Question

If there is an error of $\pm 0.04$ cm in the measurenment of the diameter of the sphere then the approximate percentage error in its volume, when the radius is 10 cm, is

• A. $\pm\,1.2$
• B. $\pm\,0.06$
• C. $\pm\,0.006$
• D. $\pm\,0.6$

Answer 1

Answer: (D)

Since there is are error of $± 0.04 \,cm$ in the measurement at diameter.
$\therefore \frac{du}{dt}= \pm 0.02$, $r=10$
Now $\upsilon=\frac{4}{3}\,\pi\,r^{3}$
$\Rightarrow log\,\upsilon=log \frac{4}{3}\,\pi+3\,log\,r$
$\Rightarrow \frac{\delta v}{\upsilon}\times100=0+3\cdot\frac{\delta r}{r}\times 100$
$=3\cdot \frac{0.02}{10}\times100= \pm0.6$