If the radius of a sphere is mentioned as \(7 \mathrm{~m}\) with an error of \(0.02 \mathrm{~m}\), then the…
If the radius of a sphere is mentioned as \(7 \mathrm{~m}\) with an error of \(0.02 \mathrm{~m}\), then the approximate error in calculating its volume is
\(1.83 \pi \mathrm{m}^3\)
\(2.25 \pi \mathrm{m}^3\)
\(4.39 \pi \mathrm{m}^3\)
\(3.92 \pi \mathrm{m}^3\)
Solution
Radius \((r)=7 \mathrm{~m}\)
Error in radius \((d r)=0.02 \mathrm{~m}\)
Volume of sphere \((v)=\frac{4}{3} \pi r^3\)
differentiate w.r. to ' \(r\) ' on both sides,
\(\begin{aligned}
& \frac{d v}{d r}=\frac{4 \pi}{3}\left(3 r^2\right) \\
& d v=4 \pi\left(r^2\right) \cdot d r=4 \pi(49) \cdot(0.02) \\
& d v=3.92 \pi m^3
\end{aligned}\)
Hence, option (d) is correct.