Q. If the radius of the sphere is $(5.3 \pm 0.1)cm.$ Then percentage error in its volume will be
A) $3 + 6.01 \times \frac{100}{5.3}$
B) $\frac{1}{3} \times 0.01 \times \frac{100}{5.3}$
C) $\left( \begin{array}{r} \frac{3 \times 0.01}{5.3} \\ \ \end{array} \right) \times 100$
D) $\frac{0.1}{5.3} \times 100$
Explanation
Volume of sphere $(V) = \frac{4}{3}\pi r^{3}$ % error in volume $= 3 \times \frac{\mathrm{\Delta}r}{r} \times 100 = \left( 3 \times \frac{0.1}{5.3} \right) \times 100$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.