Q. 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) $\pm 0.6$
Explanation
(d)
Given error in diameter=$\pm 0.04$
$\therefore$Error in radius, $\delta r = \pm 0.02$
$\therefore$percent error in the volume of sphere
$= \frac{\delta V}{V} \times 100 = \frac{\delta\left( \frac{4}{3}\pi r^{2} \right)}{\frac{4}{3}\pi r^{3}} \times 100 = \frac{3\delta}{r} \times 100$
$= \frac{3 \times \pm 0.02}{10}$ $\times 100 = \pm 0.6$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.