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 $\pm 0.04cm$ in the measurement at diameter.
$\therefore \frac{du}{dt}=\pm 0.02$, $r=10$
Now $\upsilon =\frac{4}{3}\pi r^3$
$\Rightarrow log\upsilon =log\frac{4}{3}\pi +3logr$
$\Rightarrow \frac{\delta v}{\upsilon }\times 100=0+3\cdot \frac{\delta r}{r}\times 100$
$=3\cdot \frac{0.02}{10}\times 100=\pm 0.6$