm{as}\mathrm{D}=2\mathrm{R},\frac{∆\mathrm{R}}{\mathrm{R}}=2\%,3\% \\ \frac{∆\mathrm{V}}{\mathrm{V}}=\frac{\frac{4}{3}\mathrm{\pi }\times 3{\mathrm{R}}^2\left(∆\mathrm{R}\right)}{\frac{4}{3}{\mathrm{\pi R}}^3}=3\left(\frac{∆\mathrm{R}}{\mathrm{R}}\right) \\ \mathrm{percentage}\mathrm{error}\mathrm{in}2\%\mathrm{error}\mathrm{in}\mathrm{radius}=3\left(2\%\right)=6\% \\ \mathrm{percentage}\mathrm{error}\mathrm{in}3\%\mathrm{error}\mathrm{in}\mathrm{radius}=3\left(3\%\right)=9\% \\ \mathrm{Regards} \end{gathered}$$