Question

If an error of k% is made in measuring the radius of a sphere, then percentage error in its volume is
(a) k%
(b) 3k%
(c) 2k%
(d) k/3%

Answer 1

(b) 3k%
Let x be the radius of the sphere and y be its volume.
Then,
$$\begin{gathered} \frac{∆x}{x}\times 100=k \\ \mathrm{Also},y=\frac{4}{3}\pi x^3 \\ \Rightarrow \frac{dy}{dx}=4\pi x^2 \\ \Rightarrow \frac{∆y}{y}=\frac{4\pi x^2}{y}dx=\frac{4\pi x^2}{{\displaystyle \frac{4}{3}}\pi x^3}\times \frac{kx}{100} \\ \Rightarrow \frac{∆y}{y}\times 100=3k \\ \mathrm{Hence},\mathrm{the}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{volume}\mathrm{is}3k\%. \end{gathered}$$