Question

There is an error of $\pm \,0.04 \,cm$ in the measurement of the diameter of a sphere. When the radius is $10\, cm$, the percentage error in the volume of the sphere is

• A. $ \pm 1.2 $
• B. $ \pm 1.0 $
• C. $ \pm 0.8 $
• D. $ \pm 0.6 $

Answer 1

Answer: (D)

Given, error in diameter $=\pm 0.04$
$\therefore $ Error in radius, $d r=\pm 0.02$
$\therefore $ Per cent error in the volume of sphere
$=\frac{d V}{V} \times 100=\frac{d\left(\frac{4}{3} \pi r^{3}\right)}{\frac{4}{3} \pi r^{3}} \times 100$
$=\frac{3 d r}{r} \times 100$
$=\frac{3 \times(\pm 0.02)}{10} \times 100=\pm 0.6$