Q. A wire has a mass $0.3 \pm 0.003g$, radius $0.5 \pm 0.005\ mm$ and length $6 \pm 0.06\ cm$. The maximum percentage error in the measurement of its density is
A) 1
B) 2
C) 3
D) 4
Explanation
$\because$ Density,$\ \rho = \frac{M}{V} = \frac{M}{\pi\gamma^{2}L}$ $\Rightarrow \frac{\mathrm{\Delta}\rho}{\rho} = \frac{\mathrm{\Delta}M}{M} + 2\frac{\mathrm{\Delta}r}{r} + \frac{\mathrm{\Delta}L}{L}$ $= \frac{0.003}{0.3} + 2 \times \frac{0.005}{0.5} + \frac{0.06}{6}$ $= 0.01 + 0.02 + 0.01 = 0.04$ $\therefore Percentage\ error = \frac{\mathrm{\Delta}\rho}{\rho} \times 100 = 0.04 \times 100 = 4\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.