There is an error of $\pm 0.04 \mathrm{~cm}$ in the measurement of the diameter of a sphere. When the radius…
There is an error of $\pm 0.04 \mathrm{~cm}$ in the measurement of the diameter of a sphere. When the radius is $10 \mathrm{~cm}$, the percentage error in the volume of the sphere is
$\pm 1.2$
$\pm 1.0$
$\pm 0.8$
$\pm 0.6$
Solution
Given, error in diameter $= \pm 0.04$
$\therefore \quad$ Error in radius, $d r= \pm 0.02$
$\therefore$ Per cent error in the volume of sphere
$\begin{aligned}
& =\frac{d V}{V} \times 100=\frac{d\left(\frac{4}{3} \pi r^3\right)}{\frac{4}{3} \pi r^3} \times 100=\frac{3 d r}{r} \times 100 \\
& =\frac{3 \times( \pm 0.02)}{10} \times 100= \pm 0.6
\end{aligned}$